Chapter 11 Resource Masters - KTL MATH CLASSESktlmathclass.weebly.com/uploads/2/5/7/6/25760552/alg_2_resource_ws_ch_11.pdf©Glencoe/McGraw-Hill iv Glencoe Algebra 2 Teacher’s Guide

Chapter 11Resource Masters

Consumable WorkbooksMany of the worksheets contained in the Chapter Resource Masters bookletsare available as consumable workbooks.

Study Guide and Intervention Workbook 0-07-828029-XSkills Practice Workbook 0-07-828023-0Practice Workbook 0-07-828024-9

ANSWERS FOR WORKBOOKS The answers for Chapter 11 of these workbookscan be found in the back of this Chapter Resource Masters booklet.

Copyright © by The McGraw-Hill Companies, Inc. All rights reserved.Printed in the United States of America. Permission is granted to reproduce the material contained herein on the condition that such material be reproduced only for classroom use; be provided to students, teacher, and families without charge; and be used solely in conjunction with Glencoe’s Algebra 2. Any other reproduction, for use or sale, is prohibited without prior written permission of the publisher.

Send all inquiries to:The McGraw-Hill Companies8787 Orion PlaceColumbus, OH 43240-4027

ISBN: 0-07-828014-1 Algebra 2Chapter 11 Resource Masters

1 2 3 4 5 6 7 8 9 10 066 11 10 09 08 07 06 05 04 03 02

Glencoe/McGraw-Hill

© Glencoe/McGraw-Hill iii Glencoe Algebra 2

Contents

Vocabulary Builder . . . . . . . . . . . . . . . . vii

Lesson 11-1Study Guide and Intervention . . . . . . . . 631–632Skills Practice . . . . . . . . . . . . . . . . . . . . . . . 633Practice . . . . . . . . . . . . . . . . . . . . . . . . . . . 634Reading to Learn Mathematics . . . . . . . . . . 635Enrichment . . . . . . . . . . . . . . . . . . . . . . . . . 636








Chapter 11 AssessmentChapter 11 Test, Form 1 . . . . . . . . . . . 679–680Chapter 11 Test, Form 2A . . . . . . . . . . 681–682Chapter 11 Test, Form 2B . . . . . . . . . . 683–684Chapter 11 Test, Form 2C . . . . . . . . . . 685–686Chapter 11 Test, Form 2D . . . . . . . . . . 687–688Chapter 11 Test, Form 3 . . . . . . . . . . . 689–690Chapter 11 Open-Ended Assessment . . . . . 691Chapter 11 Vocabulary Test/Review . . . . . . . 692Chapter 11 Quizzes 1 & 2 . . . . . . . . . . . . . . 693Chapter 11 Quizzes 3 & 4 . . . . . . . . . . . . . . 694Chapter 11 Mid-Chapter Test . . . . . . . . . . . . 695Chapter 11 Cumulative Review . . . . . . . . . . 696Chapter 11 Standardized Test Practice . 697–698

Standardized Test Practice Student Recording Sheet . . . . . . . . . . . . . . A1

ANSWERS . . . . . . . . . . . . . . . . . . . . . . A2–A35

© Glencoe/McGraw-Hill iv Glencoe Algebra 2

Teacher’s Guide to Using theChapter 11 Resource Masters

The Fast File Chapter Resource system allows you to conveniently file the resourcesyou use most often. The Chapter 11 Resource Masters includes the core materialsneeded for Chapter 11. These materials include worksheets, extensions, andassessment options. The answers for these pages appear at the back of this booklet.

All of the materials found in this booklet are included for viewing and printing in theAlgebra 2 TeacherWorks CD-ROM.

Vocabulary Builder Pages vii–viiiinclude a student study tool that presentsup to twenty of the key vocabulary termsfrom the chapter. Students are to recorddefinitions and/or examples for each term.You may suggest that students highlight orstar the terms with which they are notfamiliar.

WHEN TO USE Give these pages tostudents before beginning Lesson 11-1.Encourage them to add these pages to theirAlgebra 2 Study Notebook. Remind them to add definitions and examples as theycomplete each lesson.

Study Guide and InterventionEach lesson in Algebra 2 addresses twoobjectives. There is one Study Guide andIntervention master for each objective.

WHEN TO USE Use these masters asreteaching activities for students who needadditional reinforcement. These pages canalso be used in conjunction with the StudentEdition as an instructional tool for studentswho have been absent.

Skills Practice There is one master foreach lesson. These provide computationalpractice at a basic level.

WHEN TO USE These masters can be used with students who have weakermathematics backgrounds or needadditional reinforcement.

Practice There is one master for eachlesson. These problems more closely followthe structure of the Practice and Applysection of the Student Edition exercises.These exercises are of average difficulty.

WHEN TO USE These provide additionalpractice options or may be used ashomework for second day teaching of thelesson.

Reading to Learn MathematicsOne master is included for each lesson. Thefirst section of each master asks questionsabout the opening paragraph of the lessonin the Student Edition. Additionalquestions ask students to interpret thecontext of and relationships among termsin the lesson. Finally, students are asked tosummarize what they have learned usingvarious representation techniques.

WHEN TO USE This master can be usedas a study tool when presenting the lessonor as an informal reading assessment afterpresenting the lesson. It is also a helpfultool for ELL (English Language Learner)students.

Enrichment There is one extensionmaster for each lesson. These activities mayextend the concepts in the lesson, offer anhistorical or multicultural look at theconcepts, or widen students’ perspectives onthe mathematics they are learning. Theseare not written exclusively for honorsstudents, but are accessible for use with alllevels of students.

WHEN TO USE These may be used asextra credit, short-term projects, or asactivities for days when class periods areshortened.

© Glencoe/McGraw-Hill v Glencoe Algebra 2

Assessment OptionsThe assessment masters in the Chapter 11Resource Masters offer a wide range ofassessment tools for intermediate and finalassessment. The following lists describe eachassessment master and its intended use.

Chapter Assessment CHAPTER TESTS• Form 1 contains multiple-choice questions

and is intended for use with basic levelstudents.

• Forms 2A and 2B contain multiple-choicequestions aimed at the average levelstudent. These tests are similar in formatto offer comparable testing situations.

• Forms 2C and 2D are composed of free-response questions aimed at the averagelevel student. These tests are similar informat to offer comparable testingsituations. Grids with axes are providedfor questions assessing graphing skills.

• Form 3 is an advanced level test withfree-response questions. Grids withoutaxes are provided for questions assessinggraphing skills.

All of the above tests include a free-response Bonus question.

• The Open-Ended Assessment includesperformance assessment tasks that aresuitable for all students. A scoring rubricis included for evaluation guidelines.Sample answers are provided forassessment.

• A Vocabulary Test, suitable for allstudents, includes a list of the vocabularywords in the chapter and ten questionsassessing students’ knowledge of thoseterms. This can also be used in conjunc-tion with one of the chapter tests or as areview worksheet.

Intermediate Assessment• Four free-response quizzes are included

to offer assessment at appropriateintervals in the chapter.

• A Mid-Chapter Test provides an optionto assess the first half of the chapter. It iscomposed of both multiple-choice andfree-response questions.

Continuing Assessment• The Cumulative Review provides

students an opportunity to reinforce andretain skills as they proceed throughtheir study of Algebra 2. It can also beused as a test. This master includes free-response questions.

• The Standardized Test Practice offerscontinuing review of algebra concepts invarious formats, which may appear onthe standardized tests that they mayencounter. This practice includes multiple-choice, grid-in, and quantitative-comparison questions. Bubble-in andgrid-in answer sections are provided onthe master.

Answers• Page A1 is an answer sheet for the

Standardized Test Practice questionsthat appear in the Student Edition onpages 628–629. This improves students’familiarity with the answer formats theymay encounter in test taking.

• The answers for the lesson-by-lessonmasters are provided as reduced pageswith answers appearing in red.

• Full-size answer keys are provided forthe assessment masters in this booklet.

Reading to Learn MathematicsVocabulary Builder

NAME ______________________________________________ DATE ____________ PERIOD _____

1111

Voca

bula

ry B

uild

erThis is an alphabetical list of the key vocabulary terms you will learn in Chapter 11.As you study the chapter, complete each term’s definition or description. Rememberto add the page number where you found the term. Add these pages to your AlgebraStudy Notebook to review vocabulary at the end of the chapter.

Vocabulary Term Found on Page Definition/Description/Example

arithmetic mean

AR·ihth·MEH·tihk

arithmetic sequence

arithmetic series

Binomial Theorem

common difference

common ratio

factorial

Fibonacci sequence

fih·buh·NAH·chee

geometric mean

geometric sequence

(continued on the next page)

© Glencoe/McGraw-Hill vii Glencoe Algebra 2

© Glencoe/McGraw-Hill viii Glencoe Algebra 2

Vocabulary Term Found on Page Definition/Description/Example

geometric series

index of summation

inductive hypothesis

infinite geometric series

iteration

IH·tuh·RAY·shuhn

mathematical induction

partial sum

Pascal’s triangle

pas·KALZ

recursive formula

rih·KUHR·sihv

sigma notation

SIHG·muh

Reading to Learn MathematicsVocabulary Builder (continued)

NAME ______________________________________________ DATE ____________ PERIOD _____

1111

Study Guide and InterventionArithmetic Sequences

NAME ______________________________________________ DATE ____________ PERIOD _____

11-111-1

© Glencoe/McGraw-Hill 631 Glencoe Algebra 2

Less

on

11-

1

Arithmetic Sequences An arithmetic sequence is a sequence of numbers in which eachterm after the first term is found by adding the common difference to the preceding term.

nth Term of an an � a1 � (n � 1)d, where a1 is the first term, d is the common difference, Arithmetic Sequence and n is any positive integer

Find the next fourterms of the arithmetic sequence 7, 11, 15, … .Find the common difference by subtractingtwo consecutive terms.

11 � 7 � 4 and 15 � 11 � 4, so d � 4.

Now add 4 to the third term of the sequence,and then continue adding 4 until the fourterms are found. The next four terms of thesequence are 19, 23, 27, and 31.

Find the thirteenth termof the arithmetic sequence with a1 � 21and d � �6.Use the formula for the nth term of anarithmetic sequence with a1 � 21, n � 13,and d � �6.

an � a1 � (n � 1)d Formula for nth term

a13 � 21 � (13 � 1)(�6) n � 13, a1 � 21, d � �6

a13 � �51 Simplify.

The thirteenth term is �51.

Example 1Example 1 Example 2Example 2

Example 3Example 3 Write an equation for the nth term of the arithmetic sequence �14, �5, 4, 13, … .In this sequence a1 � �14 and d � 9. Use the formula for an to write an equation.

an � a1 � (n � 1)d Formula for the nth term

� �14 � (n � 1)9 a1 � �14, d � 9

� �14 � 9n � 9 Distributive Property

� 9n � 23 Simplify.

Find the next four terms of each arithmetic sequence.

1. 106, 111, 116, … 2. �28, �31, �34, … 3. 207, 194, 181, …121, 126, 131, 136 �37, �40, �43, �46 168, 155, 142, 129

Find the first five terms of each arithmetic sequence described.

4. a1 � 101, d � 9 5. a1 � �60, d � 4 6. a1 � 210, d � �40101, 110, 119, 128, 137 �60, �56, �52, �48, �44 210, 170, 130, 90, 50

Find the indicated term of each arithmetic sequence.

7. a1 � 4, d � 6, n � 14 82 8. a1 � �4, d � �2, n � 12 �269. a1 � 80, d � �8, n � 21 �80 10. a10 for 0, �3, �6, �9, … �27

Write an equation for the nth term of each arithmetic sequence.

11. 18, 25, 32, 39, … 12. �110, �85, �60, �35, … 13. 6.2, 8.1, 10.0, 11.9, …7n � 11 25n � 135 1.9n � 4.3

ExercisesExercises


Arithmetic Means The arithmetic means of an arithmetic sequence are the termsbetween any two nonsuccessive terms of the sequence.To find the k arithmetic means between two terms of a sequence, use the following steps.

Step 1 Let the two terms given be a1 and an , where n � k � 2.Step 2 Substitute in the formula an � a1 � (n � 1)d.Step 3 Solve for d, and use that value to find the k arithmetic means:

a1 � d, a1 � 2d, … , a1 � kd.

Find the five arithmetic means between 37 and 121.You can use the nth term formula to find the common difference. In the sequence,37, , , , , , 121, …, a1 is 37 and a7 is 121.

an � a1 � (n � 1)d Formula for the nth term

121 � 37 � (7 � 1)d a1 � 37, a7 � 121, n � 7

121 � 37 � 6d Simplify.

84 � 6d Subtract 37 from each side.

d � 14 Divide each side by 6.

Now use the value of d to find the five arithmetic means.37 � 51 � 65 � 79 � 93 � 107 � 121

� 14 � 14 � 14 � 14 � 14 � 14The arithmetic means are 51, 65, 79, 93, and 107.

Find the arithmetic means in each sequence.

1. 5, , , , �3 2. 18, , , , �2 3. 16, , , 373, 1, �1 13, 8, 3 23, 30

4. 108, , , , , 48 5. �14, , , , �30 6. 29, , , , 8996, 84, 72, 60 �18, �22, �26 44, 59, 74

7. 61, , , , , 116 8. 45, , , , , , 8172, 83, 94, 105 51, 57, 63, 69, 75

9. �18, , , , 14 10. �40, , , , , , �82�10, �2, 6 �47, �54, �61, �68, �75

11. 100, , , 235 12. 80, , , , , �30145, 190 58, 36, 14, �8

13. 450, , , , 570 14. 27, , , , , , 57480, 510, 540 32, 37, 42, 47, 52

15. 125, , , , 185 16. 230, , , , , , 128140, 155, 170 213, 196, 179, 162, 145

17. �20, , , , , 370 18. 48, , , , 10058, 136, 214, 292 61, 74, 87

???????

????????

????????

??????

????????

?????????

??????????

????????

?????

Study Guide and Intervention (continued)

Arithmetic Sequences

NAME ______________________________________________ DATE ____________ PERIOD _____

11-111-1

ExampleExample

ExercisesExercises

Skills PracticeArithmetic Sequences

NAME ______________________________________________ DATE ____________ PERIOD _____

11-111-1


Less

on

11-

1


1. 7, 11, 15, … 19, 23, 27, 31 2. �10, �5, 0, … 5, 10, 15, 20

3. 101, 202, 303, … 404, 505, 606, 707 4. 15, 7, �1, … �9, �17, �25, �33

5. �67, �60, �53, … 6. �12, �15, �18, …�46, �39, �32, �25 �21, �24, �27, �30


7. a1 � 6, d � 9 6, 15, 24, 33, 42 8. a1 � 27, d � 4 27, 31, 35, 39, 43

9. a1 � �12, d � 5 �12, �7, �2, 3, 8 10. a1 � 93, d � �15 93, 78, 63, 48, 33

11. a1 � �64, d � 11 12. a1 � �47, d � �20�64, �53, �42, �31, �20 �47, �67, �87, �107, �127


13. a1 � 2, d � 6, n � 12 68 14. a1 � 18, d � 2, n � 8 32

15. a1 � 23, d � 5, n � 23 133 16. a1 � 15, d � �1, n � 25 �9

17. a31 for 34, 38, 42, … 154 18. a42 for 27, 30, 33, … 150

Complete the statement for each arithmetic sequence.

19. 55 is the th term of 4, 7, 10, … . 18 20. 163 is the th term of �5, 2, 9, … . 25


21. 4, 7, 10, 13, … an � 3n � 1 22. �1, 1, 3, 5, … an � 2n � 3

23. �1, 3, 7, 11, … an � 4n � 5 24. 7, 2, �3, �8, … an � �5n � 12


25. 6, , , , 38 14, 22, 30 26. 63, , , , 147 84, 105, 126??????

??



1. 5, 8, 11, … 14, 17, 20, 23 2. �4, �6, �8, … �10, �12, �14, �16

3. 100, 93, 86, … 79, 72, 65, 58 4. �24, �19, �14, … �9, �4, 1, 6

5. , 6, , 11, … , 16, , 21 6. 4.8, 4.1, 3.4, … 2.7, 2, 1.3, 0.6


7. a1 � 7, d � 7 8. a1 � �8, d � 2

7, 14, 21, 28, 35 �8, �6, �4, �2, 0

9. a1 � �12, d � �4 10. a1 � , d �

�12, �16, �20, �24, �28 , 1, , 2,

11. a1 � � , d � � 12. a1 � 10.2, d � �5.8

� , � , � , � , � 10.2, 4.4, �1.4, �7.2, �13


13. a1 � 5, d � 3, n � 10 32 14. a1 � 9, d � 3, n � 29 93

15. a18 for �6, �7, �8, … . �23 16. a37 for 124, 119, 114, … . �56

17. a1 � , d � � , n � 10 � 18. a1 � 14.25, d � 0.15, n � 31 18.75

Complete the statement for each arithmetic sequence.

19. 166 is the th term of 30, 34, 38, … 35 20. 2 is the th term of , , 1, … 8


21. �5, �3, �1, 1, … an � 2n � 7 22. �8, �11, �14, �17, … an � �3n � 5

23. 1, �1, �3, �5, … an � �2n � 3 24. �5, 3, 11, 19, … an � 8n � 13


25. �5, , , , 11 �1, 3, 7 26. 82, , , , 18 66, 50, 34

27. EDUCATION Trevor Koba has opened an English Language School in Isehara, Japan.He began with 26 students. If he enrolls 3 new students each week, in how many weekswill he have 101 students? 26 wk

28. SALARIES Yolanda interviewed for a job that promised her a starting salary of $32,000with a $1250 raise at the end of each year. What will her salary be during her sixth yearif she accepts the job? $38,250

??????

4�5

3�5

??

18�3

�59�5

13�

11�

3�

7�

5�

1�3

5�6

5�

3�

1�

1�2

1�2

37�

27�17

�27�2

Practice (Average)

Arithmetic Sequences

NAME ______________________________________________ DATE ____________ PERIOD _____

11-111-1

Reading to Learn MathematicsArithmetic Sequences

NAME ______________________________________________ DATE ____________ PERIOD _____

11-111-1


Less

on

11-

1

Pre-Activity How are arithmetic sequences related to roofing?

Read the introduction to Lesson 11-1 at the top of page 578 in your textbook.

Describe how you would find the number of shingles needed for the fifteenthrow. (Do not actually calculate this number.) Explain why your method willgive the correct answer. Sample answer: Add 3 times 14 to 2. Thisworks because the first row has 2 shingles and 3 more areadded 14 times to go from the first row to the fifteenth row.

Reading the Lesson

1. Consider the formula an � a1 � (n � 1)d.

a. What is this formula used to find?a particular term of an arithmetic sequence

b. What do each of the following represent?

an: the nth term

a1: the first term

n: a positive integer that indicates which term you are finding

d: the common difference

2. Consider the equation an � �3n � 5.

a. What does this equation represent? Sample answer: It gives the nth term ofan arithmetic sequence with first term 2 and common difference �3.

b. Is the graph of this equation a straight line? Explain your answer. Sampleanswer: No; the graph is a set of points that fall on a line, but thepoints do not fill the line.

c. The functions represented by the equations an � �3n � 5 and f(x) � �3x � 5 arealike in that they have the same formula. How are they different? Sampleanswer: They have different domains. The domain of the first functionis the set of positive integers. The domain of the second function isthe set of all real numbers.

Helping You Remember3. A good way to remember something is to explain it to someone else. Suppose that your

classmate Shala has trouble remembering the formula an � a1 � (n � 1)d correctly. Shethinks that the formula should be an � a1 � nd. How would you explain to her that sheshould use (n � 1)d rather than nd in the formula? Sample answer: Each termafter the first in an arithmetic sequence is found by adding d to theprevious term. You would add d once to get to the second term, twice toget to the third term, and so on. So d is added n � 1 times, not n times,to get the nth term.


Fibonacci SequenceLeonardo Fibonacci first discovered the sequence of numbers named for himwhile studying rabbits. He wanted to know how many pairs of rabbits wouldbe produced in n months, starting with a single pair of newborn rabbits. Hemade the following assumptions.

1. Newborn rabbits become adults in one month.

2. Each pair of rabbits produces one pair each month.

3. No rabbits die.

Let Fn represent the number of pairs of rabbits at the end of n months. If youbegin with one pair of newborn rabbits, F0 � F1 � 1. This pair of rabbitswould produce one pair at the end of the second month, so F2 � 1 � 1, or 2.At the end of the third month, the first pair of rabbits would produce anotherpair. Thus, F3 � 2 � 1, or 3.

The chart below shows the number of rabbits each month for several months.

Solve.

1. Starting with a single pair of newborn rabbits, how many pairs of rabbitswould there be at the end of 12 months?

2. Write the first 10 terms of the sequence for which F0 � 3, F1 � 4, and Fn � Fn � 2 � Fn � 1.

3. Write the first 10 terms of the sequence for which F0 � 1, F1 � 5,Fn � Fn � 2 � Fn � 1.

Month Adult Pairs Newborn Pairs Total

F0 0 1 1

F1 1 0 1

F2 1 1 2

F3 2 1 3

F4 3 2 5

F5 5 3 8

Enrichment

NAME ______________________________________________ DATE ____________ PERIOD _____

11-111-1

Study Guide and InterventionArithmetic Series

NAME ______________________________________________ DATE ____________ PERIOD _____

11-211-2


Less

on

11-

2

Arithmetic Series An arithmetic series is the sum of consecutive terms of anarithmetic sequence.

Sum of an The sum Sn of the first n terms of an arithmetic series is given by the formulaArithmetic Series Sn � �

n2

�[2a1 � (n � 1)d ] or Sn � �n2

�(a1 � an)

Find Sn for thearithmetic series with a1 � 14,an � 101, and n � 30.Use the sum formula for an arithmeticseries.

Sn � (a1 � an) Sum formula

S30 � (14 � 101) n � 30, a1 � 14, an � 101

� 15(115) Simplify.

� 1725 Multiply.

The sum of the series is 1725.

30�2

n�2

Find the sum of allpositive odd integers less than 180.The series is 1 � 3 � 5 � … � 179.Find n using the formula for the nth term ofan arithmetic sequence.

an � a1 � (n � 1)d Formula for nth term

179 � 1 � (n � 1)2 an � 179, a1 � 1, d � 2

179 � 2n � 1 Simplify.

180 � 2n Add 1 to each side.

n � 90 Divide each side by 2.

Then use the sum formula for an arithmeticseries.


S90 � (1 � 179) n � 90, a1 � 1, an � 179

� 45(180) Simplify.

� 8100 Multiply.

The sum of all positive odd integers lessthan 180 is 8100.

90�2

n�2


ExercisesExercises

Find Sn for each arithmetic series described.

1. a1 � 12, an � 100, 2. a1 � 50, an � �50, 3. a1 � 60, an � �136,n � 12 672 n � 15 0 n � 50 �1900

4. a1 � 20, d � 4, 5. a1 � 180, d � �8, 6. a1 � �8, d � �7,an � 112 1584 an � 68 1860 an � �71 �395

7. a1 � 42, n � 8, d � 6 8. a1 � 4, n � 20, d � 2 9. a1 � 32, n � 27, d � 3

504 555 1917Find the sum of each arithmetic series.

10. 8 � 6 � 4 � … � �10 �10 11. 16 � 22 � 28 � … � 112 1088

12. �45 � (�41) � (�37) � … � 35 �105

Find the first three terms of each arithmetic series described.

13. a1 � 12, an � 174, 14. a1 � 80, an � �115, 15. a1 � 6.2, an � 12.6,Sn � 1767 12, 21, 30 Sn � �245 80, 65, 50 Sn � 84.6 6.2, 7.0, 7.8

1�2


Sigma Notation A shorthand notation for representing a series makes use of the Greek

letter Σ. The sigma notation for the series 6 � 12 � 18 � 24 � 30 is �5

n�16n.

Evaluate �18

k�1

(3k � 4).

The sum is an arithmetic series with common difference 3. Substituting k � 1 and k � 18into the expression 3k � 4 gives a1 � 3(1) � 4 � 7 and a18 � 3(18) � 4 � 58. There are 18 terms in the series, so n � 18. Use the formula for the sum of an arithmetic series.


S18 � (7 � 58) n � 18, a1 � 7, an � 58

� 9(65) Simplify.

� 585 Multiply.

So �18

k�1(3k � 4) � 585.

Find the sum of each arithmetic series.

1. �20

n�1(2n � 1) 2. �

25

n�5(x � 1) 3. �

18

k�1(2k � 7)

440 294 216

4. �75

r�10(2r � 200) 5. �

15

x�1(6x � 3) 6. �

50

t�1(500 � 6t)

�7590 765 17,350

7. �80

k�1(100 � k) 8. �

85

n�20(n � 100) 9. �

200

s�13s

4760 �3135 60,300

10. �28

m�14(2m � 50) 11. �

36

p�1(5p � 20) 12. �

32

j�12(25 � 2j)

�120 2610 �399

13. �42

n�18(4n � 9) 14. �

50

n�20(3n � 4) 15. �

44

j�5(7j � 3)

2775 3379 6740

18�2

n�2


Arithmetic Series

NAME ______________________________________________ DATE ____________ PERIOD _____

11-211-2

ExampleExample

ExercisesExercises

Skills PracticeArithmetic Series

NAME ______________________________________________ DATE ____________ PERIOD _____

11-211-2


Less

on

11-

2


1. a1 � 1, an � 19, n � 10 100 2. a1 � �5, an � 13, n � 7 28

3. a1 � 12, an � �23, n � 8 �44 4. a1 � 7, n � 11, an � 67 407

5. a1 � 5, n � 10, an � 32 185 6. a1 � �4, n � 10, an � �22 �130

7. a1 � �8, d � �5, n � 12 �426 8. a1 � 1, d � 3, n � 15 330

9. a1 � 100, d � �7, an � 37 685 10. a1 � �9, d � 4, an � 27 90

11. d � 2, n � 26, an � 42 442 12. d � �12, n � 11, an � �52 88


13. 1 � 4 � 7 � 10 � … � 43 330 14. 5 � 8 � 11 � 14 � … � 32 185

15. 3 � 5 � 7 � 9 � … � 19 99 16. �2 � (�5) � (�8) � … � (�20) �77

17. �5

n�1(2n � 3) 15 18. �

18

n�1(10 � 3n) 693

19. �10

n�2(4n � 1) 225 20. �

12

n�5(4 � 3n) �172


21. a1 � 4, an � 31, Sn � 175 4, 7, 10 22. a1 � �3, an � 41, Sn � 228 �3, 1, 5

23. n � 10, an � 41, Sn � 230 5, 9, 13 24. n � 19, an � 85, Sn � 760 �5, 0, 5



1. a1 � 16, an � 98, n � 13 741 2. a1 � 3, an � 36, n � 12 234

3. a1 � �5, an � �26, n � 8 �124 4. a1 � 5, n � 10, an � �13 �40

5. a1 � 6, n � 15, an � �22 �120 6. a1 � �20, n � 25, an � 148 1600

7. a1 � 13, d � �6, n � 21 �987 8. a1 � 5, d � 4, n � 11 275

9. a1 � 5, d � 2, an � 33 285 10. a1 � �121, d � 3, an � 5 �2494

11. d � 0.4, n � 10, an � 3.8 20 12. d � � , n � 16, an � 44 784


13. 5 � 7 � 9 � 11 � … � 27 192 14. �4 � 1 � 6 � 11 � … � 91 870

15. 13 � 20 � 27 � … � 272 5415 16. 89 � 86 � 83 � 80 � … � 20 1308

17. �4

n�1(1 � 2n) �16 18. �

6

j�1(5 � 3n) 93 19. �

5

n�1(9 � 4n) �15

20. �10

k�4(2k � 1) 105 21. �

8

n�3(5n � 10) 105 22. �

101

n�1(4 � 4n) �20,200


23. a1 � 14, an � �85, Sn � �1207 24. a1 � 1, an � 19, Sn � 100

14, 11, 8 1, 3, 5

25. n � 16, an � 15, Sn � �120 26. n � 15, an � 5 , Sn � 45

�30, �27, �24 , , 1

27. STACKING A health club rolls its towels and stacks them in layers on a shelf. Eachlayer of towels has one less towel than the layer below it. If there are 20 towels on thebottom layer and one towel on the top layer, how many towels are stacked on the shelf?210 towels

28. BUSINESS A merchant places $1 in a jackpot on August 1, then draws the name of aregular customer. If the customer is present, he or she wins the $1 in the jackpot. If thecustomer is not present, the merchant adds $2 to the jackpot on August 2 and drawsanother name. Each day the merchant adds an amount equal to the day of the month. Ifthe first person to win the jackpot wins $496, on what day of the month was her or hisname drawn? August 31

3�

1�

4�5

2�3

Practice (Average)

Arithmetic Series

NAME ______________________________________________ DATE ____________ PERIOD _____

11-211-2

Reading to Learn MathematicsArithmetic Series

NAME ______________________________________________ DATE ____________ PERIOD _____

11-211-2


Less

on

11-

2

Pre-Activity How do arithmetic series apply to amphitheaters?


Suppose that an amphitheater can seat 50 people in the first row and thateach row thereafter can seat 9 more people than the previous row. Usingthe vocabulary of arithmetic sequences, describe how you would find thenumber of people who could be seated in the first 10 rows. (Do not actuallycalculate the sum.) Sample answer: Find the first 10 terms of anarithmetic sequence with first term 50 and common difference9. Then add these 10 terms.

Reading the Lesson1. What is the relationship between an arithmetic sequence and the corresponding

arithmetic series? Sample answer: An arithmetic sequence is a list of termswith a common difference between successive terms. The correspondingarithmetic series is the sum of the terms of the sequence.

2. Consider the formula Sn � (a1 � an). Explain the meaning of this formula in words.

Sample answer: To find the sum of the first n terms of an arithmeticsequence, find half the number of terms you are adding. Multiply thisnumber by the sum of the first term and the nth term.

3. a. What is the purpose of sigma notation?Sample answer: to write a series in a concise form

b. Consider the expression �12

i�2(4i � 2).

This form of writing a sum is called .

The variable i is called the .

The first value of i is .

The last value of i is .

How would you read this expression? The sum of 4i �2 as i goes from 2 to 12.

Helping You Remember4. A good way to remember something is to relate it to something you already know. How

can your knowledge of how to find the average of two numbers help you remember the

formula Sn � (a1 � an)? Sample answer: Rewrite the formula as

Sn � n � . The average of the first and last terms is given by the

expression . The sum of the first n terms is the average of thefirst

a1 � an�2

a1 � an�2

n�2

12

2

index of summation

sigma notation

n�2


Geometric Puzzlers

For the problems on this page, you will need to use the PythagoreanTheorem and the formulas for the area of a triangle and a trapezoid.

Enrichment

NAME ______________________________________________ DATE ____________ PERIOD _____

11-211-2

1. A rectangle measures 5 by 12 units. Theupper left corner is cut off as shown inthe diagram.

a. Find the area A(x) of the shadedpentagon.

b. Find x and 2x so that A(x) is amaximum. What happens to the cut-off triangle?

3. The coordinates of the vertices of a triangle are A(0, 0), B(11, 0), and C(0, 11). A line x � k cuts the triangleinto two regions having equal area.

a. What are the coordinates of point D?

b. Write and solve an equation forfinding the value of k.

2. A triangle with sides of lengths a, a, andb is isosceles. Two triangles are cut off sothat the remaining pentagon has fiveequal sides of length x. The value of xcan be found using this equation.

(2b � a)x2 � (4a2 � b2)(2x � a) � 0

a. Find x when a � 10 and b � 12.

b. Can a be equal to 2b?

4. Inside a square are five circles with thesame radius.

a. Connect the center of the top left circleto the center of the bottom right circle.Express this length in terms of r.

b. Draw the square with vertices at thecenters of the four outside circles.Express the diagonal of this squarein terms of r and a.

ra

b

xx

x x

x

a

x

y

A B

C

D

x � k

2x

12

5

x

Study Guide and InterventionGeometric Sequences

NAME ______________________________________________ DATE ____________ PERIOD _____

11-311-3


Less

on

11-

3

Geometric Sequences A geometric sequence is a sequence in which each term afterthe first is the product of the previous term and a constant called the constant ratio.

nth Term of a an � a1 � r n � 1, where a1 is the first term, r is the common ratio, Geometric Sequence and n is any positive integer

Find the next twoterms of the geometric sequence 1200, 480, 192, ….

Since � 0.4 and � 0.4, the

sequence has a common ratio of 0.4. Thenext two terms in the sequence are192(0.4) � 76.8 and 76.8(0.4) � 30.72.

192�480

480�1200

Write an equation for thenth term of the geometric sequence 3.6, 10.8, 32.4, … .In this sequence a1 � 3.6 and r � 3. Use thenth term formula to write an equation.

an � a1 � rn � 1 Formula for nth term

� 3.6 � 3n � 1 a1 � 3.6, r � 3

An equation for the nth term is an � 3.6 � 3n � 1.


ExercisesExercises

Find the next two terms of each geometric sequence.

1. 6, 12, 24, … 2. 180, 60, 20, … 3. 2000, �1000, 500, …

48, 96 , �250, 125

4. 0.8, �2.4, 7.2, … 5. 80, 60, 45, … 6. 3, 16.5, 90.75, …

�21.6, 64.8 33.75, 25.3125 499.125, 2745.1875

Find the first five terms of each geometric sequence described.

7. a1 � , r � 3 8. a1 � 240, r � � 9. a1 � 10, r �

, , 1, 3, 9 240, �180, 135, 10, 25, 62 , 156 ,

�101 , 75 390

Find the indicated term of each geometric sequence.

10. a1 � �10, r � 4, n � 2 11. a1 � �6, r � � , n � 8 12. a3 � 9, r � �3, n � 7

�40 729

13. a4 � 16, r � 2, n � 10 14. a4 � �54, r � �3, n � 6 15. a1 � 8, r � , n � 5

1024 �486

Write an equation for the nth term of each geometric sequence.

16. 500, 350, 245, … 17. 8, 32, 128, … 18. 11, �24.2, 53.24, …500 � 0.7n�1 8 � 4n� 1 11 � (�2.2)n � 1

128�

2�3

3�

1�2

5�

15�

1�

1�

1�

1�

1�

5�2

3�4

1�9

20�

20�


Geometric Means The geometric means of a geometric sequence are the termsbetween any two nonsuccessive terms of the sequence.To find the k geometric means between two terms of a sequence, use the following steps.

Step 1 Let the two terms given be a1 and an, where n � k � 2.Step 2 Substitute in the formula an � a1 � r n � 1 (� a1 � rk � 1).Step 3 Solve for r, and use that value to find the k geometric means:

a1 � r, a1 � r2, … , a1 � rk

Find the three geometric means between 8 and 40.5.Use the nth term formula to find the value of r. In the sequence 8, , , , 40.5, a1 is 8and a5 is 40.5.

an � a1 � rn � 1 Formula for nth term

40.5 � 8 � r5 � 1 n � 5, a1 � 8, a5 � 40.5

5.0625 � r4 Divide each side by 8.

r � �1.5 Take the fourth root of each side.

There are two possible common ratios, so there are two possible sets of geometric means.Use each value of r to find the geometric means.

r � 1.5 r � �1.5a2 � 8(1.5) or 12 a2 � 8(�1.5) or �12a3 � 12(1.5) or 18 a3 � �12(�1.5) or 18a4 � 18(1.5) or 27 a4 � 18(�1.5) or �27

The geometric means are 12, 18, and 27, or �12, 18, and �27.

Find the geometric means in each sequence.

1. 5, , , , 405 2. 5, , , 20.48

�15, 45, �135 8, 12.8

3. , , , , 375 4. �24, , ,

�3, 15, �75 4, �

5. 12, , , , , , 6. 200, , , , 414.72

�6, 3, � , , � �240, 288, �345.6

7. , , , , , �12,005 8. 4, , , , 156

� , 35, �245, 1715 �10, 25, �62

9. � , , , , , , �9 10. 100, , , , 384.16

� , � , � , �1, �3 �140, 196, �274.41�

1�

1�

????????1�81

1�

35�

1�4

???????35�49

3�

3�

3�

???3�16

?????

2�

1�9

?????3�5

?????

???


Geometric Sequences

NAME ______________________________________________ DATE ____________ PERIOD _____

11-311-3

ExampleExample

ExercisesExercises

Skills PracticeGeometric Sequences

NAME ______________________________________________ DATE ____________ PERIOD _____

11-311-3


Less

on

11-

3


1. �1, �2, �4, … �8, �16 2. 6, 3, , … ,

3. �5, �15, �45, … �135, �405 4. 729, �243, 81 , … �27, 9

5. 1536, 384, 96, … 24, 6 6. 64, 160, 400, … 1000, 2500


7. a1 � 6, r � 2 8. a1 � �27, r � 3

6, 12, 24, 48, 96 �27, �81, �243, �729, �2187

9. a1 � �15, r � �1 10. a1 � 3, r � 4

�15, 15, �15, 15, �15 3, 12, 48, 192, 768

11. a1 � 1, r � 12. a1 � 216, r � �

1, , , , 216, �72, 24, �8,


13. a1 � 5, r � 2, n � 6 160 14. a1 � 18, r � 3, n � 6 4374

15. a1 � �3, r � �2, n � 5 �48 16. a1 � �20, r � �2, n � 9 �5120

17. a8 for �12, �6, �3, … � 18. a7 for 80, , , …


19. 3, 9, 27, … an � 3n 20. �1, �3, �9, … an � �1(3)n � 1

21. 2, �6, 18, … an � 2(�3)n � 1 22. 5, 10, 20, … an � 5(2)n � 1


23. 4, , , , 64 �8, 16, �32 24. 1, , , , 81 �3, 9, �27??????

80�80

�980�3

3�

8�

1�

1�

1�

1�

1�3

1�2

3�

3�3

�2



1. �15, �30, �60, … �120, �240 2. 80, 40, 20, … 10, 5

3. 90, 30, 10, … , 4. �1458, 486, �162, … 54, �18

5. , , , … , 6. 216, 144, 96, … 64,


7. a1 � �1, r � �3 8. a1 � 7, r � �4

�1, 3, �9, 27, �81 7, �28, 112, �448, 1792

9. a1 � � , r � 2 10. a1 � 12, r �

� , � , � , � , � 12, 8, , ,


11. a1 � 5, r � 3, n � 6 1215 12. a1 � 20, r � �3, n � 6 �4860

13. a1 � �4, r � �2, n � 10 2048 14. a8 for � , � , � , … �

15. a12 for 96, 48, 24, … 16. a1 � 8, r � , n � 9

17. a1 � �3125, r � � , n � 9 � 18. a1 � 3, r � , n � 8


19. 1, 4, 16, … an � (4)n � 1 20. �1, �5, �25, … an � �1(5)n � 1

21. 1, , , … an � � �n � 122. �3, �6, �12, … an � �3(2)n � 1

23. 7, �14, 28, … an � 7(�2)n � 1 24. �5, �30, �180, … an � �5(6)n � 1


25. 3, , , , 768 12, 48, 192 26. 5, , , , 1280 �20, 80, �320

27. 144, , , , 9 28. 37,500, , , , , �12

�72, 36, �18 �7500, 1500, �300, 60

29. BIOLOGY A culture initially contains 200 bacteria. If the number of bacteria doublesevery 2 hours, how many bacteria will be in the culture at the end of 12 hours? 12,800

30. LIGHT If each foot of water in a lake screens out 60% of the light above, what percent ofthe light passes through 5 feet of water? 1.024%

31. INVESTING Raul invests $1000 in a savings account that earns 5% interest compoundedannually. How much money will he have in the account at the end of 5 years? $1276.28

???????

??????

1�1

�41�2

3��1

�101

�125

1�5

1�1

�23

�

625�1

�101

�501

�250

64�

32�

16�

16�

8�

4�

2�

1�

2�3

1�3

128�

81�

27�9

�163�8

1�4

10�

10�

Practice (Average)

Geometric Sequences

NAME ______________________________________________ DATE ____________ PERIOD _____

11-311-3

Reading to Learn MathematicsGeometric Sequences

NAME ______________________________________________ DATE ____________ PERIOD _____

11-311-3


Less

on

11-

3

Pre-Activity How do geometric sequences apply to a bouncing ball?


Suppose that you drop a ball from a height of 4 feet, and that each time itfalls, it bounces back to 74% of the height from which it fell. Describe howwould you find the height of the third bounce. (Do not actually calculate theheight of the bounce.)

Sample answer: Multiply 4 by 0.74 three times.

Reading the Lesson

1. Explain the difference between an arithmetic sequence and a geometric sequence.

Sample answer: In an arithmetic sequence, each term after the first isfound by adding the common difference to the previous term. In ageometric sequence, each term after the first is found by multiplying theprevious term by the common ratio.

2. Consider the formula an � a1 � rn � 1.

a. What is this formula used to find? a particular term of a geometric sequence


an: the nth term

a1: the first term

r: the common ratio

n: a positive integer that indicates which term you are finding

3. a. In the sequence 5, 8, 11, 14, 17, 20, the numbers 8, 11, 14, and 17 are

between 5 and 20.

b. In the sequence 12, 4, , , , the numbers 4, , and are

between 12 and .

Helping You Remember

4. Suppose that your classmate Ricardo has trouble remembering the formula an � a1 � rn � 1

correctly. He thinks that the formula should be an � a1 � rn. How would you explain tohim that he should use rn � 1 rather than rn in the formula?

Sample answer: Each term after the first in a geometric sequence isfound by multiplying the previous term by r. There are n � 1 termsbefore the nth term, so you would need to multiply by r a total of n � 1times, not n times, to get the nth term.

4�27

geometric means

4�9

4�3

4�27

4�9

4�3

arithmetic means


Half the DistanceSuppose you are 200 feet from a fixed point, P. Suppose that you are able tomove to the halfway point in one minute, to the next halfway point oneminute after that, and so on.

An interesting sequence results because according to the problem, you neveractually reach the point P, although you do get arbitrarily close to it.

You can compute how long it will take to get within some specified smalldistance of the point. On a calculator, you enter the distance to be coveredand then count the number of successive divisions by 2 necessary to getwithin the desired distance.

How many minutes are needed to get within 0.1 footof a point 200 feet away?

Count the number of times you divide by 2.

Enter: 200 2 2 2 , and so on

Result: 0.0976562

You divided by 2 eleven times. The time needed is 11 minutes.

Use the method illustrated above to solve each problem.

1. If it is about 2500 miles from Los Angeles to New York, how many minutes would it take to get within 0.1 mile of New York? How far from New York are you at that time?

2. If it is 25,000 miles around Earth, how many minutes would it take to get within 0.5 mile of the full distance around Earth? How far short would you be?

3. If it is about 250,000 miles from Earth to the Moon, how many minutes would it take to get within 0.5 mile of the Moon? How far from the surface of the Moon would you be?

4. If it is about 30,000,000 feet from Honolulu to Miami, how many minutes would it take to get to within 1 foot of Miami? How far from Miami would you be at that time?

5. If it is about 93,000,000 miles to the sun, how many minutes would it take to get within 500 miles of the sun? How far from the sun would you be at that time?

ENTER�ENTER�ENTER�

100

200 feet

150 175 P

1st minute 2nd minute 3rd minute

Enrichment

NAME ______________________________________________ DATE ____________ PERIOD _____

11-311-3

ExampleExample

Study Guide and InterventionGeometric Series

NAME ______________________________________________ DATE ____________ PERIOD _____

11-411-4


Less

on

11-

4

Geometric Series A geometric series is the indicated sum of consecutive terms of ageometric sequence.

Sum of a The sum Sn of the first n terms of a geometric series is given byGeometric Series Sn � or Sn � , where r � 1.

a1 � a1rn

��1 � r

a1(1 � r n)��

1 � r

Find the sum of the firstfour terms of the geometric sequence for which a1 � 120 and r � .

Sn � Sum formula

S4 � n � 4, a1 � 120, r � �13

�

� 177.78 Use a calculator.

The sum of the series is 177.78.

120�1 � ��13��4�

��1 � �

13

�

a1(1 � rn)��1 � r

1�3

Find the sum of the

geometric series �7

j�14 � 3 j � 2.

Since the sum is a geometric series, you canuse the sum formula.

Sn � Sum formula

S7 � n � 7, a1 � �43

�, r � 3

� 1457.33 Use a calculator.


�43

�(1 � 37)�1 � 3

a1(1 � rn)��1 � r


ExercisesExercises

Find Sn for each geometric series described.

1. a1 � 2, an � 486, r � 3 2. a1 � 1200, an � 75, r � 3. a1 � , an � 125, r � 5

728 2325 156.24

4. a1 � 3, r � , n � 4 5. a1 � 2, r � 6, n � 4 6. a1 � 2, r � 4, n � 6

4.44 518 2730

7. a1 � 100, r � � , n � 5 8. a3 � 20, a6 � 160, n� 8 9. a4 � 16, a7 � 1024, n � 10

68.75 1275 87,381.25

Find the sum of each geometric series.

10. 6 � 18 � 54 � … to 6 terms 11. � � 1 � … to 10 terms

2184 255.75

12. �8

j�42 j 13. �

7

k�13 � 2k � 1

496 381

1�2

1�4

1�2

1�3

1�25

1�2


Specific Terms You can use one of the formulas for the sum of a geometric series to helpfind a particular term of the series.


Geometric Series

NAME ______________________________________________ DATE ____________ PERIOD _____

11-411-4

Find a1 in a geometricseries for which S6 � 441 and r � 2.

Sn � Sum formula

441 � S6 � 441, r � 2, n � 6

441 � Subtract.

a1 � Divide.

a1 � 7 Simplify.

The first term of the series is 7.

441�63

�63a1��1

a1(1 � 26)��1 � 2

a1(1 � rn)��1 � r

Find a1 in a geometricseries for which Sn � 244, an � 324, and r� �3.Since you do not know the value of n, use thealternate sum formula.

Sn � Alternate sum formula

244 � Sn � 244, an � 324, r � �3

244 � Simplify.

976 � a1 � 972 Multiply each side by 4.

a1 � 4 Subtract 972 from each side.

The first term of the series is 4.

a1 � 972��4

a1 � (324)(�3)��1 � (�3)

a1 � anr��1 � r


Example 3Example 3 Find a4 in a geometric series for which Sn � 796.875, r � , and n � 8.First use the sum formula to find a1.

Sn � Sum formula

796.875 � S8 � 796.875, r � , n � 8

796.875 � Use a calculator.

a1 � 400

Since a4 � a1 � r3, a4 � 400��12��3

� 50. The fourth term of the series is 50.

Find the indicated term for each geometric series described.

1. Sn � 726, an � 486, r � 3; a1 6 2. Sn � 850, an � 1280, r � �2; a1 �10

3. Sn � 1023.75, an � 512, r � 2; a1 4. Sn � 118.125, an � �5.625, r � � ; a1 180

5. Sn � 183, r � �3, n � 5; a1 3 6. Sn � 1705, r � 4, n � 5; a1 5

7. Sn � 52,084, r � �5, n � 7; a1 4 8. Sn � 43,690, r � , n � 8; a1 32, 768

9. Sn � 381, r � 2, n � 7; a4 24

1�4

1�2

1�

0.99609375a1��0.5

1�2

a1�1 � ��12��8�

��1 � �

12

�

a1(1 � rn)��1 � r

1�2

ExercisesExercises

Skills PracticeGeometric Series

NAME ______________________________________________ DATE ____________ PERIOD _____

11-411-4


Less

on

11-

4


1. a1 � 2, a5 � 162, r � 3 242 2. a1 � 4, a6 � 12,500, r � 5 15,624

3. a1 � 1, a8 � �1, r � �1 0 4. a1 � 4, an � 256, r � �2 172

5. a1 � 1, an � 729, r � �3 547 6. a1 � 2, r � �4, n � 5 410

7. a1 � �8, r � 2, n � 4 �120 8. a1 � 3, r � �2, n � 12 �4095

9. a1 � 8, r � 3, n � 5 968 10. a1 � 6, an � , r �

11. a1 � 8, r � , n � 7 12. a1 � 2, r � � , n � 6


13. 4 � 8 � 16 � … to 5 terms 124 14. �1 � 3 � 9 � … to 6 terms �364

15. 3 � 6 � 12 � … to 5 terms 93 16. �15 � 30 � 60 � … to 7 terms �645

17. �4

n�13n � 1 40 18. �

5

n�1(�2)n � 1 11

19. �4

n�1� �n � 1

20. �9

n�12(�3)n � 1 9842


21. Sn � 1275, an � 640, r � 2; a1 5 22. Sn � �40, an � �54, r � �3; a1 2

23. Sn � 99, n � 5, r � � ; a1 144 24. Sn � 39,360, n � 8, r � 3; a1 121�2

40�1

�3

21�1

�2127�1

�2

93�1

�23�8



1. a1 � 2, a6 � 64, r � 2 126 2. a1 � 160, a6 � 5, r � 315

3. a1 � �3, an � �192, r � �2 �129 4. a1 � �81, an � �16, r � � �55

5. a1 � �3, an � 3072, r � �4 2457 6. a1 � 54, a6 � , r �

7. a1 � 5, r � 3, n � 9 49,205 8. a1 � �6, r � �1, n � 21 �6

9. a1 � �6, r � �3, n � 7 �3282 10. a1 � �9, r � , n � 4 �

11. a1 � , r � 3, n � 10 12. a1 � 16, r � �1.5, n � 6 �66.5


13. 162 � 54 � 18 � … to 6 terms 14. 2 � 4 � 8 � … to 8 terms 510

15. 64 � 96 � 144 � … to 7 terms 463 16. � � 1 � … to 6 terms �

17. �8

n�1(�3)n � 1 �1640 18. �

9

n�15(�2)n � 1 855 19. �

5

n�1�1(4)n � 1 �341

20. �6

n�1� �n � 1

21. �10

n�12560� �n � 1

5115 22. �4

n�19� �n � 1


23. Sn � 1023, an � 768, r � 4; a1 3 24. Sn � 10,160, an � 5120, r � 2; a1 80

25. Sn � �1365, n � 12, r � �2; a1 1 26. Sn � 665, n � 6, r � 1.5; a1 32

27. CONSTRUCTION A pile driver drives a post 27 inches into the ground on its first hit.

Each additional hit drives the post the distance of the prior hit. Find the total distance

the post has been driven after 5 hits. 70 in.

28. COMMUNICATIONS Hugh Moore e-mails a joke to 5 friends on Sunday morning. Eachof these friends e-mails the joke to 5 of her or his friends on Monday morning, and so on.Assuming no duplication, how many people will have heard the joke by the end ofSaturday, not including Hugh? 97,655 people

1�

2�3

65�2

�31�2

63�1

�2

182�1

�31�9

728�

29,524�1

�3

65�2

�3

728�1

�32�9

2�3

1�2

Practice (Average)

Geometric Series

NAME ______________________________________________ DATE ____________ PERIOD _____

11-411-4

Reading to Learn MathematicsGeometric Series

NAME ______________________________________________ DATE ____________ PERIOD _____

11-411-4


Less

on

11-

4

Pre-Activity How is e-mailing a joke like a geometric series?


• Suppose that you e-mail the joke on Monday to five friends, rather thanthree, and that each of those friends e-mails it to five friends on Tuesday,and so on. Write a sum that shows that total number of people, includingyourself, who will have read the joke by Thursday. (Write out the sumusing plus signs rather than sigma notation. Do not actually find the sum.)1 � 5 � 25 � 125

• Use exponents to rewrite the sum you found above. (Use an exponent ineach term, and use the same base for all terms.)50 � 51 � 52 � 53

Reading the Lesson

1. Consider the formula Sn � .

a. What is this formula used to find? the sum of the first n terms of ageometric series


Sn: the sum of the first n terms

a1: the first term

r: the common ratio

c. Suppose that you want to use the formula to evaluate 3 � 1 � � � . Indicate

the values you would substitute into the formula in order to find Sn. (Do not actuallycalculate the sum.)

n � a1 � r � rn �

d. Suppose that you want to use the formula to evaluate the sum �6

n�18(�2)n � 1. Indicate

the values you would substitute into the formula in order to find Sn. (Do not actuallycalculate the sum.)

n � a1 � r � rn �


2. This lesson includes three formulas for the sum of the first n terms of a geometric series.All of these formulas have the same denominator and have the restriction r � 1. How canthis restriction help you to remember the denominator in the formulas?Sample answer: If r � 1, then r � 1 � 0. Because division by 0 isundefined, a formula with r � 1 in the denominator will not apply when r � 1.

(�2)6 or 64�286

��13

��5 or ��2143��

13

�35

1�27

1�9

1�3

a1(1 � rn)��1 � r


AnnuitiesAn annuity is a fixed amount of money payable at given intervals. For example,suppose you wanted to set up a trust fund so that $30,000 could be withdrawneach year for 14 years before the money ran out. Assume the money can beinvested at 9%.

You must find the amount of money that needs to be invested. Call thisamount A. After the third payment, the amount left is

1.09[1.09A � 30,000(1 � 1.09)] � 30,000 � 1.092A � 30,000(1 � 1.09 � 1.092).

The results are summarized in the table below.

1. Use the pattern shown in the table to find the number of dollars left afterthe fourth payment.

2. Find the amount left after the tenth payment.

The amount left after the 14th payment is 1.0913A � 30,000(1 � 1.09 �1.092 � … � 1.0913). However, there should be no money left after the 14thand final payment.

1.0913A � 30,000(1 � 1.09 � 1.092 � … � 1.0913) � 0

Notice that 1 � 1.09 � 1.092 � … � 1.0913 is a geometric series where a1 � 1, an � 1.0913, n � 14 and r � 1.09.

Using the formula for Sn,

1 � 1.09 � 1.092 � … � 1.0913 � � �11��

11.0.099

14� � �

1 ��0

1..009914

�.

3. Show that when you solve for A you get A � �300,.00090

� ��1.019.0

14

91�3

1��.

Therefore, to provide $30,000 for 14 years where the annual interest rate is 9%,

you need �300,.00090

� ��1.019.0

14

91�3

1�� dollars.

4. Use a calculator to find the value of A in problem 3.

In general, if you wish to provide P dollars for each of n years at an annualrate of r%, you need A dollars where

�1 � �10r0��n � 1 A � P�1 � �1 � �10

r0�� 1 � �10

r0��2

� … � �1 � �10r0��n � 1� � 0.

You can solve this equation for A, given P, n, and r.

a1 � a1rn

��1 � r

Payment Number Number of Dollars Left After Payment

1 A � 30,0002 1.09A � 30,000(1 � 1.09)3 1.092A � 30,000(1 � 1.09 � 1.092)

Enrichment

NAME ______________________________________________ DATE ____________ PERIOD _____

11-411-4

Study Guide and InterventionInfinite Geometric Series

NAME ______________________________________________ DATE ____________ PERIOD _____

11-511-5


Less

on

11-

5

Infinite Geometric Series A geometric series that does not end is called an infinitegeometric series. Some infinite geometric series have sums, but others do not because thepartial sums increase without approaching a limiting value.

Sum of an Infinite S � for � 1 � r � 1.Geometric Series If |r | 1, the infinite geometric series does not have a sum.

Find the sum of each infinite geometric series, if it exists.

a1�1 � r

ExampleExample

a. 75 � 15 � 3 � …

First, find the value of r to determine ifthe sum exists. a1 � 75 and a2 � 15, so

r � or . Since � � � 1, the sum

exists. Now use the formula for the sumof an infinite geometric series.

S � Sum formula

� a1 � 75, r �

� or 93.75 Simplify.


75�

�45

�

1�5

75�1 � �

15

�

a1�1 � r

1�5

1�5

15�75

b. ��

n�148�� n � 1

In this infinite geometric series, a1 � 48

and r � � .

S � Sum formula

� a1 � 48, r � �

� or 36 Simplify.

Thus �

n�148�� n � 1

� 36.1�3

48�

�43

�

1�3

48��1 � ��

13

��

a1�1 � r

1�3

1�3

ExercisesExercises


1. a1 � �7, r � 2. 1 � � � … 3. a1 � 4, r �

�18 does not exist 8

4. � � � … 5. 15 � 10 � 6 � … 6. 18 � 9 � 4 � 2 � …

1 45 12

7. � � � … 8. 1000 � 800 � 640 � … 9. 6 � 12 � 24 � 48 � …

5000 does not exist

10. �

n�150� �n � 1

11. �

k�122�� k � 1

12. �

s�124� �s � 1

250 14 57 3�

2�

7�12

1�2

4�5

1�

1�40

1�20

1�10

1�

1�4

1�2

2�3

25�162

5�27

2�9

2�

1�2

25�16

5�4

5�8


Repeating Decimals A repeating decimal represents a fraction. To find the fraction,write the decimal as an infinite geometric series and use the formula for the sum.

Write each repeating decimal as a fraction.


Infinite Geometric Series

NAME ______________________________________________ DATE ____________ PERIOD _____

11-511-5

ExampleExample

a. 0.4�2�Write the repeating decimal as a sum.

0.4�2� � 0.42424242…

� � � � …

In this series a1 � and r � .

S � Sum formula

� a1 � , r �

� Subtract.

� or Simplify.

Thus 0.4�2� � .14�33

14�33

42�99

�14020

�

��19090

�

1�100

42�100

�14020

�

�1 � �

1100�

a1�1 � r

1�100

42�100

42��1,000,000

42�10,000

42�100

b. 0.52�4�Let S � 0.52�4�.

S � 0.5242424… Write as a repeating decimal.

1000S � 524.242424… Multiply each side by 1000.

10S � 5.242424… Mulitply each side by 10.

990S � 519 Subtract the third equationfrom the second equation.

S � or Simplify.

Thus, 0.52�4� �173�330

173�330

519�990

ExercisesExercises


1. 0.2� 2. 0.8� 3. 0.3�0� 4. 0.8�7�

5. 0.1�0� 6. 0.5�4� 7. 0.7�5� 8. 0.1�8�

9. 0.6�2� 10. 0.7�2� 11. 0.07�2� 12. 0.04�5�

13. 0.06� 14. 0.01�3�8� 15. 0.0�1�3�8� 16. 0.08�1�

17. 0.24�5� 18. 0.43�6� 19. 0.54� 20. 0.86�3�19�

49�

24�

27�

9�

46�

23�

1�

1�

4�

8�

62�

2�

25�

6�

10�

29�

10�

8�

2�

Skills PracticeInfinite Geometric Series

NAME ______________________________________________ DATE ____________ PERIOD _____

11-511-5


Less

on

11-

5


1. a1 � 1, r � 2 2. a1 � 5, r � �

3. a1 � 8, r � 2 does not exist 4. a1 � 6, r � 12

5. 4 � 2 � 1 � � … 8 6. 540 � 180 � 60 � 20 � … 405

7. 5 � 10 � 20 � … does not exist 8. �336 � 84 � 21 � … �268.8

9. 125 � 25 � 5 � … 156.25 10. 9 � 1 � � …

11. � � � … does not exist 12. � � � …

13. 5 � 2 � 0.8 � … 14. 9 � 6 � 4 � … 27

15. �

n�110� �n � 1

20 16. �

n�16�� n � 1

17. �

n�115� �n � 1

25 18. �

n�1�� n � 1

�2


19. 0.4� 20. 0.8�

21. 0.2�7� 22. 0.6�7�

23. 0.5�4� 24. 0.3�7�5�

25. 0.6�4�1� 26. 0.1�7�1�57�

641�

125�

6�

67�

3�

8�

4�

1�3

4�3

2�5

9�1

�31�2

25�

1�1

�271�9

1�3

27�4

9�4

3�4

81�1

�9

1�2

1�2

25�2

�51�2



1. a1 � 35, r � 49 2. a1 � 26, r � 52

3. a1 � 98, r � � 56 4. a1 � 42, r � does not exist

5. a1 � 112, r � � 70 6. a1 � 500, r � 625

7. a1 � 135, r � � 90 8. 18 � 6 � 2 � …

9. 2 � 6 � 18 � … does not exist 10. 6 � 4 � � … 18

11. � � 1 � … does not exist 12. 10 � 1 � 0.1 � …

13. 100 � 20 � 4 � … 125 14. �270 � 135 � 67.5 � … �180

15. 0.5 � 0.25 � 0.125 � … 1 16. � � � …

17. 0.8 � 0.08 � 0.008 � … 18. � � � … does not exist

19. 3 � � � … 20. 0.3 � 0.003 � 0.00003 � …

21. 0.06 � 0.006 � 0.0006 � … 22. � 2 � 6 � … does not exist

23. �

n�13� �n � 1

4 24. �

n�1�� n � 1

25. �

n�118� �n � 1

54 26. �

n�15(�0.1)n � 1


27. 0.6� 28. 0.0�9� 29. 0.4�3� 30. 0.2�7�

31. 0.2�4�3� 32. 0.8�4� 33. 0.9�9�0� 34. 0.1�5�0�

35. PENDULUMS On its first swing, a pendulum travels 8 feet. On each successive swing,

the pendulum travels the distance of its previous swing. What is the total distance

traveled by the pendulum when it stops swinging? 40 ft

36. ELASTICITY A ball dropped from a height of 10 feet bounces back �190� of that distance.

With each successive bounce, the ball continues to reach �190� of its previous height. What is

the total vertical distance (both up and down) traveled by the ball when it stops bouncing?(Hint: Add the total distance the ball falls to the total distance it rises.) 190 ft

4�5

50�

110�

28�

9�

3�

43�

1�

2�

50�2

�3

8�3

�42�3

1�4

2�3

1�

30�

21�27

�499�7

1�3

1�6

1�12

8�

7�7

�10007

�1007

�10

100�2

�54

�25

8�3

27�1

�2

1�5

3�5

6�5

3�4

1�2

2�7

Practice (Average)

Infinite Geometric Series

NAME ______________________________________________ DATE ____________ PERIOD _____

11-511-5

Reading to Learn MathematicsInfinite Geometric Series

NAME ______________________________________________ DATE ____________ PERIOD _____

11-511-5


Less

on

11-

5

Pre-Activity How does an infinite geometric series apply to a bouncing ball?


Note the following powers of 0.6: 0.61 � 0.6; 0.62 � 0.36; 0.63 � 0.216;0.64 � 0.1296; 0.65 � 0.07776; 0.66 � 0.046656; 0.67 � 0.0279936. If a ballis dropped from a height of 10 feet and bounces back to 60% of its previousheight on each bounce, after how many bounces will it bounce back to aheight of less than 1 foot? 5 bounces

Reading the Lesson1. Consider the formula S � .

a. What is the formula used to find? the sum of an infinite geometric series


S: the sum

a1: the first term

r: the common ratio

c. For what values of r does an infinite geometric sequence have a sum? �1 � r � 1

d. Rewrite your answer for part d as an absolute value inequality. |r | � 1

2. For each of the following geometric series, give the values of a1 and r. Then statewhether the sum of the series exists. (Do not actually find the sum.)

a. � � � … a1 � r �

Does the sum exist?

b. 2 � 1 � � � … a1 � r �

Does the sum exist?

c. �

i�13i a1 � r �

Does the sum exist?


3. One good way to remember something is to relate it to something you already know. How

can you use the formula Sn � that you learned in Lesson 11-4 for finding the

sum of a geometric series to help you remember the formula for finding the sum of aninfinite geometric series? Sample answer: If �1 � r � 1, then as n gets large,rn approaches 0, so 1 � rn approaches 1. Therefore, Sn approaches

, or .a1

�1 � r

a1 � 1�1 � r

a1(1 � rn)��1 � r

no

33

yes

��12

�21�4

1�2

yes

�13

��23

�2�27

2�9

2�3

a1�1 � r


Convergence and DivergenceConvergence and divergence are terms that relate to the existence of a sum ofan infinite series. If a sum exists, the series is convergent. If not, the series is

divergent. Consider the series 12 � 3 � �34� � �1

36� � … . This is a geometric

series with r � �14�. The sum is given by the formula S � �1

a�

1

r�. Thus, the sum

is 12 � �34� or 16. This series is convergent since a sum exists. Notice that the

first two terms have a sum of 15. As more terms are added, the sum comescloser (or converges) to 16.

Recall that a geometric series has a sum if and only if �1 � r � 1. Thus, ageometric series is convergent if r is between �1 and 1, and divergent if r hasanother value. An infinite arithmetic series cannot have a sum unless all ofthe terms are equal to zero.

Determine whether each series is convergent or divergent.

a. 2 � 5 � 8 � 11 � … divergent

b. �2 � 4 � (�8) � 16 � … divergent

c. 16 � 8 � 4 � 2 � … convergent

Determine whether each series is convergent or divergent. If theseries is convergent, find the sum.

1. 5 � 10 � 15 � 20 � … 2. 16 � 8 � 4 � 2 � …

3. 1 � 0.1 � 0.01 � 0.001 � … 4. 4 � 2 � 0 � 2 � …

5. 2 � 4 � 8 � 16 � … 6. 1 � �15� � �2

15� � �1

125� � …

7. 4 � 2.4 � 1.44 � 0.864 � … 8. �18� � �

14� � �

12� � 1 � …

9. ��53� � �

190� � �

2207� � �

4801� � … 10. 48 � 12 � 3 � �

34� � …

Bonus: Is 1 � �12� � �

13� � �

14� � �

15� � … convergent or divergent?

Enrichment

NAME ______________________________________________ DATE ____________ PERIOD _____

11-511-5

ExampleExample

Study Guide and InterventionRecursion and Special Sequences

NAME ______________________________________________ DATE ____________ PERIOD _____

11-611-6


Less

on

11-

6Special Sequences In a recursive formula, each succeeding term is formulated fromone or more previous terms. A recursive formula for a sequence has two parts:

1. the value(s) of the first term(s), and2. an equation that shows how to find each term from the term(s) before it.

Find the first five terms of the sequence in which a1 � 6, a2 � 10,and an � 2an � 2 for n 3.a1 � 6a2 � 10a3 � 2a1� 2(6) � 12a4 � 2a2 � 2(10) � 20a5 � 2a3 � 2(12) � 24

The first five terms of the sequence are 6, 10, 12, 20, 24.

Find the first five terms of each sequence.

1. a1 � 1, a2 � 1, an � 2(an � 1 � an � 2), n 3 1, 1, 4, 10, 28

2. a1 � 1, an � , n 2 1, , , ,

3. a1 � 3, an � an � 1 � 2(n � 2), n 2 3, 3, 5, 9, 15

4. a1 � 5, an � an � 1 � 2, n 2 5, 7, 9, 11, 13

5. a1 � 1, an � (n � 1)an � 1, n 2 1, 1, 2, 6, 24

6. a1 � 7, an � 4an � 1 � 1, n 2 7, 27, 107, 427, 1707

7. a1 � 3, a2 � 4, an � 2an � 2 � 3an � 1, n 3 3, 4, 18, 62, 222

8. a1 � 0.5, an � an � 1 � 2n, n 2 0.5, 4.5, 10.5, 18.5, 28.5

9. a1 � 8, a2 � 10, an � , n 3 8, 10, 0.8, 12.5, 0.064

10. a1 � 100, an � , n 2 100, 50, , , 50�

50�

50�

an � 1�n

an � 2�an � 1

5�

3�

2�

1�

1��1 � an � 1

ExampleExample

ExercisesExercises


Iteration Combining composition of functions with the concept of recursion leads to theprocess of iteration. Iteration is the process of composing a function with itself repeatedly.

Find the first three iterates of f(x) � 4x � 5 for an initial value of x0 � 2.To find the first iterate, find the value of the function for x0 � 2x1 � f(x0) Iterate the function.

� f(2) x0 � 2

� 4(2) � 5 or 3 Simplify.

To find the second iteration, find the value of the function for x1 � 3.x2 � f(x1) Iterate the function.

� f(3) x1 � 3

� 4(3) � 5 or 7 Simplify.

To find the third iteration, find the value of the function for x2 � 7.x3 � f(x2) Iterate the function.

� f(7) x2 � 7

� 4(7) � 5 or 23 Simplify.

The first three iterates are 3, 7, and 23.

Find the first three iterates of each function for the given initial value.

1. f(x) � x � 1; x0 � 4 2. f(x) � x2 � 3x; x0 � 1 3. f(x) � x2 � 2x � 1; x0 � �2

3, 2, 1 �2, 10, 70 1, 4, 25

4. f(x) � 4x � 6; x0 � �5 5. f(x) � 6x � 2; x0 � 3 6. f(x) � 100 � 4x; x0 � �5

�26, �110, �446 16, 94, 562 120, �380, 1620

7. f(x) � 3x � 1; x0 � 47 8. f(x) � x3 � 5x2; x0 � 1 9. f(x) � 10x � 25; x0 � 2

140, 419, 1256 �4, �144, �3,089,664 �5, �75, �775

10. f(x) � 4x2 � 9; x0 � �1 11. f(x) � 2x2 � 5; x0 � �4 12. f(x) � ; x0 � 1

�5, 91, 33,115 37, 2743, 15,048,103 0, � , �1

13. f(x) � (x � 11); x0 � 3 14. f(x) � ; x0 � 9 15. f(x) � x � 4x2; x0 � 1

7, 9, 10 , 9, �3, �39, �6123

16. f(x) � x � ; x0 � 2 17. f(x) � x3 � 5x2 � 8x � 10; 18. f(x) � x3 � x2; x0 � �2

2.5, 2.9, about 3.245x0 � 1

�12, �1872, �6, �454, �94,610,886�6,563,711,232

1�x

1�

1�

3�x

1�2

1�

x � 1�x � 2


Recursion and Special Sequences

NAME ______________________________________________ DATE ____________ PERIOD _____

11-611-6

ExampleExample

ExercisesExercises

Skills PracticeRecursion and Special Sequences

NAME ______________________________________________ DATE ____________ PERIOD _____

11-611-6


Less

on

11-

6Find the first five terms of each sequence.

1. a1 � 4, an � 1 � an � 7 2. a1 � �2, an � 1 � an � 3

4, 11, 18, 25, 32 �2, 1, 4, 7, 10

3. a1 � 5, an � 1 � 2an 4. a1 � �4, an � 1 � 6 � an

5, 10, 20, 40, 80 �4, 10, �4, 10, �4

5. a1 � 1, an � 1 � an � n 6. a1 � �1, an � 1 � n � an

1, 2, 4, 7, 11 �1, 2, 0, 3, 1

7. a1 � �6, an � 1 � an � n � 1 8. a1 � 8, an � 1 � an � n � 2

�6, �4, �1, 3, 8 8, 5, 1, �4, �10

9. a1 � �3, an � 1 � 2an � 7 10. a1 � 4, an � 1 � �2an � 5

�3, 1, 9, 25, 57 4, �13, 21, �47, 89

11. a1 � 0, a2 � 1, an � 1 � an � an � 1 12. a1 � �1, a2 � �1, an � 1 � an � an � 1

0, 1, 1, 2, 3 �1, �1, 0, 1, 1

13. a1 � 3, a2 � �5, an � 1 � �4an � an � 1 14. a1 � �3, a2 � 2, an � 1 � an � 1 � an

3, �5, 23, �97, 411 �3, 2, �5, 7, �12


15. f(x) � 2x � 1, x0 � 3 5, 9, 17 16. f(x) � 5x � 3, x0 � 2 7, 32, 157

17. f(x) � 3x � 4, x0 � �1 1, 7, 25 18. f(x) � 4x � 7, x0 � �5 �13, �45, �173

19. f(x) � �x � 3, x0 � 10 �13, 10, �13 20. f(x) � �3x � 6, x0 � 6 �12, 42, �120

21. f(x) � �3x � 4, x0 � 2 �2, 10, �26 22. f(x) � 6x � 5, x0 � 1 1, 1, 1

23. f(x) � 7x � 1, x0 � �4 24. f(x) � x2 � 3x, x0 � 5�27, �188, �1315 10, 70, 4690



1. a1 � 3, an � 1 � an � 5 2. a1 � �7, an � 1 � an � 83, 8, 13, 18, 23 �7, 1, 9, 17, 25

3. a1 � �3, an � 1 � 3an � 2 4. a1 � �8, an � 1 � 10 � an�3, �7, �19, �55, �163 �8, 18, �8, 18, �8

5. a1 � 4, an � 1 � n � an 6. a1 � �3, an � 1 � 3an4, �3, 5, �2, 6 �3, �9, �27,�81, �243

7. a1 � 4, an � 1 � �3an � 4 8. a1 � 2, an � 1 � �4an � 54, �8, 28, �80, 244 2, �13, 47, �193, 767

9. a1 � 3, a2 � 1, an � 1 � an � an � 1 10. a1 � �1, a2 � 5, an � 1 � 4an � 1 � an3, 1, �2, �3, �1 �1, 5, �9, 29, �65

11. a1 � 2, a2 � �3, an � 1 � 5an � 8an � 1 12. a1 � �2, a2 � 1, an � 1 � �2an � 6an � 12, �3, �31, �131, �407 �2, 1, �14, 34, �152


13. f(x) � 3x � 4, x0 � �1 1, 7, 25 14. f(x) � 10x � 2, x0 � �1 �8, �78, �778

15. f(x) � 8 � 3x, x0 � 1 11, 41, 131 16. f(x) � 8 � x, x0 � �3 11, �3, 11

17. f(x) � 4x � 5, x0 � �1 1, 9, 41 18. f(x) � 5(x � 3), x0 � �2 5, 40, 215

19. f(x) � �8x � 9, x0 � 1 1, 1, 1 20. f(x) � �4x2, x0 � �1 �4; �64; �16,384

21. f(x) � x2 � 1, x0 � 3 8, 63, 3968 22. f(x) � 2x2; x0 � 5 50; 5000; 50,000,000

23. INFLATION Iterating the function c(x) � 1.05x gives the future cost of an item at aconstant 5% inflation rate. Find the cost of a $2000 ring in five years at 5% inflation.$2552.56

FRACTALS For Exercises 24–27, use the following information.Replacing each side of the square shown with thecombination of segments below it gives the figure to its right.

24. What is the perimeter of the original square?12 in.

25. What is the perimeter of the new shape? 20 in.

26. If you repeat the process by replacing each side of the new shape by a proportionalcombination of 5 segments, what will the perimeter of the third shape be? 33 in.

27. What function f(x) can you iterate to find the perimeter of each successive shape if youcontinue this process? f(x) � x5

�

1�

1 in. 1 in.

1 in.

3 in.

1 in. 1 in.

Practice (Average)

Recursion and Special Sequences

NAME ______________________________________________ DATE ____________ PERIOD _____

11-611-6

Reading to Learn MathematicsRecursion and Special Sequences

NAME ______________________________________________ DATE ____________ PERIOD _____

11-611-6


Less

on

11-

6Pre-Activity How is the Fibonacci sequence illustrated in nature?


What are the next three numbers in the sequence that gives the number ofshoots corresponding to each month? 8, 13, 21

Reading the Lesson

1. Consider the sequence in which a1 � 4 and an � 2an � 1 � 5.

a. Explain why this is a recursive formula. Sample answer: Each term is foundfrom the value of the previous term.

b. Explain in your own words how to find the first four terms of this sequence. (Do notactually find any terms after the first.) Sample answer: The first term is 4. Tofind the second term, double the first term and add 5. To find the thirdterm, double the second term and add 5. To find the fourth term,double the third term and add 5.

c. What happens to the terms of this sequence as n increases? Sample answer:They keep getting larger and larger.

2. Consider the function f(x) � 3x � 1 with an initial value of x0 � 2.

a. What does it mean to iterate this function?to compose the function with itself repeatedly

b. Fill in the blanks to find the first three iterates. The blanks that follow the letter xare for subscripts.

x1 � f(x ) � f( ) � 3( ) � 1 � � 1 �

x2 � f(x ) � f( ) � 3( ) � 1 �

x3 � f(x ) � f( ) � 3( ) � 1 �

c. As this process continues, what happens to the values of the iterates?Sample answer: They keep getting larger and larger.


3. Use a dictionary to find the meanings of the words recurrent and iterate. How can themeanings of these words help you to remember the meaning of the mathematical termsrecursive and iteration? How are these ideas related? Sample answer: Recurrentmeans happening repeatedly, while iterate means to repeat a process oroperation. A recursive formula is used repeatedly to find the value of oneterm of a sequence based on the previous term. Iteration means tocompose a function with it self repeatedly. Both ideas have to do withrepetition—doing the same thing over and over again.

4114142

14551

56220


Continued FractionsThe fraction below is an example of a continued fraction. Note that eachfraction in the continued fraction has a numerator of 1.

2 � 1��3 � �

4 �

11–5

�

Enrichment

NAME ______________________________________________ DATE ____________ PERIOD _____

11-611-6

Evaluate the continuedfraction above. Start at the bottom andwork your way up.

Step 1: 4 � �15� � �

250� � �

15� � �

251�

Step 2: � �251�

Step 3: 3 � �251� � �

6231� � �2

51� � �

6281�

Step 4: � �2618�

Step 5: 2 � �2618� � 2�

2618�

1��6281�

1��251�

Change �2151� into a

continued fraction.Follow the steps.

Step 1: �2151� � �

2121� � �1

31� � 2 � �1

31�

Step 2: �131� �

Step 3: �131� � �

93� � �

23� � 3 � �

23�

Step 4: �23� �

Step 5: �32� � �

22� � �

12� � 1 � �

12�

Stop, because the numerator is 1.

Thus, �2151� can be written as 2 �

1��3 � �

1 �

11–2

�

1��32�

1��131�


Evaluate each continued fraction.

1. 1 � 1 2. 0 �1 �

3. 2 � 1 4. 5 �4 �

Change each fraction into a continued fraction.

5. �7351� 6. �

289� 7. �

1139�

1��6 � �

8 �

11—

10

�

1��7 � �

9 �

11—11

�

1

2 � �3 �

11–3

�

1��6 � �

4 �

11–2

�

2

1 � �1 � �

12

�

1 � �2

�

Study Guide and InterventionThe Binomial Theorem

NAME ______________________________________________ DATE ____________ PERIOD _____

11-711-7


Less

on

11-

7

Pascal’s Triangle Pascal’s triangle is the pattern of coefficients of powers of binomialsdisplayed in triangular form. Each row begins and ends with 1 and each coefficient is thesum of the two coefficients above it in the previous row.

(a � b)0 1(a � b)1 1 1

Pascal’s Triangle(a � b)2 1 2 1(a � b)3 1 3 3 1(a � b)4 1 4 6 4 1(a � b)5 1 5 10 10 5 1

Use Pascal’s triangle to find the number of possible sequencesconsisting of 3 as and 2 bs.The coefficient 10 of the a3b2-term in the expansion of (a � b)5 gives the number ofsequences that result in three as and two bs.

Expand each power using Pascal’s triangle.

1. (a � 5)4 a4 � 20a3 � 150a2 � 500a � 625

2. (x � 2y)6 x6 � 12x5y � 60x4y2 � 160x3y3 � 240x2y4 � 192xy5 � 64y6

3. ( j � 3k)5 j 5 � 15j4k � 90j3k2 � 270j2k3 � 405jk4 � 243k5

4. (2s � t)7 128s7 � 448s6t � 672s5t2 � 560s4t3 � 280s3t4 � 84s2t5 � 14st 6 � t7

5. (2p � 3q)6 64p6 � 576p5q � 2160p4q2 � 4320p3q3 � 4860p2q4 � 2916pq5 � 729q6

6. �a � �4 a4 � 2a3b � a2b2 � ab3 � b4

7. Ray tosses a coin 15 times. How many different sequences of tosses could result in 4heads and 11 tails? 1365

8. There are 9 true/false questions on a quiz. If twice as many of the statements are true asfalse, how many different sequences of true/false answers are possible? 84

1�

1�

3�b

�2

ExampleExample

ExercisesExercises


The Binomial Theorem

Binomial If n is a nonnegative integer, then

Theorem (a � b)n � 1anb0 � an � 1b1 � an � 2b2 � an � 3b3 � … �1a0bn

Another useful form of the Binomial Theorem uses factorial notation and sigma notation.

Factorial If n is a positive integer, then n ! � n (n � 1)(n � 2) � … � 2 � 1.

Binomial Theorem,

(a � b)n � anb0 � an � 1b1 � an � 2b2 � … � a0bn

Factorial Form

� �n

k�0

an � kbk

Evaluate .

�

� 11 � 10 � 9 � 990

Expand (a � 3b)4.

(a � 3b)4 � �4

k�0a4 � k(�3b)k

� a4 � a3(�3b)1 � a2(�3b)2 � a(�3b)3 � (�3b)4

� a4 � 12a3b � 54a2b2 � 108ab3 � 81b4

Evaluate each expression.

1. 5! 120 2. 36 3. 210

Expand each power.

4. (a � 3)6 a6 � 18a5 � 135a4 � 540a3 � 1215a2 � 1458a � 729

5. (r � 2s)7 r7 � 14r 6s � 84r5s2 � 280r4s3 � 560r3s4 � 672r2s5 � 448rs6 � 128s7

6. (4x � y)4 256x4 � 256x3y � 96x2y2 � 16xy3 � y4

7. �2 � �5 32 � 40m � 20m2 � 5m3 � m4 � m5

Find the indicated term of each expansion.

8. third term of (3x � y)5 270x3y2 9. fifth term of (a � 1)7 35a3

10. fourth term of ( j � 2k)8 448j5k3 11. sixth term of (10 � 3t)7 �510,300t5

12. second term of �m � �9 6m8 13. seventh term of (5x � 2)11 92,400,000x52�3

1�

5�m

�2

10!�6!4!

9!�7!2!

4!�0!4!

4!�1!3!

4!�2!2!

4!�3!1!

4!�4!0!

4!��(4 � k)!k!

11 � 10 � 9 � 8 � 7 � 6 � 5 � 4 � 3 � 2 � 1��8 � 7 � 6 � 5 � 4 � 3 � 2 � 1

11!�8!

11!�8!

n!��(n � k)!k !

n!�0!n!

n!��(n � 2)!2!

n!��(n � 1)!1!

n!�n!0!

n(n � 1)(n � 2)��

1 � 2 � 3n(n � 1)��

1 � 2n�1



NAME ______________________________________________ DATE ____________ PERIOD _____

11-711-7

Example 1Example 1

Example 2Example 2

ExercisesExercises

Skills PracticeThe Binomial Theorem

NAME ______________________________________________ DATE ____________ PERIOD _____

11-711-7


Less

on

11-

7


1. 8! 40,320 2. 10! 3,628,800

3. 12! 479,001,600 4. 210

5. 120 6. 45

7. 84 8. 15,504

Expand each power.

9. (x � y)3 10. (a � b)5

x3 � 3x2y � 3xy2 � y3 a5 � 5a4b � 10a3b2 � 10a2b3 � 5ab4 � b5

12. (m � 1)4 11. (g � h)4

m4 � 4m3 � 6m2 � 4m � 1 g4 � 4g3h � 6g2h2 � 4gh3 � h4

13. (r � 4)3 14. (a � 5)4

r3 � 12r2 � 48r � 64 a4 � 20a3 � 150a2 � 500a � 625

15. ( y � 7)3 16. (d � 2)5

y3 � 21y2 � 147y � 343 d 5 � 10d4 � 40d3 � 80d 2 � 80d � 32

17. (x � 1)4 18. (2a � b)4

x4 � 4x3 � 6x2 � 4x � 1 16a4 � 32a3b � 24a2b2 � 8ab3 � b4

19. (c � 4d)3 20. (2a � 3)3

c3 � 12c2d � 48cd 2 � 64d3 8a3 � 36a2 � 54a � 27


21. fourth term of (m � n)10 120m7n3 22. seventh term of (x � y)8 28x2y6

23. third term of (b � 6)5 360b3 24. sixth term of (s � 2)9 �4032s4

25. fifth term of (2a � 3)6 4860a2 26. second term of (3x � y)7 �5103x6y

20!�15!5!

9!�3!6!

10!�2!8!

6!�3!

15!�13!



1. 7! 5040 2. 11! 39,916,800 3. 3024 4. 380

5. 28 6. 56 7. 924 8. 10,660

Expand each power.

9. (n � v)5 n5 � 5n4v � 10n3v2 � 10n2v3 � 5nv4 � v5

10. (x � y)4 x4 � 4x3y � 6x2y2 �4xy3 � y4

11. (x � y)6 x6 � 6x5y � 15x4y2 � 20x3y3 � 15x2y4 � 6xy5 � y6

12. (r � 3)5 r 5 � 15r4 � 90r3 � 270r2 � 405r � 243

13. (m � 5)5 m5 � 25m4 � 250m3 � 1250m2 � 3125m � 3125

14. (x � 4)4 x4 � 16x3 � 96x2 � 256x � 256

15. (3x � y)4 81x4 � 108x3y � 54x2y2 � 12xy3 � y4

16. (2m � y)4 16m4 � 32m3y � 24m2y2 � 8my3 � y4

17. (w � 3z)3 w3 � 9w2z � 27wz2 � 27z3

18. (2d � 3)6 64d6 � 576d5 � 2160d4 � 4320d3 � 4860d2 � 2916d � 729

19. (x � 2y)5 x5 � 10x4y � 40x3y2 � 80x2y3 � 80xy4 � 32y5

20. (2x � y)5 32x5 � 80x4y � 80x3y2 � 40x2y3 � 10xy4 � y5

21. (a � 3b)4 a4 � 12a3b � 54a2b2 � 108ab3 � 81b4

22. (3 � 2z)4 16z4 � 96z3 � 216z2 � 216z � 81

23. (3m � 4n)3 27m3 � 108m2n � 144mn2 � 64n3

24. (5x � 2y)4 625x4 � 1000x3y � 600x2y2 � 160xy3 � 16y4


25. seventh term of (a � b)10 210a4b6 26. sixth term of (m � n)10 �252m5n5

27. ninth term of (r � s)14 3003r 6s8 28. tenth term of (2x � y)12 1760x3y9

29. fourth term of (x � 3y)6 �540x3y3 30. fifth term of (2x � 1)9 4032x5

31. GEOMETRY How many line segments can be drawn between ten points, no three ofwhich are collinear, if you use exactly two of the ten points to draw each segment? 45

32. PROBABILITY If you toss a coin 4 times, how many different sequences of tosses willgive exactly 3 heads and 1 tail or exactly 1 head and 3 tails? 8

41!�3!38!

12!�6!6!

8!�5!3!

8!�6!2!

20!�18!

9!�5!

Practice (Average)


NAME ______________________________________________ DATE ____________ PERIOD _____

11-711-7

Reading to Learn MathematicsThe Binomial Theorem

NAME ______________________________________________ DATE ____________ PERIOD _____

11-711-7


Less

on

11-

7

Pre-Activity How does a power of a binomial describe the numbers of boys andgirls in a family?


• If a family has four children, list the sequences of births of girls and boysthat result in three girls and one boy. BGGG GBGG GGBG GGGB

• Describe a way to figure out how many such sequences there are withoutlisting them. Sample answer: The boy could be the first,second, third, or fourth child, so there are four sequenceswith three girls and one boy.

Reading the Lesson

1. Consider the expansion of (w � z)5.

a. How many terms does this expansion have? 6

b. In the second term of the expansion, what is the exponent of w? 4

What is the exponent of z? 1

What is the coefficient of the second term? 5

c. In the fourth term of the expansion, what is the exponent of w? 2

What is the exponent of z? 3

What is the coefficient of the fourth term? 10

d. What is the last term of this expansion? z5

2. a. State the definition of a factorial in your own words. (Do not use mathematicalsymbols in your definition.) Sample answer: The factorial of any positiveinteger is the product of that integer and all the smaller integers downto one. The factorial of zero is one.

b. Write out the product that you would use to calculate 10!. (Do not actually calculatethe product.) 10 � 9 � 8 � 7 � 6 � 5 � 4 � 3 � 2 � 1

c. Write an expression involving factorials that could be used to find the coefficient of the

third term of the expansion of (m � n)6. (Do not actually calculate the coefficient.)


3. Without using Pascal’s triangle or factorials, what is an easy way to remember the firsttwo and last two coefficients for the terms of the binomial expansion of (a � b)n?Sample answer: The first and last coefficients are always 1. The secondand next-to-last coefficients are always n, the power to which thebinomial is being raised.

6!�


Patterns in Pascal’s TriangleYou have learned that the coefficients in the expansion of (x � y)n yield anumber pyramid called Pascal’s triangle.

As many rows can be added to the bottom of the pyramid as you please.

This activity explores some of the interesting properties of this famousnumber pyramid.

1. Pick a row of Pascal’s triangle.

a. What is the sum of all the numbers in all the rows above the row you picked?

b. What is the sum of all the numbers in the row you picked?

c. How are your answers for parts a and b related?

d. Repeat parts a through c for at least three more rows of Pascal’s triangle. What generalization seems to be true?

e. See if you can prove your generalization.

2. Pick any row of Pascal’s triangle that comes after the first.

a. Starting at the left end of the row, add the first number, the third number, the fifth number, and so on. State the sum.

b. In the same row, add the second number, the fourth number, and so on.State the sum.

c. How do the sums in parts a and b compare?

d. Repeat parts a through c for at least three other rows of Pascal’s triangle. What generalization seems to be true?

Row 1Row 2Row 3Row 4Row 5Row 6Row 7 1 6 15 20 15 6 1

1 5 10 10 51 4 6 4

1 3 41 2

11

11

11

1

Enrichment

NAME ______________________________________________ DATE ____________ PERIOD _____

11-711-7

Study Guide and InterventionProof and Mathematical Induction

NAME ______________________________________________ DATE ____________ PERIOD _____

11-811-8


Less

on

11-

8

Mathematical Induction Mathematical induction is a method of proof used to provestatements about positive integers.

Step 1 Show that the statement is true for some integer n.Mathematical Step 2 Assume that the statement is true for some positive integer k where k n. Induction Proof This assumption is called the inductive hypothesis.

Step 3 Show that the statement is true for the next integer k � 1.

Prove that 5 � 11 � 17 � … � (6n � 1) � 3n2 � 2n.Step 1 When n � 1, the left side of the given equation is 6(1) � 1 � 5. The right side is

3(1)2 � 2(1) � 5. Thus the equation is true for n � 1.

Step 2 Assume that 5 � 11 � 17 � … � (6k � 1) � 3k2 � 2k for some positive integer k.

Step 3 Show that the equation is true for n � k � 1. First, add [6(k � 1) � 1] to each side.5 � 11 � 17 � … � (6k � 1) � [6(k � 1) � 1] � 3k2 � 2k � [6(k � 1) � 1]

� 3k2 � 2k � 6k � 5 Add.

� 3k2 � 6k � 3 � 2k � 2 Rewrite.

� 3(k2 � 2k � 1) � 2(k � 1) Factor.

� 3(k � 1)2 � 2(k � 1) Factor.

The last expression above is the right side of the equation to be proved, where n has beenreplaced by k � 1. Thus the equation is true for n � k � 1.This proves that 5 � 11 � 17 � … � (6n � 1) � 3n2 � 2n for all positive integers n.

Prove that each statement is true for all positive integers.

1. 3 � 7 � 11 � … � (4n � 1) � 2n2 � n.Step 1 The statement is true for n � 1 since 4(1) � 1 � 3 and 2(1)2 � 1 � 3.Step 2 Assume that 3 � 7 � 11 � … � (4k � 1) � 2k2 � k for some

positive integer k.Step 3 Adding the (k � 1)st term to each side from step 2, we get

3 � 7 � 11 � … � (4k � 1) � [4(k � 1) � 1] � 2k2 � k � [4(k � 1) � 1].Simplifying the right side of the equation gives 2(k � 1)2 � (k � 1), which isthe statement to be proved.2. 500 � 100 � 20 � … � 4 � 54 � n � 625�1 � �.

Step 1 The statement is true for n � 1, since 4 � 54 � 1 � 4 � 53 � 500 and 625�1 � � � (625) � 500.

Step 2 Assume that 500 � 100 � 20 � … � 4 � 54 � k � 625�1 � � forsome positive integer k.

Step 3 Adding the (k � 1)st term to each side from step 2 and simplifying gives 500 � 100 � 20 � … � 4 � 54 � k � 4 � 53 � k �

625�1 � � � 4 � 53 � k � 625�1 � �, which is the statement

to be proved.

1�5k � 1

1�

1�

4�

1�

1�5n

ExampleExample

ExercisesExercises


Counterexamples To show that a formula or other generalization is not true, find acounterexample. Often this is done by substituting values for a variable.

Find a counterexample for the formula 2n2 � 2n � 3 � 2n � 2 � 1.Check the first few positive integers.

n Left Side of Formula Right Side of Formula

1 2(1)2 � 2(1) � 3 � 2 � 2 �3 or 7 21 � 2 � 1 � 23 � 1 or 7 true

2 2(2)2 � 2(2) � 3 � 8 � 4 � 3 or 15 22 � 2 � 1 � 24 � 1 or 15 true

3 2(3)2 � 2(3) � 3 � 18 � 6 � 3 or 27 23 � 2 � 1 � 25 � 1 or 31 false

The value n � 3 provides a counterexample for the formula.

Find a counterexample for the statement x2 � 4 is either prime or divisible by 4.

n x2 � 4 True? n x2 � 4 True?

1 1 � 4 or 5 Prime 6 36 � 4 or 40 Div. by 4

2 4 � 4 or 8 Div. by 4 7 49 � 4 or 53 Prime

3 9 � 4 or 13 Prime 8 64 � 4 or 68 Div. by 4

4 16 � 4 or 20 Div. by 4 9 81 � 4 or 85 Neither

5 25 � 4 or 29 Prime

The value n � 9 provides a counterexample.

Find a counterexample for each statement. Sample answers are given.1. 1 � 5 � 9 � … � (4n � 3) � 4n � 3 n � 2

2. 100 � 110 � 120 � … � (10n � 90) � 5n2 � 95 n � 2

3. 900 � 300 � 100 � … � 100(33 � n) � 900 � n � 3

4. x2 � x � 1 is prime. n � 4

5. 2n � 1 is a prime number. n � 4

6. 7n � 5 is a prime number. n � 2

7. � 1 � � … � � n � n � 3

8. 5n2 � 1 is divisible by 3. n � 3

9. n2 � 3n � 1 is prime for n � 2. n � 9

10. 4n2 � 1 is divisible by either 3 or 5. n � 6

1�2

n�2

3�2

1�2

2n�n � 1


Proof by Mathematical Induction

NAME ______________________________________________ DATE ____________ PERIOD _____

11-811-8

Example 1Example 1

Example 2Example 2

ExercisesExercises

Skills PracticeProof and Mathematical Induction

NAME ______________________________________________ DATE ____________ PERIOD _____

11-811-8


Less

on

11-

8


1. 1 � 3 � 5 � … � (2n � 1) � n2

Step 1: When n � 1, 2n � 1 � 2(1) � 1 � 1 � 12. So, the equation is truefor n � 1.

Step 2: Assume that 1 � 3 � 5 � … � (2k � 1) � k2 for some positiveinteger k.

Step 3: Show that the given equation is true for n � k � 1.1 � 3 � 5 � … � (2k � 1) � [2(k � 1) � 1] � k2 � [2(k � 1) � 1]

� k2 � 2k � 1 � (k � 1)2

So, 1 � 3 � 5 � … � (2n � 1) � n2 for all positive integers n.

2. 2 � 4 � 6 � … � 2n � n2 � n

Step 1: When n � 1, 2n � 2(1) � 2 � 12 � 1. So, the equation is true for n � 1.

Step 2: Assume that 2 � 4 � 6 � … � 2k � k2 � k for some positiveinteger k.

Step 3: Show that the given equation is true for n � k � 1.2 � 4 � 6 � …. � 2k � 2(k � 1) � k2 � k � 2(k � 1)

� (k2 � 2k � 1) � (k � 1)� (k � 1)2 � (k � 1)

So, 2 � 4 � 6 � … � 2n � n2 � n for all positive integers n.

3. 6n � 1 is divisible by 5.

Step 1: When n � 1, 6n � 1 � 61 � 1 � 5. So, the statement is true for n � 1.

Step 2: Assume that 6k � 1 is divisible by 5 for some positive integer k.Then there is a whole number r such that 6k � 1 � 5r.

Step 3: Show that the statement is true for n � k � 1.6k � 1 � 5r

6k � 5r � 16(6k ) � 6(5r � 1)6k � 1 � 30r � 6

6k � 1 � 1 � 30r � 56k � 1 � 1 � 5(6r � 1)

Since r is a whole number, 6r � 1 is a whole number, and 6k � 1 � 1 isdivisible by 5. The statement is true for n � k � 1. So, 6n � 1 is divisibleby 5 for all positive integers n.

Find a counterexample for each statement.

4. 3n � 3n is divisible by 6. 5. 1 � 4 � 8 � … � 2n �

Sample answer: n � 2 Sample answer: n � 3

n(n � 1)(2n � 1)��6



1. 1 � 2 � 4 � 8 � … � 2n � 1 � 2n � 1Step 1: When n � 1, then 2n � 1 � 21 � 1 � 20 � 1 � 21 � 1.

So, the equation is true for n � 1.Step 2: Assume that 1 � 2 � 4 � 8 � … � 2k � 1 � 2k � 1 for some positive

integer k.Step 3: Show that the given equation is true for n � k � 1.

1 � 2 � 4 � 8 � … � 2k � 1 � 2(k � 1) � 1 � (2k � 1) � 2(k � 1) � 1

� 2k � 1 � 2k � 2 � 2k � 1 � 2k � 1 � 1So, 1 � 2 � 4 � 8 � … � 2n � 1 � 2n � 1 for all positive integers n.

2. 1 � 4 � 9 � … � n2 �

Step 1: When n � 1, n2 � 12 � 1 � ; true for n � 1.

Step 2: Assume that 1 � 4 � 9 � … � k2 � �k(k � 1)

6(2k � 1)� for some positive

integer k.Step 3: Show that the given equation is true for n � k � 1.

1 � 4 � 9 � … � k2 � (k � 1)2 � � (k � 1)2

� � �

� �

�

So, 1 � 4 � 9 � … � n2 � for all positive integers n.

3. 18n � 1 is a multiple of 17.Step 1: When n � 1, 18n � 1 � 18 � 1 or 17; true for n � 1.Step 2: Assume that 18k � 1 is divisible by 17 for some positive integer k. This

means that there is a whole number r such that 18k � 1 � 17r.Step 3: Show that the statement is true for n � k � 1.

18k � 1 � 17r, so 18k � 17r � 1, and 18(18k) � 18(17r � 1). This isequivalent to 18k � 1 � 306r � 18, so 18k � 1 � 1 � 306r � 17, and 18k � 1 � 1 � 17(18r � 1).

Since r is a whole number, 18r � 1 is a whole number, and 18k � 1 � 1 isdivisible by 17. The statement is true for n � k � 1. So, 18n � 1 is divisible by17 for all positive integers n.

Find a counterexample for each statement.

4. 1 � 4 � 7 � … � (3n � 2) � n3 � n2 � 1 5. 5n � 2n � 3 is divisible by 3.


6. 1 � 3 � 5 � … � (2n � 1) � 7. 13 � 23 � 33 � … � n3 � n4 � n3 � 1


n2 � 3n � 2��2

n(n � 1)(2n � 1)��

6

(k � 1)[(k � 1) � 1][2(k � 1) � 1]��

6

(k � 1)[(k � 2)(2k � 3)]��

6(k � 1)(2k2 � 7k � 6)��

6

(k � 1)[k(2k � 1) � 6(k � 1)]��

66(k � 1)2��

6k(k � 1)(2k � 1)��

6

k(k � 1)(2k � 1)��

6

1(1 � 1)(2 � 1 � 1)��

n(n � 1)(2n � 1)��6

Practice (Average)

Proof and Mathematical Induction

NAME ______________________________________________ DATE ____________ PERIOD _____

11-811-8

Reading to Learn MathematicsProof and Mathematical Induction

NAME ______________________________________________ DATE ____________ PERIOD _____

11-811-8


Less

on

11-

8

Pre-Activity How does the concept of a ladder help you prove statements aboutnumbers?


What are two ways in which a ladder could be constructed so that you couldnot reach every step of the ladder?

Sample answer: 1. The first step could be too far off theground for you to climb on it. 2. The steps could be too farapart for you to go up from one step to the next.

Reading the Lesson

1. Fill in the blanks to describe the three steps in a proof by mathematical induction.

Step 1 Show that the statement is for the number .

Step 2 Assume that the statement is for some positive k.

This assumption is called the .

Step 3 Show that the statement is for the next integer .

2. Suppose that you wanted to prove that the following statement is true for all positiveintegers.

3 � 6 � 9 � … � 3n �

a. Which of the following statements shows that the statement is true for n � 1? ii

i. 3 � �3 � 2

2� 1� ii. 3 � iii. 3 �

b. Which of the following is the statement for n � k � 1? iv

i. 3 � 6 � 9 � … � 3k �

ii. 3 � 6 � 9 � … � 3k � 1 �

iii. 3 � 6 � 9 � … � 3k � 1 � 3(k � 1)(k � 2)

iv. 3 � 6 � 9 � … � 3(k � 1) �


3. Many students confuse the roles of n and k in a proof by mathematical induction. What is agood way to remember the difference in the ways these variables are used in such a proof?Sample answer: The letter n stands for “number” and is used as avariable to represent any natural number. The letter k is used torepresent a particular value of n.

3(k � 1)(k � 2)��2

3k(k � 1)��2

3k(k � 1)��2

3 � 1 � 2��2

3 � 1 � 2�2

3n(n � 1)��2

k � 1true

inductive hypothesis

integertrue

1true


Proof by InductionMathematical induction is a useful tool when you want to prove that astatement is true for all natural numbers.

The three steps in using induction are:1. Prove that the statement is true for n � 1.2. Prove that if the statement is true for the natural number n, it must also

be true for n � 1.3. Conclude that the statement is true for all natural numbers.

Follow the steps to complete each proof.

Theorem A: The sum of the first n odd natural numbers is equal to n2.

1. Show that the theorem is true for n � 1.

2. Suppose 1 � 3 � 5 � … � (2n � 1) � n2. Show that 1 � 3 � 5 � … � (2n � 1) � (2n � 1) � (n � 1)2.

3. Summarize the results of problems 1 and 2.

Theorem B: Show that an � bn is exactly divisible by a � b for n equal to 1, 2, 3, and all natural numbers.

4. Show that the theorem is true for n � 1.

5. The expression an � 1 � bn � 1 can be rewritten as a(an � bn) � bn(a � b).Verify that this is true.

6. Suppose a � b is a factor of an � bn. Use the result in problem 5 to show that a � b must then also be a factor of an � 1 � bn � 1.

7. Summarize the results of problems 4 through 6.

.

Enrichment

NAME ______________________________________________ DATE ____________ PERIOD _____

11-811-8

Chapter 11 Test, Form 1

NAME DATE PERIOD

SCORE


Ass

essm

ent

Write the letter for the correct answer in the blank at the right of each question.

1. Find the next four terms of the arithmetic sequence 11, 15, 19, … .A. 24, 29, 34, 39 B. 22, 25, 28, 31C. 20, 21, 22, 23 D. 23, 27, 31, 35 1.

2. Find the seventh term of the arithmetic sequence in which a1 � 3 and d � 5.

A. 33 B. 38 C. 30 D. 31 2.

3. Find the two arithmetic means between 10 and 70.A. 30, 50 B. 25, 45 C. 40, 40 D. 28, 43 3.

4. Find Sn for the arithmetic series in which a1 � 4, d � 3, and an � 61.A. 20 B. 1280 C. 64 D. 650 4.

5. Find the sum of the arithmetic series 8 � 5 � 2 � (�1) � … � (�13).A. 1 B. �20 C. 50 D. 29 5.

6. Find �5

n�1(4n � 1).

A. 44 B. 60 C. 65 D. 90 6.

7. Find the next two terms of the geometric sequence 567, 189, 63 … .A. 21, 3 B. 21, 7 C. �63, �189 D. 9, 3 7.

8. Find the fifth term of a geometric sequence for which a3 � 20 and r � 2.A. 80 B. 40 C. 160 D. 24 8.

9. Find two geometric means between 1 and 8.A. �2, 6 B. �2, 4 C. 2, 6 D. 2, 4 9.

10. Find the sum of a geometric series for which a1 � 7, n � 4, and r � 3.A. 91 B. 280 C. 147 D. 189 10.

11. Find �4

n�13 � 2n�1.

A. 80 B. �80 C. 45 D. �45 11.

12. Find a1 in a geometric series for which Sn � 93, r � 2, and n � 5.

A. �3 B. 15.5 C. 3 D. �13� 12.

1111


Chapter 11 Test, Form 1 (continued)

13. Find the sum of the infinite geometric series 12 � 6 � 3 � …, if it exists.A. 24 B. 8 C. 27 D. does not exist 13.

14. Write 0.4�8� as a fraction.

A. �418�

B. �136� C. �

1225�

D. �1363�

14.

15. Find the fifth term of the sequence in which a1 � 4 and an�1 � an � 6.

A. 5184 B. 34 C. 28 D. 22 15.

16. Find the third iterate x3 of f(x) � 2x � 3 for an initial value of x0 � 2.

A. 7 B. 15 C. 17 D. 37 16.

17. Use Pascal’s triangle to expand (m � 1)3.A. m3 � 3m2 � 3m � 1 B. m2 � 2m � 1C. m3 � 1 D. m3 � 2m2 � 2m � 1 17.

18. Evaluate �46!2!!�.

A. 1 B. 15 C. �34� D. 30 18.

19. n � 1 is a counterexample to which statement?A. 2 � 4 � 6 � … � 2n � n(n � 1) B. 4n � 1 is divisible by 3.

C. 1 � 2 � 3 � … � n � �n(n

2� 1)� D. 2n � 1 is divisible by 2. 19.

20. Which is not a step in an induction proof?A. Assume that the statement is true for some positive integer k.B. Show that the statement is true for some integer n.C. Show that the statement is true for some positive integer k.D. Show that the statement is true for the next integer k � 1. 20.

Bonus Express the series 3 � 6 � 12 � 24 � 48 using sigma notation. B:

NAME DATE PERIOD

1111

Chapter 11 Test, Form 2A

NAME DATE PERIOD

SCORE


Ass

essm

ent


1. Find the 20th term of the arithmetic sequence in which a1 � 5 and d � 4.A. 81 B. 85 C. 96 D. 105 1.

2. Write an equation for the nth term of the arithmetic sequence �7, �2, 3, 8, … .A. an � n � 5 B. an � 5n � 12C. an � �7n � 12 D. an � �7(n � 5) 2.


4. Find Sn for the arithmetic series in which a1 � 3, d � �12�, and an � �

127�.

A. 27 B. 54 C. �1329

� D. 69 4.

5. Find �22

n�18(50 � 2n).

A. 20 B. 40 C. 50 D. 100 5.

6. Find the sixth term of the geometric sequence for which a1 � 4 and r � 3.A. 247 B. 972 C. 733 D. 2916 6.

7. Write an equation for the nth term of the geometric sequence

�10, 5, � �52�, … .

A. an � �10��12��

n�1B. an � 10��

12��

n�1

C. an � �10��12��

n�1D. an � �10��

12��

�n�17.

8. Find four geometric means between 486 and 2.A. 162, 54, 18, 6 B. 389.2, 292.4, 195.6, 98.8C. 242, 121, 81, 16 D. �162, 54, �18, 6 8.

9. Find the sum of the geometric series 81 � 27 � 9 � … to 6 terms.

A. ��13� B. 121 C. 4941 D. �

1832

� 9.

10. Find �7

n�14(�3)n�1.

A. �2186 B. 2188 C. �728 D. 2916 10.

11. Find a1 in a geometric series for which Sn � 210, r � �2, and n � 6.

A. 10 B. �10 C. �110�

D. �130� 11.

1111


Chapter 11 Test, Form 2A (continued)

For Questions 12 and 13, find the sum of each infinite geometric series, if it exists.

12. ��

n�110��

15��

n�1

A. �235� B. 8 C. �

225� D. does not exist 12.

13. 5 � 4 � �156� � …

A. 20 B. 25 C. �245� D. does not exist 13.


A. �171�

B. �16030�

C. �23� D. 6�

13� 14.

15. Find the fifth term of the sequence in which a1 � �1 and an�1 � 2an � n.

A. �1 B. 25 C. 3 D. 10 15.

16. Find the third iterate x3 of f(x) � x2 � 3 for an initial value of x0 � 2.

A. 2 B. �2 C. 1 D. �1 16.

17. Use Pascal’s triangle to expand (m � n)5.A. m5 � 4m4n � 6m3n2 � (�6m2n3) � 4mn4 � n5

B. m5 � 5m4n � 10m3n2 � 10m2n3 � 5mn4 � n5

C. m5 � 4m4n � 6m3n2 � 6m2n3 � 4mn4 � n5

D. m5 � 5m4n � 10m3n2 � 10m2n3 � 5mn4 � n5 17.

18. Use the Binomial Theorem to find the third term in the expansion of (x � 3y)6.A. 15x4y2 B. 135x4y2 C. 540x3y3 D. 20x3y3 18.

19. Which is not a counterexample to the formula 22 � 42 � 62 � … � (2n)2 � 4n(2n � 1)?A. n � 4 B. n � 2 C. n � 3 D. n � 1 19.

20. In an induction proof of the statement 3 � 7 � 11 � … � (4n � 1) � n(2n � 1),the first step is to show that the statement is true for some integer n.Note: 4(1) � 1 � 1[2(1) � 1] is true. Select the steps required to complete the proof.A. Show that the statement is true for any real number k.

Show that the statement is true for k � 1.B. Assume that the statement is true for some positive integer k � 1.

Show that the statement is true for k.C. Assume that the statement is true for some positive integer k.

Show that the statement is true for k � 1.D. Show that the statement is true for some positive integer k.

Give a counterexample. 20.

Bonus Write the series 8 � 16 � 32 � 64 � 128 using sigma notation. B:

NAME DATE PERIOD

1111

Chapter 11 Test, Form 2B

NAME DATE PERIOD

SCORE


Ass

essm

ent


1. Find the 20th term of the arithmetic sequence in which a1 � 3 and d � 7.A. 143 B. 136 C. 140 D. 133 1.

2. Write an equation for the nth term of the arithmetic sequence �3, 3, 9, 15, … .A. an � n � 6 B. an � 6n � 9 C. an � 6n � 9 D. an � n � 3 2.


4. Find Sn for the arithmetic series in which a1 � 3, d � �12�, and an � 15.

A. 225 B. 9 C. 45 D. 210 4.

5. Find �8

k�3(40 � 3k).

A. 45 B. 282 C. �90 D. 141 5.


7. Write an equation for the nth term of the geometric sequence

�12, 4, ��43�, … .

A. an � �12��13��

n�1B. an � 12��

13��

n�1

C. an � �12��13��

�n�1D. an � �12��

13��

n�17.

8. Find four geometric means between 5 and 1215.A. �15, 45, �135, 405 B. 15, 45, 135, 405C. 247, 489, 731, 973 D. �247, 489, �731, 973 8.

9. Find the sum of the geometric series 128 � 64 � 32 � … to 8 terms.

A. 85 B. 255 C. 86 D. �825� 9.

10. Find �6

n�15(�4)n�1.

A. 6825 B. �4095 C. �1023 D. �5120 10.

11. Find a1 in a geometric series for which Sn � 300, r � �3, and n � 4.

A. 15 B. �125� C. �15 D. �1

15�

11.

For Questions 12 and 13, find the sum of each infinite geometric series,if it exists.

12. ��

n�120��

14��

n�1

A. 25 B. �830� C. 16 D. does not exist 12.

1111


Chapter 11 Test, Form 2B (continued)

13. 4 � 3 � �94� � …

A. �176� B. 16 C. �12 D. does not exist 13.


A. �79� B. �1

81�

C. �1285�

D. 7�29� 14.

15. Find the fifth term of the sequence in which a1 � �3 and an�1 � 3an � n.

A. �301 B. �99 C. �193 D. �341 15.

16. Find the third iterate x3 of f(x) � x2 � 4 for an initial value of x0 � 2.

A. �4 B. 4 C. 12 D. �12 16.

17. Use Pascal’s triangle to expand (w � x)5.A. w5 � 4w4x � 6w3x2 � 6w2x3 � 4wx4 � x5

B. w5 � 5w4x � 10w3x2 � 10w2x3 � 5wx4 � x5

C. w5 � 4w4x � 6w3x2 � 6w2x3 � 4wx4 � x5

D. w5 � 5w4x � 10w3x2 � 10w2x3 � 5wx4 � x5 17.

18. Use the Binomial Theorem to find the third term in the expansion of (n � 2p)6.A. 60n4p2 B. 120n3p3 C. �12n5p D. �160n2p4 18.

19. Which is not a counterexample to the formula

12 � 32 � 52 � … � (2n � 1)2 � �n(2n

3� 1)�?

A. n � 3 B. n � 2 C. n � 1 D. n � 4 19.

20. In an induction proof of the statement 4 � 7 � 10 � … � (3n � 1) � �n(3n

2� 5)�,

the first step is to show that the statement is true for some integer n.

Note: 3(1) � 1 � �1[3(1

2) � 5]� is true. Select the steps required to complete

the proof.A. Show that the statement is true for any real number k.

Show that the statement is true for k � 1.B. Assume that the statement is true for some positive integer k.

Show that the statement is true for k � 1.C. Show that the statement is true for some positive integer k.

Give a counterexample.D. Assume that the statement is true for some positive integer k � 1.

Show that the statement is true for k. 20.

Bonus Write the series 6 � 18 � 54 � 162 � 486 using sigma notation. B:

NAME DATE PERIOD

1111

Chapter 11 Test, Form 2C


1. Find the next four terms of the arithmetic sequence 1.18, 13, 8, … .

2. Find the 15th term of the arithmetic sequence in which 2.a1 � 10 and d � 4.

3. Write an equation for the nth term of the arithmetic 3.sequence 17, 8, �1, �10, … .

4. Find the four arithmetic means between �8 and 17. 4.

5. Find Sn for the arithmetic series in which 5.a1 � 5, an � 104, and n � 34.

6. Find the sum of the arithmetic series 6.7 � 4 � 1 � … � (�32).

7. Find �7

n�3(2n � 4). 7.

8. Find the fifth term of the geometric sequence for which 8.

a1 � 80 and r � �32�.

9. Find the next two terms of the geometric sequence 9.9, 6, 4, … .

10. Write an equation for the nth term of the geometric 10.

sequence 12, �3, �34�, … .

11. Find four geometric means between 2430 and 10. 11.

12. Find the sum of the geometric series �14� � �

12� � 1 � … 12.

to 7 terms.

13. Find �6

n�15 � 3n�1. 13.

NAME DATE PERIOD

SCORE 1111

Ass

essm

ent


Chapter 11 Test, Form 2C (continued)

14. Find a1 in a geometric series for which Sn � 242, r � 3, 14.and n � 5.

For Questions 15 and 16, find the sum of each infinite geometric series, if it exists. 15.

15. ��

n�115��

45��

n�116. 3 � 4 � �

136� � … 16.

17. Write 0.2�4� as a fraction. 17.

For Questions 18 and 19, find the first five terms of each sequence.

18. a1 � 11, an�1 � an � 2n 18.

19. a1 � 4, a2 � �3, an�2 � an�1 � an 19.

20. Find the first three iterates x1, x2, x3 of f(x) � x2 � 5 for an 20.initial value of x0 � �1.

21. Use Pascal’s triangle to expand (g � 3)4. 21.

22. Use the Binomial Theorem to find the second term in the 22.expansion of (3u � v)5.

23. Find a counterexample to the statement 5n � 2 is prime. 23.

24. A rock climber climbs 90 feet of steep rock face in the first 24.half-hour of climbing. In each succeeding half-hour, the climber achieves only 80% of the height achieved in the previous half-hour. Find the total height climbed.

25. Prove that the statement 25.

5 � 8 � 11 � � (3n � 2) � �n(3n

2� 7)� is true for all positive

integers n. Write your proof on a separate piece of paper.

Bonus The first term of a binomial expansion is x4 and the second term is 20x3y. What is the binomial, and to what power is it raised? B:

NAME DATE PERIOD

1111

Chapter 11 Test, Form 2D



2. Find the 13th term of the arithmetic sequence in which 2.a1 � 7 and d � 3.

3. Write an equation for the nth term of the arithmetic 3.sequence 15, 7, �1, �9, … .

4. Find the four arithmetic means between �6 and 9. 4.

5. Find Sn for the arithmetic series in which a1 � 7, d � 11, 5.and n � 20.

6. Find the sum of the arithmetic series 6.13 � 11 � 9 � … � (�25).

7. Find �5

n�1(2n � 9). 7.

8. Find the fifth term of the geometric sequence for which 8.

a1 � 1458 and r � �23�.

9. Find the next two terms of the geometric sequence 9.

10, 4, �85�, … .

10. Write an equation for the nth term of the geometric 10.sequence 27, �9, 3, … .


12. Find the sum of the geometric series �19� � �

13� � 1 � … 12.

to 6 terms.

13. Find �7

n�16 � 2n�1. 13.

14. Find a1 in a geometric series for which Sn � 105, r � �2, 14.and n � 6.

NAME DATE PERIOD

SCORE 1111

Ass

essm

ent


Chapter 11 Test, Form 2D (continued)


15. �74� � �

94� � �

8218�

� … 15.

16. ��

n�136��

45��

n�116.


For Questions 18 and 19, find the first five terms of each 18.sequence.

18. a1 � �4, an�1 � 5an � n

19. a1 � 13, a2 � 5, an�2 � an�1 � an

19.

20. Find the first three iterates x1, x2, x3 of 20.f(x) � 5 � x2 for an initial value of x0 � 2.

21. Use Pascal’s triangle to expand (c � 3)5.21.

22. Use the Binomial Theorem to find the fifth term in the 22.expansion of (4k � m)6.

23. Find a counterexample to the statement 3n � 2 is prime. 23.

24. A rock climber climbs 75 feet in the first half-hour of 24.climbing. In each succeeding half-hour, the climber achieves only 90% of the height achieved in the previous half-hour. Find the total height climbed.

25. Prove that the statement �12� � �2

12� � �2

13� � … � �2

1n� � 1 � �2

1n� 25.

is true for all positive integers n. Write your proof on a separate piece of paper.

Bonus The first term of a binomial expansion is x5 and the second term is 15x4y. What is the binomial, and to what power is it raised? B:

NAME DATE PERIOD

1111

Chapter 11 Test, Form 3


1. Find a25 for the arithmetic sequence �12�, �

15�, ��1

10�

, … . 1.

2. Write an equation for the nth term of the arithmetic 2.sequence �6.5, �5.1, �3.7, … .

3. �139� is the ? th term of ��

133�, ��

131�, �3, … . 3.

4. Find the three arithmetic means between �15� and �

1135�

. 4.

5. Find Sn for the arithmetic series in which 5.

a1 � �14�, an � �

4132�

, and d � �23�.

6. Find the first three terms of the arithmetic sequence in 6.which n � 39, an � 134.4, and Sn � 5538.

7. Find �200

k�30(12 � 4k). 7.

8. Find the next two terms of the geometric sequence 8.

�34�, ��

12�, �

13�, … .

9. Find the sixth term of the geometric sequence for which 9.a3 � 0.02 and r � 0.8.

10. Write an equation for the nth term of the geometric 10.sequence �6561, 1458, �324, … .

11. Find three geometric means between 12.8 and 0.8. 11.

12. Find the sum of a geometric series for which a4 � �80, 12.a7 � 640, and n � 12.

13. Find �10

n�196��

12��

n�1. 13.

14. Find a2 in a geometric series for which Sn � 65.984, r � 0.4, 14.and n � 5.

NAME DATE PERIOD

SCORE 1111

Ass

essm

ent


Chapter 11 Test, Form 3 (continued)


15. ��32� � �

12� � �

16� � … 15.

16. ��

n�1(2.4)(�0.8)n�1 16.

17. The first term of an infinite geometric series is �6 and its 17.sum is �5. Find the first four terms of the series.


For Questions 19 and 20, find the first five terms of each sequence.

19. a1 � �25�, an�1 � �n �

n2� � an 19.

20. a1 � �141�, a2 � �

72�, an�2 � 5an�1 � 2nan 20.

21. Find the first three iterates x1, x2, x3 of f(x) � 4x2 � 2x � 1 21.

for an initial value of x0 � �14�.

22. Use Pascal’s triangle to expand �a � �25��

6.

22.

23. Use the Binomial Theorem to find the fifth term in the 23.expansion of �2 � �4

x��

8.

24. Find a counterexample to the statement n2 � n � 1 is prime. 24.

25. A homeowner is planning a brick walkway that fans out from the back steps of the house. The first row uses four 25.bricks and each subsequent row uses five more bricks than the previous row. Prove that the number of bricks needed

for n rows is �n(5n2� 3)�. Write your proof on a separate piece

of paper.

Bonus Find all the values of x and y for which 3, x, y is an arithmetic sequence and x, y, 8 is a geometric sequence. B:

NAME DATE PERIOD

1111

Chapter 11 Open-Ended Assessment


Demonstrate your knowledge by giving a clear, concise solutionto each problem. Be sure to include all relevant drawings andjustify your answers. You may show your solution in more thanone way or investigate beyond the requirements of the problem.

1. While studying bacteria growth, Marita and Owen note that theirculture doubles in size every six hours. The initial bacteriapopulation was 500. They need to record in a table the number ofbacteria present at the end of each day for five days. They consulta math text to find a formula to help them with their calculations,but are not sure where to find the formula they need.a. From the lessons listed below, indicate the lesson(s) of the

math text to which you would refer Marita and Owen.Explain your choice(s).

11-1 Arithmetic Sequences11-2 Arithmetic Series11-3 Geometric Sequences11-4 Geometric Series

b. What formula should be used to determine the number ofbacteria present at any given time? (Note: 500 is the initialpopulation, but consider the number present at the end of the1st six-hour period as a1.)

c. Explain how to find the number of bacteria present at the endof the 5th day.

d. What might the final table prepared by Marita and Owen looklike?

2. Lucas was asked to find the 8th term in the expansion of (2x � y)11.a. Explain how Lucas could expand the binomial using Pascal’s

triangle.b. Explain how he could expand the binomial using the Binomial

Theorem.c. For this expansion, which method would you prefer to use?

Explain your choice. Then use the method you selected to findthe desired term.

3. Each student in your algebra class is given an assignment toprepare a portion of a practice test that the entire class will usefor review of Chapter 11. Your assignment: list the termsarithmetic sequence, arithmetic series, geometric sequence,geometric series, infinite geometric series, binomial expansion, andcounterexample in a column. Then list an example of each in asecond column. These examples should be scrambled to create amatching test. At least one example should use sigma notation.Once your test is created, prepare an answer key.

4. Explain the difference between �6

n�12 � 3n�1 and 2 � �

6

n�11 � 3n�1.

Write and evaluate the series represented by each, then discussthe results.

NAME DATE PERIOD

SCORE 1111

Ass

essm

ent


Chapter 11 Vocabulary Test/Review

Write whether each sentence is true or false. If false, replace the underlined word or words to make a true sentence.

1. In a geometric sequence, there is a common difference 1.between successive terms.

2. An inductive hypothesis is used in Pascal’s triangle. 2.

3. Sigma notation is a shorthand way to write a sequence. 3.

4. The variable that indicates the starting and ending values to 4.be used in sigma notation is called the index of summation.

5. The formula S � �1

a

�1

r� is used to find the sum of a(n) 5.

Fibonacci sequence.

6. If a sequence is defined by a formula that uses the previous 6.term to find each term after the first, the sequence is defined by a partial sum.

7. The coefficients of the terms in a binomial expansion can be 7.found by using factorials or Pascal’s triangle.

8. The process of repeatedly composing a function with itself 8.is called mathematical induction.

9. Each number in a sequence is called a factorial. 9.

10. The sum of the terms in a sequence in which there is a 10.common ratio between consecutive terms is called a geometric series.

In your own words—Define each term.

11. Fibonacci sequence

12. geometric mean

arithmetic meanarithmetic sequencearithmetic seriesBinomial Theoremcommon differencecommon ratio

factorialFibonacci sequencegeometric meangeometric sequencegeometric seriesindex of summation

inductive hypothesisinfinite geometric seriesiterationmathematical inductionpartial sumPascal’s triangle

recursive formulasequenceseriessigma notationterm

NAME DATE PERIOD

SCORE 1111

Chapter 11 Quiz (Lessons 11–1 and 11–2)

1111



2. Find the first five terms of the arithmetic sequence in which 2.a1 � 4 and d � 7.

3. Find the tenth term of the arithmetic sequence in which 3.a1 � 6 and d � 5.

4. Write an equation for the nth term of the arithmetic sequence 4.4, 1, �2, �5, … .

5. Find the three arithmetic means between �4 and 16. 5.6. Find Sn for the arithmetic series in which a1 � �5, n � 8, 6.

and d � 6.7. Find the sum of the arithmetic series 12 � 8 � 4 � … � (�20). 7.8. Find the first three terms of the arithmetic series in which 8.

a1 � 2, an � �25, and Sn � �115.Find the sum of each arithmetic series. 9.

9. �7

j�3(4 � j) 10. �

15

k�11(3k � 2) 10.

NAME DATE PERIOD

SCORE


1. Standardized Test Practice Find the missing term in the geometric sequence 64, 96, 144, 216, __?___.

A. 72 B. 1024 C. 324 D. 360 1.

2. Find the first five terms of the geometric sequence for which 2.a1 � 3 and r � �2.


4. Find the sum of a geometric series for which a1 � 3125, 4.

an � 1, and r � �15�.

5. Find a1 in a geometric series for which Sn � 3045, 5.

r � �25�, and an � 120.

NAME DATE PERIOD

SCORE 1111

Ass

essm

ent



1. ��

n�127��

23��

n�12. �

�

n�14��

52��

n�1

3. a1 � 24, r � ��35� 4. �

92� � �

98� � �3

92�

� …


5. 0.6� 6. 0.4�5�


7. a1 � 3, an�1 � an � 2n 8. a1 � �2, an�1 � 1 � 2an

Find the first three interates of each function for the given initial value.

9. f(x) � 5x � 4, x0 � �3 10. f(x) � 2x2 � 7, x0 � 2


Use Pascal’s triangle to expand each power.

1. (x � y)5 2. (m � 6)3

Evaluate.

3. �56!1!!� 4. �2

9!7!!�

Use the Binomial Theorem to expand each power.

5. (3r � 2)4 6. (x � 2y)6

For Questions 7 and 8, find the indicated term of each expansion.

7. fifth term of (a � 1)7 8. seventh term of (3x � y)9

9. Find a counterexample to the statement 4n � 4 is divisible by 8.

10. Prove that the statement 1 � 3 � 5 � … � (2n � 1) � n2 is true for all positive integers n. Write your proof on a separate piece of paper.

NAME DATE PERIOD

SCORE


1111

NAME DATE PERIOD

SCORE

1111

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

Chapter 11 Mid-Chapter Test (Lessons 11–1 through 11–4)



1. Find the tenth term of the arithmetic sequence in which a1 � 5 and d � 4.A. 37 B. 44 C. 41 D. 20 1.

2. A water tank is emptied at a constant rate. At the end of the first hour,36,000 gallons of water were in the tank. After six hours, 21,000 gallons remained. How many gallons of water were in the tank at the end of the fourth hour?A. 30,000 gal B. 24,000 gal C. 28,500 gal D. 27,000 gal 2.

3. Find Sn for the arithmetic series in which a1 � 37, n � 11, and d � �3.A. 45 B. 235 C. 242 D. 572 3.


5. Find a1 in a geometric series for which Sn � �728, r � 3, and n � 6.A. �2 B. 1456 C. �4 D. 4 5.

6. Evaluate �15

n�7(3n � 5).

A. 252 B. 285 C. 342 D. 435 6.

7. Find the sum of the geometric series 512 � 256 � 128 � … to 6 terms.A. 992 B. 1000 C. 896 D. 1008 7.

8. Find the sum of the arithmetic series 23 � 18 � 13 � 8 � … � (�82).A. �590 B. 590 C. �649 D. 649 8.

9. Find two geometric means between 9 and �243. 9.

10. Write an equation for the nth term of the arithmetic sequence 10.5, 3, 1, �1, … .

11. Find the four arithmetic means between 22 and 2. 11.

12. Write an equation for the nth term of the geometric sequence 40, 20, 10, … .

13. Evaluate �6

n�12 � 3n�1. 13.

14. Find the first three terms of the arithmetic series in which 14.n � 12, an � �41, and Sn � �228.

15. Find the first five terms of the geometric sequence for which 15.

a1 � 20 and r � ��12�.

16. Find the next four terms of the arithmetic sequence 7, 4, 1, … . 16.

Part II

Part I

NAME DATE PERIOD

SCORE 1111

Ass

essm

ent

12.


Chapter 11 Cumulative Review (Chapters 1–11)

1. Find the slope of the line that passes through (�2, 5) and 1.(�6, �2). (Lesson 2-3)

2. Use Cramer’s Rule to solve the system of equations.2x � 3y � z � �23x � y � z � �14x � 3y � 5z � 14 (Lesson 4-6)

3. Simplify (4x � 3)(3x � 4). (Lesson 5-2)

4. Graph y � �x � 4�. Then state the domain and range of the function. (Lesson 7-9)

5. Write 4x2 � y � �24x � 37 in standard form. Then state whether the graph of the equation is a parabola, circle,ellipse, or hyperbola. (Lesson 8-6)

6. Solve the system of inequalities y � x2 � 8x � 16x � y 4

by graphing. (Lessons 8-2 and 8-7)

7. Determine the equations of any vertical asymptotes and the values of x for any holes in the graph of the rational

function f(x) � �x2 �x �

7x1� 8�. (Lesson 9-3)

8. If y varies inversely as x and y � 5 when x � 5, find ywhen x � 35. (Lesson 9-4)

9. Simplify 8�11� � 85�11�. (Lesson 10-1)

10. Use log5 2 � 0.4307 and log5 3 � 0.6826 to approximate

the value of log5 �29�. (Lesson 10-2)

11. A radioactive compound decays according to the equation y � ae�0.0935t, where t is in days. Find the half-life of the substance. (Lesson 10-6)

12. Find the first five terms of the arithmetic sequence in which a1 � 7 and d � �3. (Lesson 11-1)

13. Find the sum of a geometric series for which a1 � 80,r � �

12�, and n � 5. (Lesson 11-4)

14. Find the sum of the infinite geometric series 90 � 60 � 40 � …, if it exists. (Lesson 11-5)

15. Use the Binomial Theorem to expand (x � 3y)4. (Lesson 11-7)

NAME DATE PERIOD

1111

1.

2.

3.4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

y

xO

y

xO

Standardized Test Practice (Chapters 1–11)


1. Let represent one of the four arithmetic operations.If, for any real number except zero, it is true that a1 � aand ab � ba, then which operation(s) might represent?I. addition II. subtraction III. multiplication IV. divisionA. I and III only B. IV onlyC. III only D. II and IV only 1.

2. If b � 16 � a2 and 2 � a � 4, what is the largest possible value of b?E. 8 F. 16 G. 0 H. 12 2.

3. If 23x�2 � 44, what is the value of x?

A. �23� B. �1 C. 2 D. �

43� 3.

4. How many prime numbers p are in the interval 11 � p � 53?E. eleven F. ten G. twelve H. thirteen 4.

5. A store sells 16-ounce jars of tomato sauce for $1.89 each.A 24-ounce jar of the same sauce costs $2.69. How much money is saved purchasing 96 ounces of sauce in 24-ounce jars rather than in 16-ounce jars?A. $0.58 B. $0.80 C. $1.20 D. $0.42 5.

For Questions 6 and 7, the bar graph represents the approximate annual incomes for the households in one city.

6. How many households had annual incomes between $25,000 and $100,000?E. 1000 F. 600G. 1200 H. 1800 6.

7. What percent of these households had annual incomes greater than $50,000?A. 80% B. 40% C. 60% D. 50% 7.

8. If 6 � 8x � 2x � 11 and �y �3

1� � �1

72�

, then x � y � _______.

E. �54� F. ��

14� G. �

76� H. �

14� 8.

9. If m2 � am � 9b2 � 3ab, what is a in terms of m and b?

A. m � 3b B. m � 3b C. 9b2 � m D. �3mb�

9. DCBA

HGFE

DCBA

HGFE

DCBA

HGFE

DCBA

HGFE

DCBA

NAME DATE PERIOD

1111

Ass

essm

ent

Part 1: Multiple Choice

Instructions: Fill in the appropriate oval for the best answer.

200300

1000

500600700

400

Num

ber o

f hou

seho

lds

Annual household salary(in thousands of dollars)

25 50 75 100150 over150


Standardized Test Practice (continued)

10. In the figure, the area of 10. 11.circle O is 16π and the circumference of circle Pis 16π. What is the length of O�P�?

11. What is the length of a line segment whose endpoints are at (3, 3�6�) and (�2, 5�6�)?

12. If A � {1, 2, 3, …, 25} and B � {0, 5, 10, …, 20}, 12. 13.how many elements are in the set A � B?

13. If the letters A, H, M, and T are randomly arranged,what is the probability that the arrangement willresult in the word MATH?

Column A Column B

14. 2r � 24, 24 � 2s 14.

15. p q 15.

16. 16.

40a � b

DCBA

a˚ a˚

b˚ 5b˚

2b˚6a˚

(p � q)2(p � q)2

DCBA

sr

DCBA

Part 3: Quantitative Comparison

Instructions: Compare the quantities in columns A and B. Shade in if the quantity in column A is greater; if the quantity in column B is greater; if the quantities are equal; or if the relationship cannot be determined from the information given.

0 0 0

.. ./ /

.

99 9 987654321

87654321

87654321

87654321

0 0 0

.. ./ /

.

99 9 987654321

87654321

87654321

87654321

0 0 0

.. ./ /

.

99 9 987654321

87654321

87654321

87654321

0 0 0

.. ./ /

.

99 9 987654321

87654321

87654321

87654321

NAME DATE PERIOD

1111

NAME DATE PERIOD

Part 2: Grid In

Instructions: Enter your answer by writing each digit of the answer in a column boxand then shading in the appropriate oval that corresponds to that entry.

O P

A

D

C

B

Standardized Test PracticeStudent Record Sheet (Use with pages 628–629 of the Student Edition.)

© Glencoe/McGraw-Hill A1 Glencoe Algebra 2

NAME DATE PERIOD

1111

An

swer

s

Select the best answer from the choices given and fill in the corresponding oval.

1 4 7 9

2 5 8 10

3 6

Solve the problem and write your answer in the blank.

Also enter your answer by writing each number or symbol in a box. Then fill inthe corresponding oval for that number or symbol.

11 13 15 17

12 14 16

Select the best answer from the choices given and fill in the corresponding oval.

18 20 22

19 21 DCBADCBA

DCBADCBADCBA

0 0 0

.. ./ /

.

99 9 987654321

87654321

87654321

87654321

0 0 0

.. ./ /

.

99 9 987654321

87654321

87654321

87654321

0 0 0

.. ./ /

.

99 9 987654321

87654321

87654321

87654321

0 0 0

.. ./ /

.

99 9 987654321

87654321

87654321

87654321

0 0 0

.. ./ /

.

99 9 987654321

87654321

87654321

87654321

0 0 0

.. ./ /

.

99 9 987654321

87654321

87654321

87654321

0 0 0

.. ./ /

.

99 9 987654321

87654321

87654321

87654321

DCBADCBA

DCBADCBADCBADCBA

DCBADCBADCBADCBA

Part 2 Short Response/Grid InPart 2 Short Response/Grid In

Part 1 Multiple ChoicePart 1 Multiple Choice

Part 3 Quantitative ComparisonPart 3 Quantitative Comparison


Answers (Lesson 11-1)

Stu

dy G

uid

e a

nd I

nte

rven

tion

Ari

thm

etic

Seq

uen

ces

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

11-1

11-1

©G

lenc

oe/M

cGra

w-H

ill63

1G

lenc

oe A

lgeb

ra 2

Lesson 11-1

Ari

thm

etic

Seq

uen

ces

An

arit

hm

etic

seq

uen

ce is

a s

eque

nce

of n

umbe

rs in

whi

ch e

ach

term

afte

r th

e fi

rst

term

is f

ound

by

addi

ng t

he c

omm

on d

iffe

ren

ce t

o th

e pr

eced

ing

term

.

nth

Ter

m o

f an

a n

�a 1

�(n

�1)

d, w

here

a1

is t

he f

irst

term

, d

is t

he c

omm

on d

iffer

ence

, A

rith

met

ic S

equ

ence

and

nis

any

pos

itive

inte

ger

Fin

d t

he

nex

t fo

ur

term

s of

th

e ar

ith

met

ic s

equ

ence

7,

11,1

5,…

.F

ind

the

com

mon

dif

fere

nce

by

subt

ract

ing

two

con

secu

tive

ter

ms.

11 �

7 �

4 an

d 15

�11

�4,

so d

�4.

Now

add

4 t

o th

e th

ird

term

of

the

sequ

ence

,an

d th

en c

onti

nu

e ad

din

g 4

un

til

the

fou

rte

rms

are

fou

nd.

Th

e n

ext

fou

r te

rms

of t

he

sequ

ence

are

19,

23,2

7,an

d 31

.

Fin

d t

he

thir

teen

th t

erm

of t

he

arit

hm

etic

seq

uen

ce w

ith

a1

�21

and

d�

�6.

Use

the

for

mul

a fo

r th

e nt

h te

rm o

f an

arit

hmet

ic s

eque

nce

wit

h a 1

�21

,n�

13,

and

d�

�6.

a n�

a 1�

(n�

1)d

For

mul

a fo

r nt

h te

rm

a 13

�21

�(1

3 �

1)(�

6)n

�13

, a 1

�21

, d

��

6

a 13

��

51S

impl

ify.

Th

e th

irte

enth

ter

m i

s �

51.

Exam

ple1

Exam

ple1

Exam

ple2

Exam

ple2

Exam

ple

3Ex

ampl

e 3

Wri

te a

n e

qu

atio

n f

or t

he

nth

ter

m o

f th

e ar

ith

met

ic s

equ

ence

�

14,�

5,4,

13,…

.In

th

is s

equ

ence

a1

��

14 a

nd

d�

9.U

se t

he

form

ula

for

an

to w

rite

an

equ

atio

n.

a n�

a 1�

(n�

1)d

For

mul

a fo

r th

e nt

h te

rm

��

14 �

(n�

1)9

a 1�

�14

, d

�9

��

14 �

9n�

9D

istr

ibut

ive

Pro

pert

y

�9n

�23

Sim

plify

.

Fin

d t

he

nex

t fo

ur

term

s of

eac

h a

rith

met

ic s

equ

ence

.

1.10

6,11

1,11

6,…

2.

�28

,�31

,�34

,…

3.20

7,19

4,18

1,…

121,

126,

131,

136

�37

,�40

,�43

,�46

168,

155,

142,

129

Fin

d t

he

firs

t fi

ve t

erm

s of

eac

h a

rith

met

ic s

equ

ence

des

crib

ed.

4.a 1

�10

1,d

�9

5.a 1

��

60,d

�4

6.a 1

�21

0,d

��

4010

1,11

0,11

9,12

8,13

7�

60,�

56,�

52,�

48,�

4421

0,17

0,13

0,90

,50

Fin

d t

he

ind

icat

ed t

erm

of

each

ari

thm

etic

seq

uen

ce.

7.a 1

�4,

d�

6,n

�14

828.

a 1�

�4,

d�

�2,

n�

12�

269.

a 1�

80,d

��

8,n

�21

�80

10.a

10fo

r 0,

�3,

�6,

�9,

…�

27

Wri

te a

n e

qu

atio

n f

or t

he

nth

ter

m o

f ea

ch a

rith

met

ic s

equ

ence

.

11.1

8,25

,32,

39,…

12

.�11

0,�

85,�

60,�

35,…

13

.6.2

,8.1

,10.

0,11

.9,…

7n�

1125

n�

135

1.9n

�4.

3

Exer

cises

Exer

cises

©G

lenc

oe/M

cGra

w-H

ill63

2G

lenc

oe A

lgeb

ra 2

Ari

thm

etic

Mea

ns

Th

e ar

ith

met

ic m

ean

sof

an

ari

thm

etic

seq

uen

ce a

re t

he

term

sbe

twee

n a

ny

two

non

succ

essi

ve t

erm

s of

th

e se

quen

ce.

To

fin

d th

e k

arit

hm

etic

mea

ns

betw

een

tw

o te

rms

of a

seq

uen

ce,u

se t

he

foll

owin

g st

eps.

Ste

p 1

Let

the

two

term

s gi

ven

be a

1an

d a n

, w

here

n�

k�

2.S

tep

2S

ubst

itute

in t

he f

orm

ula

a n�

a 1�

(n�

1)d.

Ste

p 3

Sol

ve f

or d

, an

d us

e th

at v

alue

to

find

the

kar

ithm

etic

mea

ns:

a 1�

d, a

1�

2d,

… ,

a1

�kd

.

Fin

d t

he

five

ari

thm

etic

mea

ns

bet

wee

n 3

7 an

d 1

21.

You

can

use

th

e n

th t

erm

for

mu

la t

o fi

nd

the

com

mon

dif

fere

nce

.In

th

e se

quen

ce,

37,

,,

,,

,121

,…,a

1is

37

and

a 7is

121

.

a n�

a 1�

(n�

1)d

For

mul

a fo

r th

e nt

h te

rm

121

�37

�(7

�1)

da 1

�37

, a 7

�12

1, n

�7

121

�37

�6d

Sim

plify

.

84 �

6dS

ubtr

act

37 f

rom

eac

h si

de.

d�

14D

ivid

e ea

ch s

ide

by 6

.

Now

use

th

e va

lue

of d

to f

ind

the

five

ari

thm

etic

mea

ns.

37 �

51 �

65 �

79 �

93 �

107 �

121

�14

�

14

�14

�

14

�14

�

14T

he

arit

hm

etic

mea

ns

are

51,6

5,79

,93,

and

107.

Fin

d t

he

arit

hm

etic

mea

ns

in e

ach

seq

uen

ce.

1.5,

,,

,�3

2.18

,,

,,�

23.

16,

,,3

73,

1,�

113

,8,3

23,3

0

4.10

8,,

,,

,48

5.�

14,

,,

,�30

6.29

,,

,,8

996

,84,

72,6

0�

18,�

22,�

2644

,59,

74

7.61

,,

,,

,116

8.45

,,

,,

,,8

172

,83,

94,1

0551

,57,

63,6

9,75

9.�

18,

,,

,14

10.�

40,

,,

,,

,�82

�10

,�2,

6�

47,�

54,�

61,�

68,�

75

11.1

00,

,,2

3512

.80,

,,

,,�

3014

5,19

058

,36,

14,�

8

13.4

50,

,,

,570

14.2

7,,

,,

,,5

748

0,51

0,54

032

,37,

42,4

7,52

15.1

25,

,,

,185

16.2

30,

,,

,,

,128

140,

155,

170

213,

196,

179,

162,

145

17.�

20,

,,

,,3

7018

.48,

,,

,100

58,1

36,2

14,2

9261

,74,

87?

??

??

??

??

??

??

??

??

??

??

??

??

??

??

??

??

??

??

??

??

??

??

?

??

??

??

??

??

??

??

??

??

??

??

?

Stu

dy G

uid

e a

nd I

nte

rven

tion

(c

onti

nued

)

Ari

thm

etic

Seq

uen

ces

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

11-1

11-1

Exam

ple

Exam

ple

Exer

cises

Exer

cises


An

swer

s


Skil

ls P

ract

ice

Ari

thm

etic

Seq

uen

ces

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

11-1

11-1

©G

lenc

oe/M

cGra

w-H

ill63

3G

lenc

oe A

lgeb

ra 2

Lesson 11-1

Fin

d t

he

nex

t fo

ur

term

s of

eac

h a

rith

met

ic s

equ

ence

.

1.7,

11,1

5,…

19,2

3,27

,31

2.�

10,�

5,0,

…5,

10,1

5,20

3.10

1,20

2,30

3,…

404,

505,

606,

707

4.15

,7,�

1,…

�9,

�17

,�25

,�33

5.�

67,�

60,�

53,…

6.�

12,�

15,�

18,…

�46

,�39

,�32

,�25

�21

,�24

,�27

,�30

Fin

d t

he

firs

t fi

ve t

erm

s of

eac

h a

rith

met

ic s

equ

ence

des

crib

ed.

7.a 1

�6,

d�

96,

15,2

4,33

,42

8.a 1

�27

,d�

427

,31,

35,3

9,43

9.a 1

��

12,d

�5

�12

,�7,

�2,

3,8

10.a

1�

93,d

��

1593

,78,

63,4

8,33

11.a

1�

�64

,d�

1112

.a1

��

47,d

��

20�

64,�

53,�

42,�

31,�

20�

47,�

67,�

87,�

107,

�12

7

Fin

d t

he

ind

icat

ed t

erm

of

each

ari

thm

etic

seq

uen

ce.

13.a

1�

2,d

�6,

n�

1268

14.a

1�

18,d

�2,

n�

832

15.a

1�

23,d

�5,

n�

2313

316

.a1

�15

,d�

�1,

n�

25�

9

17.a

31fo

r 34

,38,

42,…

154

18.a

42fo

r 27

,30,

33,…

150

Com

ple

te t

he

stat

emen

t fo

r ea

ch a

rith

met

ic s

equ

ence

.

19.5

5 is

th

e th

ter

m o

f 4,

7,10

,….

1820

.163

is

the

th t

erm

of

�5,

2,9,

….

25

Wri

te a

n e

qu

atio

n f

or t

he

nth

ter

m o

f ea

ch a

rith

met

ic s

equ

ence

.

21.4

,7,1

0,13

,…a n

�3n

�1

22.�

1,1,

3,5,

…a n

�2n

�3

23.�

1,3,

7,11

,…a n

�4n

�5

24.7

,2,�

3,�

8,…

a n�

�5n

�12

Fin

d t

he

arit

hm

etic

mea

ns

in e

ach

seq

uen

ce.

25.6

,,

,,3

814

,22,

3026

.63,

,,

,147

84,1

05,1

26?

??

??

?

??

©G

lenc

oe/M

cGra

w-H

ill63

4G

lenc

oe A

lgeb

ra 2

Fin

d t

he

nex

t fo

ur

term

s of

eac

h a

rith

met

ic s

equ

ence

.

1.5,

8,11

,…14

,17,

20,2

32.

�4,

�6,

�8,

…�

10,�

12,�

14,�

16

3.10

0,93

,86,

…79

,72,

65,5

84.

�24

,�19

,�14

,…�

9,�

4,1,

6

5.,6

,,1

1,…

,16,

,21

6.4.

8,4.

1,3.

4,…

2.7,

2,1.

3,0.

6

Fin

d t

he

firs

t fi

ve t

erm

s of

eac

h a

rith

met

ic s

equ

ence

des

crib

ed.

7.a 1

�7,

d�

78.

a 1�

�8,

d�

2

7,14

,21,

28,3

5�

8,�

6,�

4,�

2,0

9.a 1

��

12,d

��

410

.a1

�,d

�

�12

,�16

,�20

,�24

,�28

,1,

,2,

11.a

1�

�,d

��

12.a

1�

10.2

,d�

�5.

8

�,�

,�,�

,�10

.2,4

.4,�

1.4,

�7.

2,�

13

Fin

d t

he

ind

icat

ed t

erm

of

each

ari

thm

etic

seq

uen

ce.

13.a

1�

5,d

�3,

n�

1032

14.a

1�

9,d

�3,

n�

2993

15.a

18fo

r �

6,�

7,�

8,…

.�

2316

.a37

for

124,

119,

114,

….

�56

17.a

1�

,d�

�,n

�10

�18

.a1

�14

.25,

d�

0.15

,n�

3118

.75

Com

ple

te t

he

stat

emen

t fo

r ea

ch a

rith

met

ic s

equ

ence

.

19.1

66 i

s th

e th

ter

m o

f 30

,34,

38,…

3520

.2 i

s th

e th

ter

m o

f ,

,1,…

8

Wri

te a

n e

qu

atio

n f

or t

he

nth

ter

m o

f ea

ch a

rith

met

ic s

equ

ence

.

21.�

5,�

3,�

1,1,

…a n

�2n

�7

22.�

8,�

11,�

14,�

17,…

a n�

�3n

�5

23.1

,�1,

�3,

�5,

…a n

��

2n�

324

.�5,

3,11

,19,

…a n

�8n

�13

Fin

d t

he

arit

hm

etic

mea

ns

in e

ach

seq

uen

ce.

25.�

5,,

,,1

1�

1,3,

726

.82,

,,

,18

66,5

0,34

27.E

DU

CA

TIO

NT

revo

r K

oba

has

ope

ned

an

En

glis

h L

angu

age

Sch

ool

in I

seh

ara,

Japa

n.

He

bega

n w

ith

26

stu

den

ts.I

f h

e en

roll

s 3

new

stu

den

ts e

ach

wee

k,in

how

man

y w

eeks

wil

l h

e h

ave

101

stu

den

ts?

26 w

k

28.S

ALA

RIE

SYo

lan

da i

nte

rvie

wed

for

a jo

b th

at p

rom

ised

her

a s

tart

ing

sala

ry o

f $3

2,00

0w

ith

a $

1250

rai

se a

t th

e en

d of

eac

h y

ear.

Wh

at w

ill

her

sal

ary

be d

uri

ng

her

six

th y

ear

if s

he

acce

pts

the

job?

$38,

250

??

??

??

4 � 53 � 5

??

18 � 53 � 5

9 � 5

13 � 611 � 6

3 � 27 � 6

5 � 6

1 � 35 � 6

5 � 23 � 2

1 � 2

1 � 21 � 2

37 � 227 � 2

17 � 27 � 2

Pra

ctic

e (

Ave

rag

e)

Ari

thm

etic

Seq

uen

ces

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

11-1

11-1



Readin

g t

o L

earn

Math

em

ati

csA

rith

met

ic S

equ

ence

s

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

11-1

11-1

©G

lenc

oe/M

cGra

w-H

ill63

5G

lenc

oe A

lgeb

ra 2

Lesson 11-1

Pre-

Act

ivit

yH

ow a

re a

rith

met

ic s

equ

ence

s re

late

d t

o ro

ofin

g?

Rea

d th

e in

trod

ucti

on t

o L

esso

n 11

-1 a

t th

e to

p of

pag

e 57

8 in

you

r te

xtbo

ok.

Des

crib

e ho

w y

ou w

ould

fin

d th

e nu

mbe

r of

shi

ngle

s ne

eded

for

the

fif

teen

thro

w.(

Do

not

act

ual

ly c

alcu

late

th

is n

um

ber.

) E

xpla

in w

hy

you

r m

eth

od w

ill

give

th

e co

rrec

t an

swer

.S

amp

le a

nsw

er:

Ad

d 3

tim

es 1

4 to

2.T

his

wo

rks

bec

ause

th

e fi

rst

row

has

2 s

hin

gle

s an

d 3

mo

re a

read

ded

14

tim

es t

o g

o f

rom

th

e fi

rst

row

to

th

e fi

ftee

nth

ro

w.

Rea

din

g t

he

Less

on

1.C

onsi

der

the

form

ula

an

�a 1

�(n

�1)

d.

a.W

hat

is

this

for

mu

la u

sed

to f

ind?

a p

arti

cula

r te

rm o

f an

ari

thm

etic

seq

uen

ce

b.

Wh

at d

o ea

ch o

f th

e fo

llow

ing

repr

esen

t?

a n:

the

nth

ter

m

a 1:

the

firs

t te

rm

n:

a p

osi

tive

inte

ger

th

at in

dic

ates

wh

ich

ter

m y

ou

are

fin

din

g

d:

the

com

mo

n d

iffe

ren

ce

2.C

onsi

der

the

equ

atio

n a

n�

�3n

�5.

a.W

hat

doe

s th

is e

quat

ion

rep

rese

nt?

Sam

ple

an

swer

:It

giv

es t

he

nth

ter

m o

fan

ari

thm

etic

seq

uen

ce w

ith

fir

st t

erm

2 a

nd

co

mm

on

dif

fere

nce

�3.

b.

Is t

he

grap

h o

f th

is e

quat

ion

a s

trai

ght

lin

e? E

xpla

in y

our

answ

er.

Sam

ple

answ

er:

No

;th

e g

rap

h is

a s

et o

f p

oin

ts t

hat

fal

l on

a li

ne,

but

the

po

ints

do

no

t fi

ll th

e lin

e.c.

Th

e fu

nct

ion

s re

pres

ente

d by

th

e eq

uat

ion

s a n

��

3n�

5 an

d f(

x) �

�3x

�5

are

alik

e in

th

at t

hey

hav

e th

e sa

me

form

ula

.How

are

th

ey d

iffe

ren

t?S

amp

lean

swer

:Th

ey h

ave

dif

fere

nt

do

mai

ns.

Th

e d

om

ain

of

the

firs

t fu

nct

ion

is t

he

set

of

po

siti

ve in

teg

ers.

Th

e d

om

ain

of

the

seco

nd

fu

nct

ion

isth

e se

t o

f al

l rea

l nu

mb

ers.

Hel

pin

g Y

ou

Rem

emb

er3.

A g

ood

way

to

rem

embe

r so

met

hin

g is

to

expl

ain

it

to s

omeo

ne

else

.Su

ppos

e th

at y

our

clas

smat

e S

hal

a h

as t

rou

ble

rem

embe

rin

g th

e fo

rmu

la a

n�

a 1�

(n�

1)d

corr

ectl

y.S

he

thin

ks t

hat

th

e fo

rmu

la s

hou

ld b

e a n

�a 1

�n

d.H

ow w

ould

you

exp

lain

to

her

th

at s

he

shou

ld u

se (

n�

1)d

rath

er t

han

nd

in t

he

form

ula

?S

amp

le a

nsw

er:

Eac

h t

erm

afte

r th

e fi

rst

in a

n a

rith

met

ic s

equ

ence

is f

ou

nd

by

add

ing

dto

th

ep

revi

ou

s te

rm.Y

ou

wo

uld

ad

d d

on

ce t

o g

et t

o t

he

seco

nd

ter

m,t

wic

e to

get

to

th

e th

ird

ter

m,a

nd

so

on

.So

dis

ad

ded

n�

1 ti

mes

,no

t n

tim

es,

to g

et t

he

nth

ter

m.

©G

lenc

oe/M

cGra

w-H

ill63

6G

lenc

oe A

lgeb

ra 2

Fib

on

acci

Seq

uen

ceL

eon

ardo

Fib

onac

ci f

irst

dis

cove

red

the

sequ

ence

of

nu

mbe

rs n

amed

for

him

wh

ile

stu

dyin

g ra

bbit

s.H

e w

ante

d to

kn

ow h

ow m

any

pair

s of

rab

bits

wou

ldbe

pro

duce

d in

nm

onth

s,st

arti

ng

wit

h a

sin

gle

pair

of

new

born

rab

bits

.He

mad

e th

e fo

llow

ing

assu

mpt

ion

s.

1.N

ewbo

rn r

abbi

ts b

ecom

e ad

ult

s in

on

e m

onth

.

2.E

ach

pai

r of

rab

bits

pro

duce

s on

e pa

ir e

ach

mon

th.

3.N

o ra

bbit

s di

e.

Let

Fn

repr

esen

t th

e n

um

ber

of p

airs

of

rabb

its

at t

he

end

of n

mon

ths.

If y

oube

gin

wit

h o

ne

pair

of

new

born

rab

bits

,F0

�F

1�

1.T

his

pai

r of

rab

bits

wou

ld p

rodu

ce o

ne

pair

at

the

end

of t

he

seco

nd

mon

th,s

o F

2�

1 �

1,or

2.

At

the

end

of t

he

thir

d m

onth

,th

e fi

rst

pair

of

rabb

its

wou

ld p

rodu

ce a

not

her

pair

.Th

us,

F3

�2

�1,

or 3

.

Th

e ch

art

belo

w s

how

s th

e n

um

ber

of r

abbi

ts e

ach

mon

th f

or s

ever

al m

onth

s.

Sol

ve.

1.S

tart

ing

wit

h a

sin

gle

pair

of

new

born

rab

bits

,how

man

y pa

irs

of r

abbi

tsw

ould

th

ere

be a

t th

e en

d of

12

mon

ths?

233

2.W

rite

th

e fi

rst

10 t

erm

s of

th

e se

quen

ce f

or w

hic

h F

0�

3,F

1�

4,an

d F n

�F n

�2

�F n

�1.

3,4,

7,11

,18,

29,4

7,76

,123

,199

,322

3.W

rite

th

e fi

rst

10 t

erm

s of

th

e se

quen

ce f

or w

hic

h F

0�

1,F

1�

5,F n

�F n

�2

�F n

�1.

1,5,

6,11

,17,

28,4

5,73

,118

,191

,309

Mo

nth

Ad

ult

Pai

rsN

ewb

orn

Pai

rsTo

tal

F0

01

1

F1

10

1

F2

11

2

F3

21

3

F4

32

5

F5

53

8

En

rich

men

t

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

11-1

11-1


An

swer

s


Stu

dy G

uid

e a

nd I

nte

rven

tion

Ari

thm

etic

Ser

ies

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

11-2

11-2

©G

lenc

oe/M

cGra

w-H

ill63

7G

lenc

oe A

lgeb

ra 2

Lesson 11-2

Ari

thm

etic

Ser

ies

An

ari

thm

etic

ser

ies

is t

he

sum

of

con

secu

tive

ter

ms

of a

nar

ith

met

ic s

equ

ence

.

Su

m o

f an

T

he s

um S

nof

the

firs

t n

term

s of

an

arith

met

ic s

erie

s is

giv

en b

y th

e fo

rmul

aA

rith

met

ic S

erie

sS

n�

�n 2� [2a

1�

(n�

1)d

] or

Sn

��n 2� (

a 1�

a n)

Fin

d S

nfo

r th

ear

ith

met

ic s

erie

s w

ith

a1

�14

,a

n�

101,

and

n�

30.

Use

th

e su

m f

orm

ula

for

an

ari

thm

etic

seri

es.

Sn

�(a

1�

a n)

Sum

for

mul

a

S30

�(1

4 �

101)

n�

30,

a 1�

14,

a n�

101

�15

(115

)S

impl

ify.

�17

25M

ultip

ly.

Th

e su

m o

f th

e se

ries

is

1725

.

30 � 2n � 2

Fin

d t

he

sum

of

all

pos

itiv

e od

d i

nte

gers

les

s th

an 1

80.

Th

e se

ries

is

1 �

3 �

5 �

… �

179.

Fin

d n

usi

ng

the

form

ula

for

th

e n

th t

erm

of

an a

rith

met

ic s

equ

ence

.

a n�

a 1�

(n�

1)d

For

mul

a fo

r nt

h te

rm

179

�1

�(n

�1)

2a n

�17

9, a

1�

1, d

�2

179

�2n

�1

Sim

plify

.

180

�2n

Add

1 t

o ea

ch s

ide.

n�

90D

ivid

e ea

ch s

ide

by 2

.

Th

en u

se t

he

sum

for

mu

la f

or a

n a

rith

met

icse

ries

.

Sn

�(a

1�

a n)

Sum

for

mul

a

S90

�(1

�17

9)n

�90

, a 1

�1,

an

�17

9

�45

(180

)S

impl

ify.

�81

00M

ultip

ly.

Th

e su

m o

f al

l po

siti

ve o

dd i

nte

gers

les

sth

an 1

80 i

s 81

00.

90 � 2n � 2

Exam

ple1

Exam

ple1

Exam

ple2

Exam

ple2

Exer

cises

Exer

cises

Fin

d S

nfo

r ea

ch a

rith

met

ic s

erie

s d

escr

ibed

.

1.a 1

�12

,an

�10

0,2.

a 1�

50,a

n�

�50

,3.

a 1�

60,a

n�

�13

6,n

�12

67

2n

�15

0n

�50

�19

004.

a 1�

20,d

�4,

5.a 1

�18

0,d

��

8,6.

a 1�

�8,

d�

�7,

a n�

112

1584

a n�

6818

60a n

��

71�

395

7.a 1

�42

,n�

8,d

�6

8.a 1

�4,

n�

20,d

�2

9.a 1

�32

,n�

27,d

�3

504

555

1917

Fin

d t

he

sum

of

each

ari

thm

etic

ser

ies.

10.8

�6

�4

�…

��

10�

1011

.16

�22

�28

�…

�11

210

88

12.�

45 �

(�41

) �

(�37

) �

… �

35�

105

Fin

d t

he

firs

t th

ree

term

s of

eac

h a

rith

met

ic s

erie

s d

escr

ibed

.

13.a

1�

12,a

n�

174,

14.a

1�

80,a

n�

�11

5,15

.a1

�6.

2,a n

�12

.6,

Sn

�17

6712

,21,

30S

n�

�24

580

,65,

50S

n�

84.6

6.2,

7.0,

7.8

1 � 2

©G

lenc

oe/M

cGra

w-H

ill63

8G

lenc

oe A

lgeb

ra 2

Sig

ma

No

tati

on

A s

hor

than

d n

otat

ion

for

rep

rese

nti

ng

a se

ries

mak

es u

se o

f th

e G

reek

lett

er Σ

.Th

e si

gma

not

atio

nfo

r th

e se

ries

6 �

12 �

18 �

24 �

30 i

s �5

n�

16n.

Eva

luat

e �1

8

k�

1

(3k

�4)

.

Th

e su

m i

s an

ari

thm

etic

ser

ies

wit

h c

omm

on d

iffe

ren

ce 3

.Su

bsti

tuti

ng

k�

1 an

d k

�18

into

th

e ex

pres

sion

3k

�4

give

s a 1

�3(

1) �

4 �

7 an

d a 1

8�

3(18

) �

4 �

58.T

her

e ar

e 18

ter

ms

in t

he

seri

es,s

o n

�18

.Use

th

e fo

rmu

la f

or t

he

sum

of

an a

rith

met

ic s

erie

s.

Sn

�(a

1�

a n)

Sum

for

mul

a

S18

�(7

�58

)n

�18

, a 1

�7,

an

�58

�9(

65)

Sim

plify

.

�58

5M

ultip

ly.

So

�18

k�

1(3

k�

4) �

585.

Fin

d t

he

sum

of

each

ari

thm

etic

ser

ies.

1.�2

0

n�

1(2

n�

1)

2.�2

5

n�

5(x

�1)

3.

�18

k�

1(2

k�

7)

440

294

216

4.�7

5

r�10(2

r�

200)

5.

�15

x�1(6

x�

3)

6.�5

0

t�1(5

00 �

6t)

�75

9076

517

,350

7.�8

0

k�

1(1

00 �

k)8.

�85

n�

20(n

�10

0)

9.�200

s�13s

4760

�31

3560

,300

10.

�28

m�

14(2

m�

50)

11.

�36

p�

1(5

p�

20)

12.

�32

j�12(2

5 �

2j)

�12

026

10�

399

13.

�42

n�

18(4

n�

9)

14.

�50

n�

20(3

n�

4)

15.

�44

j�5(7

j�

3)

2775

3379

6740

18 � 2n � 2

Stu

dy G

uid

e a

nd I

nte

rven

tion

(c

onti

nued

)

Ari

thm

etic

Ser

ies

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

11-2

11-2

Exam

ple

Exam

ple

Exer

cises

Exer

cises



Skil

ls P

ract

ice

Ari

thm

etic

Ser

ies

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

11-2

11-2

©G

lenc

oe/M

cGra

w-H

ill63

9G

lenc

oe A

lgeb

ra 2

Lesson 11-2

Fin

d S

nfo

r ea

ch a

rith

met

ic s

erie

s d

escr

ibed

.

1.a 1

�1,

a n�

19,n

�10

100

2.a 1

��

5,a n

�13

,n�

728

3.a 1

�12

,an

��

23,n

�8

�44

4.a 1

�7,

n�

11,a

n�

6740

7

5.a 1

�5,

n�

10,a

n�

3218

56.

a 1�

�4,

n�

10,a

n�

�22

�13

0

7.a 1

��

8,d

��

5,n

�12

�42

68.

a 1�

1,d

�3,

n�

1533

0

9.a 1

�10

0,d

��

7,a n

�37

685

10.a

1�

�9,

d�

4,a n

�27

90

11.d

�2,

n�

26,a

n�

4244

212

.d�

�12

,n�

11,a

n�

�52

88

Fin

d t

he

sum

of

each

ari

thm

etic

ser

ies.

13.1

�4

�7

�10

�…

�43

330

14.5

�8

�11

�14

�…

�32

185

15.3

�5

�7

�9

�…

�19

9916

.�2

�(�

5) �

(�8)

�…

�(�

20)

�77

17. �5

n�

1(2

n�

3)15

18. �1

8

n�

1(1

0 �

3n)

693

19. �1

0

n�

2(4

n�

1)22

520

. �12

n�

5(4

�3n

)�

172

Fin

d t

he

firs

t th

ree

term

s of

eac

h a

rith

met

ic s

erie

s d

escr

ibed

.

21.a

1�

4,a n

�31

,Sn

�17

54,

7,10

22.a

1�

�3,

a n�

41,S

n�

228

�3,

1,5

23.n

�10

,an

�41

,Sn

�23

05,

9,13

24.n

�19

,an

�85

,Sn

�76

0�

5,0,

5

©G

lenc

oe/M

cGra

w-H

ill64

0G

lenc

oe A

lgeb

ra 2

Fin

d S

nfo

r ea

ch a

rith

met

ic s

erie

s d

escr

ibed

.

1.a 1

�16

,an

�98

,n�

1374

12.

a 1�

3,a n

�36

,n�

1223

4

3.a 1

��

5,a n

��

26,n

�8

�12

44.

a 1�

5,n

�10

,an

��

13�

40

5.a 1

�6,

n�

15,a

n�

�22

�12

06.

a 1�

�20

,n�

25,a

n�

148

1600

7.a 1

�13

,d�

�6,

n�

21�

987

8.a 1

�5,

d�

4,n

�11

275

9.a 1

�5,

d�

2,a n

�33

285

10.a

1�

�12

1,d

�3,

a n�

5�

2494

11.d

�0.

4,n

�10

,an

�3.

820

12.d

��

,n�

16,a

n�

4478

4

Fin

d t

he

sum

of

each

ari

thm

etic

ser

ies.

13.5

�7

�9

�11

�…

�27

192

14.�

4 �

1 �

6 �

11 �

… �

9187

0

15.1

3 �

20 �

27 �

… �

272

5415

16.8

9 �

86 �

83 �

80 �

… �

2013

08

17. �4

n�

1(1

�2n

)�

1618

. �6

j�1(5

�3n

)93

19. �5

n�

1(9

�4n

)�

15

20. �1

0

k�

4(2

k�

1)10

521

. �8

n�

3(5

n�

10)

105

22. �1

01

n�

1(4

�4n

)�

20,2

00

Fin

d t

he

firs

t th

ree

term

s of

eac

h a

rith

met

ic s

erie

s d

escr

ibed

.

23.a

1�

14,a

n�

�85

,Sn

��

1207

24.a

1�

1,a n

�19

,Sn

�10

0

14,1

1,8

1,3,

5

25.n

�16

,an

�15

,Sn

��

120

26.n

�15

,an

�5

,Sn

�45

�30

,�27

,�24

,,1

27.S

TAC

KIN

GA

hea

lth

clu

b ro

lls

its

tow

els

and

stac

ks t

hem

in

lay

ers

on a

sh

elf.

Eac

hla

yer

of t

owel

s h

as o

ne

less

tow

el t

han

th

e la

yer

belo

w i

t.If

th

ere

are

20 t

owel

s on

th

ebo

ttom

lay

er a

nd

one

tow

el o

n t

he

top

laye

r,h

ow m

any

tow

els

are

stac

ked

on t

he

shel

f?21

0 to

wel

s

28.B

USI

NES

SA

mer

chan

t pl

aces

$1

in a

jack

pot

on A

ugu

st 1

,th

en d

raw

s th

e n

ame

of a

regu

lar

cust

omer

.If

the

cust

omer

is

pres

ent,

he

or s

he

win

s th

e $1

in

th

e ja

ckpo

t.If

th

ecu

stom

er i

s n

ot p

rese

nt,

the

mer

chan

t ad

ds $

2 to

th

e ja

ckpo

t on

Au

gust

2 a

nd

draw

san

oth

er n

ame.

Eac

h d

ay t

he

mer

chan

t ad

ds a

n a

mou

nt

equ

al t

o th

e da

y of

th

e m

onth

.If

the

firs

t pe

rson

to

win

th

e ja

ckpo

t w

ins

$496

,on

wh

at d

ay o

f th

e m

onth

was

her

or

his

nam

e dr

awn

?A

ug

ust

31

3 � 51 � 5

4 � 5

2 � 3

Pra

ctic

e (

Ave

rag

e)

Ari

thm

etic

Ser

ies

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

11-2

11-2


An

swer

s


Readin

g t

o L

earn

Math

em

ati

csA

rith

met

ic S

erie

s

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

11-2

11-2

©G

lenc

oe/M

cGra

w-H

ill64

1G

lenc

oe A

lgeb

ra 2

Lesson 11-2

Pre-

Act

ivit

yH

ow d

o ar

ith

met

ic s

erie

s ap

ply

to

amp

hit

hea

ters

?

Rea

d th

e in

trod

ucti

on t

o L

esso

n 11

-2 a

t th

e to

p of

pag

e 58

3 in

you

r te

xtbo

ok.

Su

ppos

e th

at a

n a

mph

ith

eate

r ca

n s

eat

50 p

eopl

e in

th

e fi

rst

row

an

d th

atea

ch r

ow t

her

eaft

er c

an s

eat

9 m

ore

peop

le t

han

th

e pr

evio

us

row

.Usi

ng

the

voca

bula

ry o

f ar

ith

met

ic s

equ

ence

s,de

scri

be h

ow y

ou w

ould

fin

d th

en

um

ber

of p

eopl

e w

ho

cou

ld b

e se

ated

in

th

e fi

rst

10 r

ows.

(Do

not

act

ual

lyca

lcu

late

th

e su

m.)

Sam

ple

an

swer

:F

ind

th

e fi

rst

10 t

erm

s o

f an

arit

hm

etic

seq

uen

ce w

ith

fir

st t

erm

50

and

co

mm

on

dif

fere

nce

9.T

hen

ad

d t

hes

e 10

ter

ms.

Rea

din

g t

he

Less

on

1.W

hat

is

the

rela

tion

ship

bet

wee

n a

n a

rith

met

ic s

equ

ence

an

d th

e co

rres

pon

din

gar

ith

met

ic s

erie

s?S

amp

le a

nsw

er:

An

ari

thm

etic

seq

uen

ce is

a li

st o

f te

rms

wit

h a

co

mm

on

dif

fere

nce

bet

wee

n s

ucc

essi

ve t

erm

s.T

he

corr

esp

on

din

gar

ith

met

ic s

erie

s is

th

e su

m o

f th

e te

rms

of

the

seq

uen

ce.

2.C

onsi

der

the

form

ula

Sn

�(a

1�

a n).

Exp

lain

th

e m

ean

ing

of t

his

for

mu

la i

n w

ords

.

Sam

ple

an

swer

:To

fin

d t

he

sum

of

the

firs

t n

term

s o

f an

ari

thm

etic

seq

uen

ce,f

ind

hal

f th

e n

um

ber

of

term

s yo

u a

re a

dd

ing

.Mu

ltip

ly t

his

nu

mb

er b

y th

e su

m o

f th

e fi

rst

term

an

d t

he

nth

ter

m.

3.a.

Wh

at i

s th

e pu

rpos

e of

sig

ma

not

atio

n?

Sam

ple

an

swer

:to

wri

te a

ser

ies

in a

co

nci

se f

orm

b.

Con

side

r th

e ex

pres

sion

�12

i�2(4

i�

2).

Th

is f

orm

of

wri

tin

g a

sum

is

call

ed

.

Th

e va

riab

le i

is c

alle

d th

e .

Th

e fi

rst

valu

e of

iis

.

Th

e la

st v

alu

e of

iis

.

How

wou

ld y

ou r

ead

this

exp

ress

ion?

Th

e su

m o

f 4i

�2

as i

go

es f

rom

2 t

o 1

2.

Hel

pin

g Y

ou

Rem

emb

er4.

A g

ood

way

to

rem

embe

r so

met

hin

g is

to

rela

te i

t to

som

eth

ing

you

alr

eady

kn

ow.H

owca

n y

our

know

ledg

e of

how

to

fin

d th

e av

erag

e of

tw

o n

um

bers

hel

p yo

u r

emem

ber

the

form

ula

Sn

�(a

1�

a n)?

Sam

ple

an

swer

:R

ewri

te t

he

form

ula

as

Sn

�n

�.T

he

aver

age

of

the

firs

t an

d la

st t

erm

s is

giv

en b

y th

e

exp

ress

ion

.T

he

sum

of

the

firs

t n

term

s is

th

e av

erag

e o

f th

e fi

rst

and

last

ter

ms

mu

ltip

lied

by

the

nu

mb

er o

f te

rms.

a 1�

a n�

2

a 1�

a n�

2n � 2

122

ind

ex o

f su

mm

atio

n

sig

ma

no

tati

on

n � 2

©G

lenc

oe/M

cGra

w-H

ill64

2G

lenc

oe A

lgeb

ra 2

Geo

met

ric

Pu

zzle

rs

For

th

e p

rob

lem

s on

th

is p

age,

you

wil

l n

eed

to

use

th

e P

yth

agor

ean

Th

eore

m a

nd

th

e fo

rmu

las

for

the

area

of

a tr

ian

gle

and

a t

rap

ezoi

d.

En

rich

men

t

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

11-2

11-2

1.A

rec

tan

gle

mea

sure

s 5

by 1

2 u

nit

s.T

he

upp

er l

eft

corn

er i

s cu

t of

f as

sh

own

in

the

diag

ram

.

a.F

ind

the

area

A(x

) of

th

e sh

aded

pen

tago

n.

A(x

) �

60 �

(5 �

x)(6

�x)

b.

Fin

d x

and

2xso

th

at A

(x)

is a

max

imu

m.W

hat

hap

pen

s to

th

e cu

t-of

f tr

ian

gle?

x�

5 an

d 2

x�

10;

the

tria

ng

lew

ill n

ot

exis

t.

3.T

he

coor

din

ates

of

the

vert

ices

of

a tr

ian

gle

are

A(0

,0),

B(1

1,0)

,an

d C

(0,1

1).A

lin

e x

�k

cuts

th

e tr

ian

gle

into

tw

o re

gion

s ha

ving

equ

al a

rea.

a.W

hat

are

the

coor

dina

tes

of p

oint

D?

(k,1

1 �

k)

b.

Wri

te a

nd

solv

e an

equ

atio

n f

orfi

ndi

ng

the

valu

e of

k.

�1 2� k(1

1 �

11 �

k)

�22

;

k�

11 �

�77�

2.A

tri

angl

e w

ith

sid

es o

f le

ngt

hs

a,a,

and

bis

iso

scel

es.T

wo

tria

ngle

s ar

e cu

t of

f so

that

th

e re

mai

nin

g pe

nta

gon

has

fiv

eeq

ual

sid

es o

f le

ngt

h x

.Th

e va

lue

of x

can

be

fou

nd

usi

ng

this

equ

atio

n.

(2b

�a)

x2�

(4a2

�b2

)(2x

�a)

�0

a.F

ind

xw

hen

a�

10 a

nd

b�

12.

x �

4.46

b.

Can

abe

equ

al t

o 2b

?Ye

s,bu

t it

wo

uld

no

t b

ep

oss

ible

to

hav

e a

pen

tag

on

of

the

typ

e d

escr

ibed

.

4.In

side

a s

quar

e ar

e fi

ve c

ircl

es w

ith

th

esa

me

radi

us.

a.C

onne

ct t

he c

ente

r of

the

top

left

cir

cle

to t

he c

ente

r of

the

bot

tom

rig

ht c

ircl

e.E

xpre

ss t

his

len

gth

in

ter

ms

of r

.4r

b.

Dra

w t

he

squ

are

wit

h v

erti

ces

at t

he

cen

ters

of

the

fou

r ou

tsid

e ci

rcle

s.E

xpre

ss t

he

diag

onal

of

this

squ

are

in t

erm

s of

ran

d a.

(a�

2r)�

2�

ra

bxx

xx

x

a

x

y

AB

C

D

x �

k

2x

12

5

x



Stu

dy G

uid

e a

nd I

nte

rven

tion

Geo

met

ric

Seq

uen

ces

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

11-3

11-3

©G

lenc

oe/M

cGra

w-H

ill64

3G

lenc

oe A

lgeb

ra 2

Lesson 11-3

Geo

met

ric

Seq

uen

ces

A g

eom

etri

c se

qu

ence

is a

seq

uen

ce i

n w

hic

h e

ach

ter

m a

fter

the

firs

t is

th

e pr

odu

ct o

f th

e pr

evio

us

term

an

d a

con

stan

t ca

lled

th

e co

nst

ant

rati

o.

nth

Ter

m o

f a

a n�

a 1�

rn�

1 , w

here

a1

is t

he f

irst

term

, r

is t

he c

omm

on r

atio

, G

eom

etri

c S

equ

ence

and

nis

any

pos

itive

inte

ger

Fin

d t

he

nex

t tw

ote

rms

of t

he

geom

etri

c se

qu

ence

12

00,4

80,1

92,…

.

Sin

ce

�0.

4 an

d �

0.4,

the

sequ

ence

has

a c

omm

on r

atio

of

0.4.

Th

en

ext

two

term

s in

th

e se

quen

ce a

re19

2(0.

4) �

76.8

an

d 76

.8(0

.4)

�30

.72.

192

� 480

480

� 1200

Wri

te a

n e

qu

atio

n f

or t

he

nth

ter

m o

f th

e ge

omet

ric

seq

uen

ce

3.6,

10.8

,32.

4,…

.In

th

is s

equ

ence

a1

�3.

6 an

d r

�3.

Use

th

en

th t

erm

for

mu

la t

o w

rite

an

equ

atio

n.

a n�

a 1�

rn�

1F

orm

ula

for

nth

term

�3.

6 �

3n�

1a 1

�3.

6, r

�3

An

equa

tion

for

the

nth

ter

m is

an

�3.

6 �

3n�

1 .

Exam

ple1

Exam

ple1

Exam

ple2

Exam

ple2

Exer

cises

Exer

cises

Fin

d t

he

nex

t tw

o te

rms

of e

ach

geo

met

ric

seq

uen

ce.

1.6,

12,2

4,…

2.

180,

60,2

0,…

3.

2000

,�10

00,5

00,…

48,9

6,

�25

0,12

5

4.0.

8,�

2.4,

7.2,

…

5.80

,60,

45,…

6.

3,16

.5,9

0.75

,…

�21

.6,6

4.8

33.7

5,25

.312

549

9.12

5,27

45.1

875

Fin

d t

he

firs

t fi

ve t

erm

s of

eac

h g

eom

etri

c se

qu

ence

des

crib

ed.

7.a 1

�,r

�3

8.a 1

�24

0,r

��

9.a 1

�10

,r�

,,1

,3,9

240,

�18

0,13

5,10

,25,

62,1

56,

�10

1,7

539

0

Fin

d t

he

ind

icat

ed t

erm

of

each

geo

met

ric

seq

uen

ce.

10.a

1�

�10

,r�

4,n

�2

11.a

1�

�6,

r�

�,n

�8

12.a

3�

9,r

��

3,n

�7

�40

729

13.a

4�

16,r

�2,

n�

1014

.a4

��

54,r

��

3,n

�6

15.a

1�

8,r

�,n

�5

1024

�48

6

Wri

te a

n e

qu

atio

n f

or t

he

nth

ter

m o

f ea

ch g

eom

etri

c se

qu

ence

.

16.5

00,3

50,2

45,…

17

.8,3

2,12

8,…

18

.11,

�24

.2,5

3.24

,…50

0 �

0.7n

�1

8 �

4n�

111

�(�

2.2)

n�

1

128

� 81

2 � 3

3 � 64

1 � 2

5 � 815 � 16

1 � 4

1 � 41 � 2

1 � 31 � 9

5 � 23 � 4

1 � 9

20 � 920 � 3

©G

lenc

oe/M

cGra

w-H

ill64

4G

lenc

oe A

lgeb

ra 2

Geo

met

ric

Mea

ns

Th

e ge

omet

ric

mea

ns

of a

geo

met

ric

sequ

ence

are

th

e te

rms

betw

een

an

y tw

o n

onsu

cces

sive

ter

ms

of t

he

sequ

ence

.T

o fi

nd

the

kge

omet

ric

mea

ns

betw

een

tw

o te

rms

of a

seq

uen

ce,u

se t

he

foll

owin

g st

eps.

Ste

p 1

Let

the

two

term

s gi

ven

be a

1an

d a n

, w

here

n�

k�

2.S

tep

2S

ubst

itute

in t

he f

orm

ula

a n�

a 1�

rn�

1(�

a 1�

rk�

1 ).

Ste

p 3

Sol

ve f

or r

, an

d us

e th

at v

alue

to

find

the

kge

omet

ric m

eans

:a 1

�r,

a1

�r2

, …

, a

1�

rk

Fin

d t

he

thre

e ge

omet

ric

mea

ns

bet

wee

n 8

an

d 4

0.5.

Use

th

e n

th t

erm

for

mu

la t

o fi

nd

the

valu

e of

r.I

n t

he

sequ

ence

8,

,,

,40.

5,a 1

is 8

and

a 5is

40.

5.a n

�a 1

�rn

�1

For

mul

a fo

r nt

h te

rm

40.5

�8

�r5

�1

n�

5, a

1�

8, a

5�

40.5

5.06

25 �

r4D

ivid

e ea

ch s

ide

by 8

.

r�

�1.

5Ta

ke t

he f

ourt

h ro

ot o

f ea

ch s

ide.

Th

ere

are

two

poss

ible

com

mon

rat

ios,

so t

her

e ar

e tw

o po

ssib

le s

ets

of g

eom

etri

c m

ean

s.U

se e

ach

val

ue

of r

to f

ind

the

geom

etri

c m

ean

s.

r�

1.5

r�

�1.

5a 2

�8(

1.5)

or

12a 2

�8(

�1.

5) o

r �

12a 3

�12

(1.5

) or

18

a 3�

�12

(�1.

5) o

r 18

a 4�

18(1

.5)

or 2

7a 4

�18

(�1.

5) o

r �

27

Th

e ge

omet

ric

mea

ns

are

12,1

8,an

d 27

,or

�12

,18,

and

�27

.

Fin

d t

he

geom

etri

c m

ean

s in

eac

h s

equ

ence

.

1.5,

,,

,405

2.5,

,,2

0.48

�15

,45,

�13

58,

12.8

3.,

,,

,375

4.�

24,

,,

�3,

15,�

754,

�

5.12

,,

,,

,,

6.20

0,,

,,4

14.7

2

�6,

3,�

,,�

�24

0,28

8,�

345.

6

7.,

,,

,,�

12,0

058.

4,,

,,1

56

�,3

5,�

245,

1715

�10

,25,

�62

9.�

,,

,,

,,�

910

.100

,,

,,3

84.1

6

�,�

,�,�

1,�

3�

140,

196,

�27

4.4

1 � 31 � 9

1 � 27

??

??

??

??

1 � 81

1 � 235 � 7

1 � 4?

??

??

??

35 � 49

3 � 83 � 4

3 � 2

??

?3 � 16

??

??

?

2 � 3

1 � 9?

??

??

3 � 5

??

??

?

??

?

Stu

dy G

uid

e a

nd I

nte

rven

tion

(c

onti

nued

)

Geo

met

ric

Seq

uen

ces

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

11-3

11-3

Exam

ple

Exam

ple

Exer

cises

Exer

cises


An

swer

s


Skil

ls P

ract

ice

Geo

met

ric

Seq

uen

ces

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

11-3

11-3

©G

lenc

oe/M

cGra

w-H

ill64

5G

lenc

oe A

lgeb

ra 2

Lesson 11-3

Fin

d t

he

nex

t tw

o te

rms

of e

ach

geo

met

ric

seq

uen

ce.

1.�

1,�

2,�

4,…

�8,

�16

2.6,

3,,…

,

3.�

5,�

15,�

45,…

�13

5,�

405

4.72

9,�

243,

81 ,

…�

27,9

5.15

36,3

84,9

6,…

24,6

6.64

,160

,400

,…10

00,2

500

Fin

d t

he

firs

t fi

ve t

erm

s of

eac

h g

eom

etri

c se

qu

ence

des

crib

ed.

7.a 1

�6,

r�

28.

a 1�

�27

,r�

3

6,12

,24,

48,9

6�

27,�

81,�

243,

�72

9,�

2187

9.a 1

��

15,r

��

110

.a1

�3,

r�

4

�15

,15,

�15

,15,

�15

3,12

,48,

192,

768

11.a

1�

1,r

�12

.a1

�21

6,r

��

1,,

,,

216,

�72

,24,

�8,

Fin

d t

he

ind

icat

ed t

erm

of

each

geo

met

ric

seq

uen

ce.

13.a

1�

5,r

�2,

n�

616

014

.a1

�18

,r�

3,n

�6

4374

15.a

1�

�3,

r�

�2,

n�

5�

4816

.a1

��

20,r

��

2,n

�9

�51

20

17.a

8fo

r �

12,�

6,�

3,…

�18

.a7

for

80,

,,…

Wri

te a

n e

qu

atio

n f

or t

he

nth

ter

m o

f ea

ch g

eom

etri

c se

qu

ence

.

19.3

,9,2

7,…

a n�

3n20

.�1,

�3,

�9,

…a n

��

1(3)

n�

1

21.2

,�6,

18,…

a n�

2(�

3)n

�1

22.5

,10,

20,…

a n�

5(2)

n�

1

Fin

d t

he

geom

etri

c m

ean

s in

eac

h s

equ

ence

.

23.4

,,

,,6

4�

8,16

,�32

24.1

,,

,,8

1�

3,9,

�27

??

??

??

80 � 729

80 � 980 � 3

3 � 32

8 � 31 � 16

1 � 81 � 4

1 � 2

1 � 31 � 2

3 � 83 � 4

3 � 2

©G

lenc

oe/M

cGra

w-H

ill64

6G

lenc

oe A

lgeb

ra 2

Fin

d t

he

nex

t tw

o te

rms

of e

ach

geo

met

ric

seq

uen

ce.

1.�

15,�

30,�

60,…

�12

0,�

240

2.80

,40,

20,…

10,5

3.90

,30,

10,…

,4.

�14

58,4

86,�

162,

…54

,�18

5.,

,,…

,6.

216,

144,

96,…

64,

Fin

d t

he

firs

t fi

ve t

erm

s of

eac

h g

eom

etri

c se

qu

ence

des

crib

ed.

7.a 1

��

1,r

��

38.

a 1�

7,r

��

4

�1,

3,�

9,27

,�81

7,�

28,1

12,�

448,

1792

9.a 1

��

,r�

210

.a1

�12

,r�

�,�

,�,�

,�12

,8,

,,

Fin

d t

he

ind

icat

ed t

erm

of

each

geo

met

ric

seq

uen

ce.

11.a

1�

5,r

�3,

n�

612

1512

.a1

�20

,r�

�3,

n�

6�

4860

13.a

1�

�4,

r�

�2,

n�

1020

4814

.a8

for

�,�

,�,…

�

15.a

12fo

r 96

,48,

24,…

16.a

1�

8,r

�,n

�9

17.a

1�

�31

25,r

��

,n�

9�

18.a

1�

3,r

�,n

�8

Wri

te a

n e

qu

atio

n f

or t

he

nth

ter

m o

f ea

ch g

eom

etri

c se

qu

ence

.

19.1

,4,1

6,…

a n�

(4)n

�1

20.�

1,�

5,�

25,…

a n�

�1(

5)n

�1

21.1

,,

,…a n

��

�n�

122

.�3,

�6,

�12

,…a n

��

3(2)

n�

1

23.7

,�14

,28,

…a n

�7(

�2)

n�

124

.�5,

�30

,�18

0,…

a n�

�5(

6)n

�1

Fin

d t

he

geom

etri

c m

ean

s in

eac

h s

equ

ence

.

25.3

,,

,,7

6812

,48,

192

26.5

,,

,,1

280

�20

,80,

�32

0

27.1

44,

,,

,928

.37,

500,

,,

,,�

12

�72

,36,

�18

�75

00,1

500,

�30

0,60

29.B

IOLO

GY

A c

ult

ure

in

itia

lly

con

tain

s 20

0 ba

cter

ia.I

f th

e n

um

ber

of b

acte

ria

dou

bles

ever

y 2

hou

rs,h

ow m

any

bact

eria

wil

l be

in

th

e cu

ltu

re a

t th

e en

d of

12

hou

rs?

12,8

00

30.L

IGH

TIf

eac

h f

oot

of w

ater

in

a l

ake

scre

ens

out

60%

of

the

ligh

t ab

ove,

wh

at p

erce

nt

ofth

e li

ght

pass

es t

hro

ugh

5 f

eet

of w

ater

?1.

024%

31.I

NV

ESTI

NG

Rau

l in

vest

s $1

000

in a

sav

ings

acc

ount

tha

t ea

rns

5% i

nter

est

com

poun

ded

ann

ual

ly.H

ow m

uch

mon

ey w

ill

he

hav

e in

th

e ac

cou

nt

at t

he

end

of 5

yea

rs?

$127

6.28

??

??

??

?

??

??

??

1 � 21 � 4

1 � 2

3�

�10

,000

,000

1 � 101

� 125

1 � 5

1 � 321 � 2

3 � 64

625

�2

1 � 101 � 50

1� 25

0

64 � 2732 � 9

16 � 316 � 3

8 � 34 � 3

2 � 31 � 3

2 � 31 � 3

128

�3

81 � 6427 � 32

9 � 163 � 8

1 � 4

10 � 910 � 3

Pra

ctic

e (

Ave

rag

e)

Geo

met

ric

Seq

uen

ces

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

11-3

11-3



Readin

g t

o L

earn

Math

em

ati

csG

eom

etri

c S

equ

ence

s

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

11-3

11-3

©G

lenc

oe/M

cGra

w-H

ill64

7G

lenc

oe A

lgeb

ra 2

Lesson 11-3

Pre-

Act

ivit

yH

ow d

o ge

omet

ric

seq

uen

ces

app

ly t

o a

bou

nci

ng

bal

l?

Rea

d th

e in

trod

ucti

on t

o L

esso

n 11

-3 a

t th

e to

p of

pag

e 58

8 in

you

r te

xtbo

ok.

Su

ppos

e th

at y

ou d

rop

a ba

ll f

rom

a h

eigh

t of

4 f

eet,

and

that

eac

h t

ime

itfa

lls,

it b

oun

ces

back

to

74%

of

the

hei

ght

from

wh

ich

it

fell

.Des

crib

e h

oww

ould

you

fin

d th

e h

eigh

t of

th

e th

ird

bou

nce

.(D

o n

ot a

ctu

ally

cal

cula

te t

he

heig

ht o

f the

bou

nce.

)

Sam

ple

answ

er:M

ultip

ly 4

by

0.74

thr

ee t

imes

.

Rea

din

g t

he

Less

on

1.E

xpla

in t

he

diff

eren

ce b

etw

een

an

ari

thm

etic

seq

uen

ce a

nd

a ge

omet

ric

sequ

ence

.

Sam

ple

an

swer

:In

an

ari

thm

etic

seq

uen

ce,e

ach

ter

m a

fter

th

e fi

rst

isfo

un

d b

y ad

din

g t

he

com

mo

n d

iffe

ren

ce t

o t

he

pre

vio

us

term

.In

ag

eom

etri

c se

qu

ence

,eac

h t

erm

aft

er t

he

firs

t is

fo

un

d b

y m

ult

iply

ing

th

ep

revi

ou

s te

rm b

y th

e co

mm

on

rat

io.

2.C

onsi

der

the

form

ula

an

�a 1

�rn

�1 .

a.W

hat

is

this

for

mu

la u

sed

to f

ind?

a p

arti

cula

r te

rm o

f a

geo

met

ric

seq

uen

ce

b.

Wh

at d

o ea

ch o

f th

e fo

llow

ing

repr

esen

t?

a n:

the

nth

ter

m

a 1:

the

firs

t te

rm

r:th

e co

mm

on

rat

io

n:

a p

osi

tive

inte

ger

th

at in

dic

ates

wh

ich

ter

m y

ou

are

fin

din

g

3.a.

In t

he

sequ

ence

5,8

,11,

14,1

7,20

,th

e n

um

bers

8,1

1,14

,an

d 17

are

betw

een

5 a

nd

20.

b.

In t

he

sequ

ence

12,

4,,

,,t

he

nu

mbe

rs 4

,,a

nd

are

betw

een

12

and

.

Hel

pin

g Y

ou

Rem

emb

er

4.S

uppo

se t

hat

your

cla

ssm

ate

Ric

ardo

has

tro

uble

rem

embe

ring

the

for

mul

a a n

�a 1

�rn

�1

corr

ectl

y.H

e th

inks

th

at t

he

form

ula

sh

ould

be

a n�

a 1�

rn.H

ow w

ould

you

exp

lain

to

him

th

at h

e sh

ould

use

rn

�1

rath

er t

han

rn

in t

he

form

ula

?

Sam

ple

an

swer

:E

ach

ter

m a

fter

th

e fi

rst

in a

geo

met

ric

seq

uen

ce is

fou

nd

by

mu

ltip

lyin

g t

he

pre

vio

us

term

by

r.T

her

e ar

e n

�1

term

s b

efo

reth

e n

th t

erm

,so

yo

u w

ou

ld n

eed

to

mu

ltip

ly b

y r

a to

tal o

f n

�1

tim

es,

no

t n

tim

es,t

o g

et t

he

nth

ter

m.

4 � 27g

eom

etri

c m

ean

s

4 � 94 � 3

4 � 274 � 9

4 � 3

arit

hm

etic

mea

ns

©G

lenc

oe/M

cGra

w-H

ill64

8G

lenc

oe A

lgeb

ra 2

Hal

f th

e D

ista

nce

Su

ppos

e yo

u a

re 2

00 f

eet

from

a f

ixed

poi

nt,

P.S

upp

ose

that

you

are

abl

e to

mov

e to

th

e h

alfw

ay p

oin

t in

on

e m

inu

te,t

o th

e n

ext

hal

fway

poi

nt

one

min

ute

aft

er t

hat

,an

d so

on

.

An

in

tere

stin

g se

quen

ce r

esu

lts

beca

use

acc

ordi

ng

to t

he

prob

lem

,you

nev

erac

tual

ly r

each

th

e po

int

P,a

lth

ough

you

do

get

arbi

trar

ily

clos

e to

it.

You

can

com

pute

how

lon

g it

wil

l ta

ke t

o ge

t w

ith

in s

ome

spec

ifie

d sm

all

dist

ance

of

the

poin

t.O

n a

cal

cula

tor,

you

en

ter

the

dist

ance

to

be c

over

edan

d th

en c

oun

t th

e n

um

ber

of s

ucc

essi

ve d

ivis

ion

s by

2 n

eces

sary

to

get

wit

hin

th

e de

sire

d di

stan

ce.

How

man

y m

inu

tes

are

nee

ded

to

get

wit

hin

0.1

foo

tof

a p

oin

t 20

0 fe

et a

way

?

Cou

nt

the

nu

mbe

r of

tim

es y

ou d

ivid

e by

2.

En

ter:

200

22

2,a

nd

so o

n

Res

ult

:0.

0976

562

You

div

ided

by

2 el

even

tim

es.T

he

tim

e n

eede

d is

11

min

ute

s.

Use

th

e m

eth

od i

llu

stra

ted

ab

ove

to s

olve

eac

h p

rob

lem

.

1.If

it

is a

bou

t 25

00 m

iles

fro

m L

os A

nge

les

to N

ew Y

ork,

how

man

y m

inu

tes

wou

ld i

t ta

ke t

o ge

t w

ith

in 0

.1 m

ile

of N

ew Y

ork?

How

far

fro

m

New

Yor

k ar

e yo

u a

t th

at t

ime?

15 m

inu

tes,

0.07

6293

4 m

ile

2.If

it

is 2

5,00

0 m

iles

aro

un

d E

arth

,how

man

y m

inu

tes

wou

ld i

t ta

ke t

o ge

t w

ith

in 0

.5 m

ile

of t

he

full

dis

tan

ce a

rou

nd

Ear

th?

How

far

sh

ort

wou

ld

you

be?

16 m

inu

tes;

0.38

1469

7 m

ile

3.If

it

is a

bou

t 25

0,00

0 m

iles

fro

m E

arth

to

the

Moo

n,h

ow m

any

min

ute

s w

ould

it

take

to

get

wit

hin

0.5

mil

e of

th

e M

oon

? H

ow f

ar f

rom

th

e su

rfac

e of

th

e M

oon

wou

ld y

ou b

e?19

min

ute

s,0.

4768

372

mile

4.If

it

is a

bou

t 30

,000

,000

fee

t fr

om H

onol

ulu

to

Mia

mi,

how

man

y m

inu

tes

wou

ld i

t ta

ke t

o ge

t to

wit

hin

1 f

oot

of M

iam

i? H

ow f

ar f

rom

Mia

mi

wou

ld

you

be

at t

hat

tim

e?25

min

ute

s,0.

8940

697

foo

t

5.If

it

is a

bou

t 93

,000

,000

mil

es t

o th

e su

n,h

ow m

any

min

ute

s w

ould

it

take

to

get

wit

hin

500

mil

es o

f th

e su

n?

How

far

fro

m t

he

sun

wou

ld y

ou b

e at

th

at t

ime?

18 m

inu

tes,

354.

7668

46 m

iles

ENTE

R�

ENTE

R�

ENTE

R�

100

200

feet

150

175

P

1st m

inut

e2n

d m

inut

e3r

d m

inut

e

En

rich

men

t

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

11-3

11-3

Exam

ple

Exam

ple


An

swer

s


Stu

dy G

uid

e a

nd I

nte

rven

tion

Geo

met

ric

Ser

ies

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

11-4

11-4

©G

lenc

oe/M

cGra

w-H

ill64

9G

lenc

oe A

lgeb

ra 2

Lesson 11-4

Geo

met

ric

Seri

esA

geo

met

ric

seri

esis

th

e in

dica

ted

sum

of

con

secu

tive

ter

ms

of a

geom

etri

c se

quen

ce.

Su

m o

f a

The

sum

Sn

of t

he f

irst

nte

rms

of a

geo

met

ric s

erie

s is

giv

en b

yG

eom

etri

c S

erie

sS

n�

or S

n�

, w

here

r�

1.a 1

�a 1r

n

��

1 �

r

a 1(1

�rn

)�

�1

�r

Fin

d t

he

sum

of

the

firs

tfo

ur

term

s of

th

e ge

omet

ric

seq

uen

ce

for

wh

ich

a1

�12

0 an

d r

�.

Sn

�S

um f

orm

ula

S4

�n

�4,

a1

�12

0, r

��1 3�

�17

7.78

Use

a c

alcu

lato

r.

Th

e su

m o

f th

e se

ries

is

177.

78.

120 �1

��1 3� �4

��

�1

��1 3�

a 1(1

�rn

)�

�1

�r

1 � 3

Fin

d t

he

sum

of

the

geom

etri

c se

ries

�7

j�14

�3

j�

2 .

Sin

ce t

he

sum

is

a ge

omet

ric

seri

es,y

ou c

anu

se t

he

sum

for

mu

la.

Sn

�S

um f

orm

ula

S7

�n

�7,

a1

��4 3� ,

r�

3

�14

57.3

3U

se a

cal

cula

tor.

Th

e su

m o

f th

e se

ries

is

1457

.33.

�4 3� (1 �

37 )�

1 �

3

a 1(1

�rn

)�

�1

�r

Exam

ple1

Exam

ple1

Exam

ple2

Exam

ple2

Exer

cises

Exer

cises

Fin

d S

nfo

r ea

ch g

eom

etri

c se

ries

des

crib

ed.

1.a 1

�2,

a n�

486,

r�

32.

a 1�

1200

,an

�75

,r�

3.a 1

�,a

n�

125,

r�

5

728

2325

156.

24

4.a 1

�3,

r�

,n�

45.

a 1�

2,r

�6,

n�

46.

a 1�

2,r

�4,

n�

6

4.44

518

2730

7.a 1

�10

0,r

��

,n�

58.

a 3�

20,a

6�

160,

n�

89.

a 4�

16,a

7�

1024

,n�

10

68.7

512

7587

,381

.25

Fin

d t

he

sum

of

each

geo

met

ric

seri

es.

10.6

�18

�54

�…

to

6 te

rms

11.

��

1 �

… t

o 10

ter

ms

2184

255.

75

12. �8

j�42

j13

. �7

k�

13

�2k

�1

496

381

1 � 21 � 4

1 � 2

1 � 3

1 � 251 � 2

©G

lenc

oe/M

cGra

w-H

ill65

0G

lenc

oe A

lgeb

ra 2

Spec

ific

Ter

ms

You

can

use

on

e of

th

e fo

rmu

las

for

the

sum

of

a ge

omet

ric

seri

es t

o h

elp

fin

d a

part

icu

lar

term

of

the

seri

es.

Stu

dy G

uid

e a

nd I

nte

rven

tion

(c

onti

nued

)

Geo

met

ric

Ser

ies

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

11-4

11-4

Fin

d a

1in

a g

eom

etri

cse

ries

for

wh

ich

S6

�44

1 an

d r

�2.

Sn

�S

um f

orm

ula

441

�S

6�

441,

r�

2, n

�6

441

�S

ubtr

act.

a 1�

Div

ide.

a 1�

7S

impl

ify.

Th

e fi

rst

term

of

the

seri

es i

s 7.

441

� 63�63

a 1�

�1

a 1(1

�26 )

��

1 �

2

a 1(1

�rn

)�

�1

�r

Fin

d a

1in

a g

eom

etri

cse

ries

for

wh

ich

Sn

�24

4,a

n�

324,

and

r�

�3.

Sin

ce y

ou d

o n

ot k

now

th

e va

lue

of n

,use

th

eal

tern

ate

sum

for

mu

la.

Sn

�A

ltern

ate

sum

for

mul

a

244

�S

n�

244,

an

�32

4, r

��

3

244

�S

impl

ify.

976

�a 1

�97

2M

ultip

ly e

ach

side

by

4.

a 1�

4S

ubtr

act

972

from

eac

h si

de.

Th

e fi

rst

term

of

the

seri

es i

s 4.

a 1�

972

�� 4

a 1�

(324

)(�

3)�

�1

�(�

3)

a 1�

a nr�

�1

�r

Exam

ple1

Exam

ple1

Exam

ple2

Exam

ple2

Exam

ple

3Ex

ampl

e 3

Fin

d a

4in

a g

eom

etri

c se

ries

for

wh

ich

Sn

�79

6.87

5,r

�,a

nd

n�

8.F

irst

use

th

e su

m f

orm

ula

to

fin

d a 1

.

S n�

Sum

for

mul

a

796.

875

�S

8�

796.

875,

r�

, n

�8

796.

875

�U

se a

cal

cula

tor.

a 1�

400

Sin

ce a

4�

a 1�

r3,a

4�

400 ��1 2� �3

�50

.Th

e fo

urt

h t

erm

of

the

seri

es i

s 50

.

Fin

d t

he

ind

icat

ed t

erm

for

eac

h g

eom

etri

c se

ries

des

crib

ed.

1.S

n�

726,

a n�

486,

r�

3;a 1

62.

Sn

�85

0,a n

�12

80,r

��

2;a 1

�10

3.S

n�

1023

.75,

a n�

512,

r�

2;a 1

4.S

n�

118.

125,

a n�

�5.

625,

r�

�;a

118

0

5.S

n�

183,

r�

�3,

n�

5;a 1

36.

Sn

�17

05,r

�4,

n�

5;a 1

5

7.S

n�

52,0

84,r

��

5,n

�7;

a 14

8.S

n�

43,6

90,r

�,n

�8;

a 132

,768

9.S

n�

381,

r�

2,n

�7;

a 424

1 � 4

1 � 21 � 4

0.99

6093

75a 1

��

0.5

1 � 2

a 1�1 �

��1 2� �8 ��

�1

��1 2�

a 1(1

�rn

)�

�1

�r

1 � 2

Exer

cises

Exer

cises



Skil

ls P

ract

ice

Geo

met

ric

Ser

ies

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

11-4

11-4

©G

lenc

oe/M

cGra

w-H

ill65

1G

lenc

oe A

lgeb

ra 2

Lesson 11-4

Fin

d S

nfo

r ea

ch g

eom

etri

c se

ries

des

crib

ed.

1.a 1

�2,

a 5�

162,

r�

324

22.

a 1�

4,a 6

�12

,500

,r�

515

,624

3.a 1

�1,

a 8�

�1,

r�

�1

04.

a 1�

4,a n

�25

6,r

��

217

2

5.a 1

�1,

a n�

729,

r�

�3

547

6.a 1

�2,

r�

�4,

n�

541

0

7.a 1

��

8,r

�2,

n�

4�

120

8.a 1

�3,

r�

�2,

n�

12�

4095

9.a 1

�8,

r�

3,n

�5

968

10.a

1�

6,a n

�,r

�

11.a

1�

8,r

�,n

�7

12.a

1�

2,r

��

,n�

6

Fin

d t

he

sum

of

each

geo

met

ric

seri

es.

13.4

�8

�16

�…

to

5 te

rms

124

14.�

1 �

3 �

9 �

… t

o 6

term

s�

364

15.3

�6

�12

�…

to

5 te

rms

9316

.�15

�30

�60

�…

to

7 te

rms

�64

5

17. �4

n�

13n

�1

4018

. �5

n�

1(�

2)n

�1

11

19. �4

n�

1�

�n�

120

. �9

n�

12(

�3)

n�

198

42

Fin

d t

he

ind

icat

ed t

erm

for

eac

h g

eom

etri

c se

ries

des

crib

ed.

21.S

n�

1275

,an

�64

0,r

�2;

a 15

22.S

n�

�40

,an

��

54,r

��

3;a 1

2

23.S

n�

99,n

�5,

r�

�;a

114

424

.Sn

�39

,360

,n�

8,r

�3;

a 112

1 � 2

40 � 271 � 3

21 � 161 � 2

127

�8

1 � 2

93 � 81 � 2

3 � 8

©G

lenc

oe/M

cGra

w-H

ill65

2G

lenc

oe A

lgeb

ra 2

Fin

d S

nfo

r ea

ch g

eom

etri

c se

ries

des

crib

ed.

1.a 1

�2,

a 6�

64,r

�2

126

2.a 1

�16

0,a 6

�5,

r�

315

3.a 1

��

3,a n

��

192,

r�

�2

�12

94.

a 1�

�81

,an

��

16,r

��

�55

5.a 1

��

3,a n

�30

72,r

��

424

576.

a 1�

54,a

6�

,r�

7.a 1

�5,

r�

3,n

�9

49,2

058.

a 1�

�6,

r�

�1,

n�

21�

6

9.a 1

��

6,r

��

3,n

�7

�32

8210

.a1

��

9,r

�,n

�4

�

11.a

1�

,r�

3,n

�10

12.a

1�

16,r

��

1.5,

n�

6�

66.5

Fin

d t

he

sum

of

each

geo

met

ric

seri

es.

13.1

62 �

54 �

18 �

… t

o 6

term

s14

.2 �

4 �

8 �

… t

o 8

term

s51

0

15.6

4 �

96 �

144

�…

to

7 te

rms

463

16.

��

1 �

… t

o 6

term

s�

17. �8

n�

1(�

3)n

�1

�16

4018

. �9

n�

15(

�2)

n�

185

519

. �5

n�

1�

1(4)

n�

1�

341

20. �6

n�

1�

�n�

121

. �10

n�

125

60�

�n�

151

1522

. �4

n�

19 �

�n�

1

Fin

d t

he

ind

icat

ed t

erm

for

eac

h g

eom

etri

c se

ries

des

crib

ed.

23.S

n�

1023

,an

�76

8,r

�4;

a 13

24.S

n�

10,1

60,a

n�

5120

,r�

2;a 1

80

25.S

n�

�13

65,n

�12

,r�

�2;

a 11

26.S

n�

665,

n�

6,r

�1.

5;a 1

32

27.C

ON

STR

UC

TIO

NA

pil

e dr

iver

dri

ves

a po

st 2

7 in

ches

in

to t

he

grou

nd

on i

ts f

irst

hit

.

Eac

h ad

diti

onal

hit

dri

ves

the

post

th

e di

stan

ce o

f th

e pr

ior

hit.

Fin

d th

e to

tal

dist

ance

the

post

has

bee

n d

rive

n a

fter

5 h

its.

70in

.

28.C

OM

MU

NIC

ATI

ON

SH

ugh

Moo

re e

-mai

ls a

joke

to

5 fr

ien

ds o

n S

un

day

mor

nin

g.E

ach

of t

hes

e fr

ien

ds e

-mai

ls t

he

joke

to

5 of

her

or

his

fri

ends

on

Mon

day

mor

nin

g,an

d so

on

.A

ssu

min

g n

o du

plic

atio

n,h

ow m

any

peop

le w

ill

hav

e h

eard

th

e jo

ke b

y th

e en

d of

Sat

urd

ay,n

ot i

ncl

udi

ng

Hu

gh?

97,6

55 p

eop

le

1 � 3

2 � 3

65 � 32 � 3

1 � 263 � 32

1 � 2

182

�9

1 � 31 � 9

728

�3

29,5

24�

31 � 3

65 � 32 � 3

728

�9

1 � 32 � 9

2 � 3

1 � 2

Pra

ctic

e (

Ave

rag

e)

Geo

met

ric

Ser

ies

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

11-4

11-4


An

swer

s


Readin

g t

o L

earn

Math

em

ati

csG

eom

etri

c S

erie

s

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

11-4

11-4

©G

lenc

oe/M

cGra

w-H

ill65

3G

lenc

oe A

lgeb

ra 2

Lesson 11-4

Pre-

Act

ivit

yH

ow i

s e-

mai

lin

g a

jok

e li

ke

a ge

omet

ric

seri

es?

Rea

d th

e in

trod

ucti

on t

o L

esso

n 11

-4 a

t th

e to

p of

pag

e 59

4 in

you

r te

xtbo

ok.

•S

upp

ose

that

you

e-m

ail

the

joke

on

Mon

day

to f

ive

frie

nds

,rat

her

th

anth

ree,

and

that

eac

h o

f th

ose

frie

nds

e-m

ails

it

to f

ive

frie

nds

on

Tu

esda

y,an

d so

on

.Wri

te a

su

m t

hat

sh

ows

that

tot

al n

um

ber

of p

eopl

e,in

clu

din

gyo

urs

elf,

wh

o w

ill

hav

e re

ad t

he

joke

by

Th

urs

day.

(Wri

te o

ut

the

sum

usin

g pl

us s

igns

rat

her

than

sig

ma

nota

tion

.Do

not

actu

ally

fin

d th

e su

m.)

1 �

5 �

25 �

125

•U

se e

xpon

ents

to

rew

rite

th

e su

m y

ou f

oun

d ab

ove.

(Use

an

exp

onen

t in

each

ter

m,a

nd

use

th

e sa

me

base

for

all

ter

ms.

)50

�51

�52

�53

Rea

din

g t

he

Less

on

1.C

onsi

der

the

form

ula

Sn

�.

a.W

hat

is

this

for

mu

la u

sed

to f

ind?

the

sum

of

the

firs

t n

term

s o

f a

geo

met

ric

seri

es

b.

Wh

at d

o ea

ch o

f th

e fo

llow

ing

repr

esen

t?

Sn:

the

sum

of

the

firs

t n

term

s

a 1:

the

firs

t te

rm

r:th

e co

mm

on

rat

io

c.S

upp

ose

that

you

wan

t to

use

th

e fo

rmu

la t

o ev

alu

ate

3 �

1 �

��

.In

dica

te

the

valu

es y

ou w

ould

su

bsti

tute

in

to t

he

form

ula

in

ord

er t

o fi

nd

Sn.(

Do

not

act

ual

lyca

lcu

late

th

e su

m.)

n�

a 1�

r�

rn�

d.

Su

ppos

e th

at y

ou w

ant

to u

se t

he

form

ula

to

eval

uat

e th

e su

m �6

n�

18(

�2)

n�

1 .In

dica

te

the

valu

es y

ou w

ould

su

bsti

tute

in

to t

he

form

ula

in

ord

er t

o fi

nd

Sn.(

Do

not

act

ual

lyca

lcu

late

th

e su

m.)

n�

a 1�

r�

rn�

Hel

pin

g Y

ou

Rem

emb

er

2.T

his

les

son

in

clu

des

thre

e fo

rmu

las

for

the

sum

of

the

firs

t n

term

s of

a g

eom

etri

c se

ries

.A

ll o

f th

ese

form

ulas

hav

e th

e sa

me

deno

min

ator

and

hav

e th

e re

stri

ctio

n r

�1.

How

can

this

res

tric

tion

hel

p yo

u t

o re

mem

ber

the

den

omin

ator

in

th

e fo

rmu

las?

Sam

ple

an

swer

:If

r�

1,th

en r

�1

�0.

Bec

ause

div

isio

n b

y 0

isu

nd

efin

ed,a

fo

rmu

la w

ith

r�

1 in

th

e d

eno

min

ato

r w

ill n

ot

app

ly

wh

en r

�1.

(�2)

6o

r 64

�2

86

��1 3� �5

or

�� 21 43�

��1 3�

35

1 � 271 � 9

1 � 3

a 1(1

�rn

)�

�1

�r

©G

lenc

oe/M

cGra

w-H

ill65

4G

lenc

oe A

lgeb

ra 2

An

nu

itie

sA

n an

nuit

y is

a fi

xed

amou

nt o

f mon

ey p

ayab

le a

t gi

ven

inte

rval

s.Fo

r ex

ampl

e,su

ppos

e yo

u w

ante

d to

set

up

a tr

ust

fund

so

that

$30

,000

cou

ld b

e w

ithd

raw

nea

ch y

ear

for

14 y

ears

bef

ore

the

mon

ey r

an o

ut.

Ass

um

e th

e m

oney

can

be

inve

sted

at

9%.

You

mu

st f

ind

the

amou

nt

of m

oney

th

at n

eeds

to

be i

nve

sted

.Cal

l th

isam

oun

t A

.Aft

er t

he

thir

d pa

ymen

t,th

e am

oun

t le

ft i

s

1.09

[1.0

9A�

30,0

00(1

�1.

09)]

�30

,000

�1.

092 A

�30

,000

(1 �

1.09

�1.

092 )

.

Th

e re

sult

s ar

e su

mm

ariz

ed i

n t

he

tabl

e be

low

.

1.U

se t

he

patt

ern

sh

own

in

th

e ta

ble

to f

ind

the

nu

mbe

r of

dol

lars

lef

t af

ter

the

fou

rth

pay

men

t.1.

093 A

�30

,000

(1 �

1.09

�1.

092

�1.

093 )

2.F

ind

the

amou

nt

left

aft

er t

he

ten

th p

aym

ent.

1.09

9 A�

30,0

00(1

�1.

09 �

1.09

2�

1.09

3�

… �

1.09

9 )

Th

e am

oun

t le

ft a

fter

th

e 14

th p

aym

ent

is 1

.091

3 A�

30,0

00(1

�1.

09 �

1.09

2�

… �

1.09

13).

How

ever

,th

ere

shou

ld b

e n

o m

oney

lef

t af

ter

the

14th

and

fin

al p

aym

ent.

1.09

13A

�30

,000

(1 �

1.09

�1.

092

�…

�1.

0913

) �

0

Not

ice

that

1 �

1.09

�1.

092

�…

�1.

0913

is a

geo

met

ric

seri

es w

her

e a 1

�1,

a n�

1.09

13,n

�14

an

d r

�1.

09.

Usi

ng

the

form

ula

for

Sn,

1 �

1.09

�1.

092

�…

�1.

0913

��

�1 1� �1 1.0 .09 914�

��1

� �01 .. 00 9914�

.

3.S

how

th

at w

hen

you

sol

ve f

or A

you

get

A�

�30 0, .0 00 90�

��1.0 19 .014 91� 3

1�

�.1.

0913

A�

30,0

00��1

� �01 .. 00 9914�

��0

resu

lts

in s

tate

d e

xpre

ssio

n f

or

A.

The

refo

re,t

o pr

ovid

e $3

0,00

0 fo

r 14

yea

rs w

here

the

ann

ual i

nter

est

rate

is

9%,

you

nee

d �30 0, .0 00 90

��1.

0 19 .014 91� 31

��d

olla

rs.

4.U

se a

cal

cula

tor

to f

ind

the

valu

e of

Ain

pro

blem

3.

$254

,607

In g

ener

al,i

f yo

u w

ish

to

prov

ide

Pdo

llar

s fo

r ea

ch o

f n

year

s at

an

an

nu

alra

te o

f r%

,you

nee

d A

doll

ars

wh

ere

�1 �

� 10r 0�

�n�

1A

�P�1

��1

�� 10

r 0��

�1 �

� 10r 0�

�2�

… �

�1 �

� 10r 0�

�n�

1 ��0.

You

can

sol

ve t

his

equ

atio

n f

or A

,giv

en P

,n,a

nd

r.

a 1�

a 1rn

��

1�

r

Pay

men

t N

um

ber

Nu

mb

er o

f D

olla

rs L

eft

Aft

er P

aym

ent

1A

�30

,000

21.

09A

�30

,000

(1 �

1.09

)3

1.09

2 A�

30,0

00(1

�1.

09 �

1.09

2 )

En

rich

men

t

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

11-4

11-4



Stu

dy G

uid

e a

nd I

nte

rven

tion

Infi

nit

e G

eom

etri

c S

erie

s

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

11-5

11-5

©G

lenc

oe/M

cGra

w-H

ill65

5G

lenc

oe A

lgeb

ra 2

Lesson 11-5

Infi

nit

e G

eom

etri

c Se

ries

A g

eom

etri

c se

ries

th

at d

oes

not

en

d is

cal

led

an i

nfi

nit

ege

omet

ric

seri

es.S

ome

infi

nit

e ge

omet

ric

seri

es h

ave

sum

s,bu

t ot

her

s do

not

bec

ause

th

ep

arti

al s

um

sin

crea

se w

ith

out

appr

oach

ing

a li

mit

ing

valu

e.

Su

m o

f an

Infi

nit

eS

�fo

r �

1 �

r�

1.G

eom

etri

c S

erie

sIf

|r|

1, t

he in

finite

geo

met

ric s

erie

s do

es n

ot h

ave

a su

m.

Fin

d t

he

sum

of

each

in

fin

ite

geom

etri

c se

ries

,if

it e

xist

s.

a 1� 1

�r

Exam

ple

Exam

ple

a.75

�15

�3

�…

Fir

st,f

ind

the

valu

e of

rto

det

erm

ine

ifth

e su

m e

xist

s.a 1

�75

an

d a 2

�15

,so

r�

or

.Sin

ce �

��1,

the

sum

exis

ts.N

ow u

se t

he

form

ula

for

th

e su

mof

an

in

fin

ite

geom

etri

c se

ries

.

S�

Sum

for

mul

a

�a 1

�75

, r

�

�or

93.

75S

impl

ify.

Th

e su

m o

f th

e se

ries

is

93.7

5.

75 � �4 5�

1 � 5

75� 1

��1 5�

a 1� 1

�r

1 � 51 � 5

15 � 75

b.��

n�

148

��n

�1

In t

his

in

fin

ite

geom

etri

c se

ries

,a1

�48

and

r�

�.

S�

Sum

for

mul

a

�a 1

�48

, r

��

�or

36

Sim

plify

.

Th

us

�

n�

148��

�n�

1�

36.

1 � 3

48 � �4 3�

1 � 3

48�

�1

��

�1 3� �

a 1� 1

�r

1 � 3

1 � 3

Exer

cises

Exer

cises

Fin

d t

he

sum

of

each

in

fin

ite

geom

etri

c se

ries

,if

it e

xist

s.

1.a 1

��

7,r

�2.

1 �

��

…

3.a 1

�4,

r�

�18

do

es n

ot

exis

t8

4.�

��

…

5.15

�10

�6

�…

6.

18 �

9 �

4�

2�

…

145

12

7.�

��

…

8.10

00 �

800

�64

0 �

…

9.6

�12

�24

�48

�…

5000

do

es n

ot

exis

t

10. �

n�

150

��n

�1

11. �

k�

122

��k

�1

12. �

s�124

��s

�1

250

1457

3 � 52 � 3

7 � 121 � 2

4 � 5

1 � 5

1 � 401 � 20

1 � 10

1 � 3

1 � 41 � 2

2 � 325 � 16

25 � 27

2 � 9

2 � 3

1 � 225 � 16

5 � 45 � 8

©G

lenc

oe/M

cGra

w-H

ill65

6G

lenc

oe A

lgeb

ra 2

Rep

eati

ng

Dec

imal

sA

rep

eati

ng

deci

mal

rep

rese

nts

a f

ract

ion

.To

fin

d th

e fr

acti

on,

wri

te t

he

deci

mal

as

an i

nfi

nit

e ge

omet

ric

seri

es a

nd

use

th

e fo

rmu

la f

or t

he

sum

.

Wri

te e

ach

rep

eati

ng

dec

imal

as

a fr

acti

on.

Stu

dy G

uid

e a

nd I

nte

rven

tion

(c

onti

nued

)

Infi

nit

e G

eom

etri

c S

erie

s

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

11-5

11-5

Exam

ple

Exam

ple

a.0.

4�2�W

rite

th

e re

peat

ing

deci

mal

as

a su

m.

0.4�2�

�0.

4242

4242

…

��

��

…

In t

his

ser

ies

a 1�

and

r�

.

S�

Sum

for

mul

a

�a 1

�,

r�

�S

ubtr

act.

�or

S

impl

ify.

Th

us

0.4�2�

�.

14 � 33

14 � 3342 � 99� 14 02 0�

� � 19 09 0�

1� 10

042 � 10

0

� 14 02 0�

� 1 �

� 11 00�

a 1� 1

�r

1� 10

042 � 10

0

42�

�1,

000,

000

42� 10

,000

42 � 100

b.

0.52�

4�L

et S

�0.

52�4�.

S�

0.52

4242

4…W

rite

as a

rep

eatin

g de

cim

al.

1000

S�

524.

2424

24…

Mul

tiply

eac

h si

de b

y 10

00.

10S

�5.

2424

24…

Mul

itply

eac

h si

de b

y 10

.

990S

�51

9S

ubtr

act

the

third

equ

atio

nfr

om t

he s

econ

d eq

uatio

n.

S�

or

Sim

plify

.

Th

us,

0.52�

4��

173

� 33017

3� 33

051

9� 99

0

Exer

cises

Exer

cises

Wri

te e

ach

rep

eati

ng

dec

imal

as

a fr

acti

on.

1.0.

2�2.

0.8�

3.0.

3�0�4.

0.8�7�

5.0.

1�0�6.

0.5�4�

7.0.

7�5�8.

0.1�8�

9.0.

6�2�10

.0.7�

2�11

.0.0

7�2�12

.0.0

4�5�

13.0

.06�

14.0

.01�3�

8�15

.0.0�

1�3�8�

16.0

.08�1�

17.0

.24�5�

18.0

.43�6�

19.0

.54�

20.0

.86�3�

19 � 2249 � 90

24 � 5527 � 11

0

9� 11

046

� 3333

23� 16

651 � 15

1 � 224 � 55

8 � 1162 � 99

2 � 1125 � 33

6 � 1110 � 99

29 � 3310 � 33

8 � 92 � 9


An

swer

s


Skil

ls P

ract

ice

Infi

nit

e G

eom

etri

c S

erie

s

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

11-5

11-5

©G

lenc

oe/M

cGra

w-H

ill65

7G

lenc

oe A

lgeb

ra 2

Lesson 11-5

Fin

d t

he

sum

of

each

in

fin

ite

geom

etri

c se

ries

,if

it e

xist

s.

1.a 1

�1,

r�

22.

a 1�

5,r

��

3.a 1

�8,

r�

2d

oes

no

t ex

ist

4.a 1

�6,

r�

12

5.4

�2

�1

��

…8

6.54

0 �

180

�60

�20

�…

405

7.5

�10

�20

�…

do

es n

ot

exis

t8.

�33

6 �

84 �

21 �

…�

268.

8

9.12

5 �

25 �

5 �

…15

6.25

10.9

�1

��

…

11.

��

�…

do

es n

ot

exis

t12

.�

��

…

13.5

�2

�0.

8 �

…14

.9 �

6 �

4 �

…27

15. �

n�

110

��n

�1

2016

. �

n�

16 ��

�n�

1

17. �

n�

115

��n

�1

2518

. �

n�

1��

��n

�1

�2

Wri

te e

ach

rep

eati

ng

dec

imal

as

a fr

acti

on.

19.0

.4�20

.0.8�

21.0

.2�7�

22.0

.6�7�

23.0

.5�4�

24.0

.3�7�5�

25.0

.6�4�1�

26.0

.1�7�1�

57 � 333

641

� 999

125

� 333

6 � 11

67 � 993 � 11

8 � 94 � 9

1 � 34 � 3

2 � 5

9 � 21 � 3

1 � 2

25 � 3

1 � 21 � 27

1 � 91 � 3

27 � 49 � 4

3 � 4

81 � 101 � 9

1 � 2

1 � 2

25 � 72 � 5

1 � 2

©G

lenc

oe/M

cGra

w-H

ill65

8G

lenc

oe A

lgeb

ra 2

Fin

d t

he

sum

of

each

in

fin

ite

geom

etri

c se

ries

,if

it e

xist

s.

1.a 1

�35

,r�

492.

a 1�

26,r

�52

3.a 1

�98

,r�

�56

4.a 1

�42

,r�

do

es n

ot

exis

t

5.a 1

�11

2,r

��

706.

a 1�

500,

r�

625

7.a 1

�13

5,r

��

908.

18 �

6 �

2 �

…

9.2

�6

�18

�…

do

es n

ot

exis

t10

.6 �

4 �

�…

18

11.

��

1 �

…d

oes

no

t ex

ist

12.1

0 �

1 �

0.1

�…

13.1

00 �

20 �

4 �

…12

514

.�27

0 �

135

�67

.5 �

…�

180

15.0

.5 �

0.25

�0.

125

�…

116

.�

��

…

17.0

.8 �

0.08

�0.

008

�…

18.

��

�…

do

es n

ot

exis

t

19.3

��

�…

20.0

.3 �

0.00

3 �

0.00

003

�…

21.0

.06

�0.

006

�0.

0006

�…

22.

�2

�6

�…

do

es n

ot

exis

t

23. �

n�

13 �

�n�

14

24. �

n�

1��

�n�

1

25. �

n�

118

��n

�1

5426

. �

n�

15(

�0.

1)n

�1

Wri

te e

ach

rep

eati

ng

dec

imal

as

a fr

acti

on.

27.0

.6�28

.0.0�

9�29

.0.4�

3�30

.0.2�

7�

31.0

.2�4�3�

32.0

.8�4�

33.0

.9�9�0�

34.0

.1�5�0�

35.P

END

ULU

MS

On

its

fir

st s

win

g,a

pen

dulu

m t

rave

ls 8

fee

t.O

n e

ach

su

cces

sive

sw

ing,

the

pen

dulu

m t

rave

ls

the

dist

ance

of

its

prev

iou

s sw

ing.

Wh

at i

s th

e to

tal

dist

ance

trav

eled

by

the

pen

dulu

m w

hen

it

stop

s sw

ingi

ng?

40 f

t

36.E

LAST

ICIT

YA

bal

l dr

oppe

d fr

om a

hei

ght

of 1

0 fe

et b

oun

ces

back

� 19 0�of

th

at d

ista

nce

.

Wit

h ea

ch s

ucce

ssiv

e bo

unce

,the

bal

l con

tinu

es t

o re

ach

� 19 0�of

its

prev

ious

hei

ght.

Wha

t is

the

tota

l ver

tica

l dis

tanc

e (b

oth

up a

nd d

own)

tra

vele

d by

the

bal

l whe

n it

sto

ps b

ounc

ing?

(Hin

t:A

dd t

he

tota

l di

stan

ce t

he

ball

fal

ls t

o th

e to

tal

dist

ance

it

rise

s.)

190

ft

4 � 5

50 � 333

110

� 111

28 � 339 � 37

3 � 1143 � 99

1 � 112 � 3

50 � 112 � 3

8 � 213 � 4

2 � 31 � 4

2 � 31 � 15

30 � 101

21 � 427 � 49

9 � 7

1 � 31 � 6

1 � 128 � 9

7 � 97

� 1000

7� 10

07 � 10

100

�9

2 � 54 � 25

8 � 3

27 � 21 � 2

1 � 53 � 5

6 � 53 � 4

1 � 22 � 7

Pra

ctic

e (

Ave

rag

e)

Infi

nit

e G

eom

etri

c S

erie

s

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

11-5

11-5



Readin

g t

o L

earn

Math

em

ati

csIn

fin

ite

Geo

met

ric

Ser

ies

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

11-5

11-5

©G

lenc

oe/M

cGra

w-H

ill65

9G

lenc

oe A

lgeb

ra 2

Lesson 11-5

Pre-

Act

ivit

yH

ow d

oes

an i

nfi

nit

e ge

omet

ric

seri

es a

pp

ly t

o a

bou

nci

ng

bal

l?

Rea

d th

e in

trod

ucti

on t

o L

esso

n 11

-5 a

t th

e to

p of

pag

e 59

9 in

you

r te

xtbo

ok.

Not

e th

e fo

llow

ing

pow

ers

of 0

.6:0

.61

�0.

6;0.

62�

0.36

;0.6

3�

0.21

6;0.

64�

0.12

96;0

.65

�0.

0777

6;0.

66�

0.04

6656

;0.6

7�

0.02

7993

6.If

a b

all

is d

ropp

ed f

rom

a h

eigh

t of

10

feet

an

d bo

un

ces

back

to

60%

of

its

prev

iou

sh

eigh

t on

eac

h b

oun

ce,a

fter

how

man

y bo

un

ces

wil

l it

bou

nce

bac

k to

ah

eigh

t of

les

s th

an 1

foo

t?5

bo

un

ces

Rea

din

g t

he

Less

on

1.C

onsi

der

the

form

ula

S�

.

a.W

hat

is

the

form

ula

use

d to

fin

d?th

e su

m o

f an

infi

nit

e g

eom

etri

c se

ries

b.

Wh

at d

o ea

ch o

f th

e fo

llow

ing

repr

esen

t?

S:

the

sum

a 1:

the

firs

t te

rm

r:th

e co

mm

on

rat

io

c.F

or w

hat

val

ues

of

rdo

es a

n i

nfi

nit

e ge

omet

ric

sequ

ence

hav

e a

sum

?�

1 �

r�

1

d.

Rew

rite

you

r an

swer

for

par

t d

as a

n a

bsol

ute

val

ue

ineq

ual

ity.

|r|�

1

2.F

or e

ach

of

the

foll

owin

g ge

omet

ric

seri

es,g

ive

the

valu

es o

f a 1

and

r.T

hen

sta

tew

het

her

th

e su

m o

f th

e se

ries

exi

sts.

(Do

not

act

ual

ly f

ind

the

sum

.)

a.�

��

…a 1

�r

�

Doe

s th

e su

m e

xist

?

b.

2 �

1 �

��

…a 1

�r

�

Doe

s th

e su

m e

xist

?

c.�

i�13i

a 1�

r�

Doe

s th

e su

m e

xist

?

Hel

pin

g Y

ou

Rem

emb

er

3.O

ne g

ood

way

to

rem

embe

r so

met

hing

is

to r

elat

e it

to

som

ethi

ng y

ou a

lrea

dy k

now

.How

can

you

use

th

e fo

rmu

la S

n�

that

you

lea

rned

in

Les

son

11-

4 fo

r fi

ndi

ng

the

sum

of

a ge

omet

ric

seri

es t

o h

elp

you

rem

embe

r th

e fo

rmu

la f

or f

indi

ng

the

sum

of

anin

fin

ite

geom

etri

c se

ries

?S

amp

le a

nsw

er:

If �

1 �

r�

1,th

en a

s n

get

s la

rge,

rnap

pro

ach

es 0

,so

1 �

rnap

pro

ach

es 1

.Th

eref

ore

,Sn

app

roac

hes

,or

.a 1

� 1 �

ra 1

�1

� 1 �

r

a 1(1

�rn

)�

�1

�r

no

33

yes�

�1 2�2

1 � 41 � 2

yes�1 3�

�2 3�2 � 27

2 � 92 � 3

a 1� 1

�r

©G

lenc

oe/M

cGra

w-H

ill66

0G

lenc

oe A

lgeb

ra 2

Co

nver

gen

ce a

nd

Div

erg

ence

Con

verg

ence

and

div

erge

nce

are

term

s th

at r

elat

e to

the

exi

sten

ce o

f a

sum

of

an i

nfi

nit

e se

ries

.If

a su

m e

xist

s,th

e se

ries

is

con

verg

ent.

If n

ot,t

he

seri

es i

s

dive

rgen

t.C

onsi

der

the

seri

es 1

2 �

3 �

�3 4��

� 13 6��

….T

his

is

a ge

omet

ric

seri

es w

ith

r�

�1 4� .T

he

sum

is

give

n b

y th

e fo

rmu

la S

�� 1

a �1

r�

.T

hu

s,th

e su

m

is 1

2 �

�3 4�or

16.

Th

is s

erie

s is

con

verg

ent

sin

ce a

su

m e

xist

s.N

otic

e th

at t

he

firs

t tw

o te

rms

hav

e a

sum

of

15.A

s m

ore

term

s ar

e ad

ded,

the

sum

com

escl

oser

(or

con

verg

es)

to 1

6.

Rec

all

that

a g

eom

etri

c se

ries

has

a s

um

if

and

only

if

�1

�r

�1.

Th

us,

age

omet

ric

seri

es i

s co

nver

gent

if

ris

bet

wee

n �

1 an

d 1,

and

dive

rgen

t if

rha

san

oth

er v

alu

e.A

n i

nfi

nit

e ar

ith

met

ic s

erie

s ca

nn

ot h

ave

a su

m u

nle

ss a

ll o

fth

e te

rms

are

equ

al t

o ze

ro.

Det

erm

ine

wh

eth

er e

ach

ser

ies

is c

onve

rgen

t or

div

erge

nt.

a.2

�5

�8

�11

�…

dive

rgen

t

b.

�2

�4

�(�

8) �

16 �

…di

verg

ent

c.16

�8

�4

�2

�…

con

verg

ent

Det

erm

ine

wh

eth

er e

ach

ser

ies

is c

onve

rgen

t or

div

erge

nt.

If t

he

seri

es i

s co

nve

rgen

t,fi

nd

th

e su

m.

1.5

�10

�15

�20

�…

2.16

�8

�4

�2

�…

div

erg

ent

conv

erg

ent;

32

3.1

�0.

1 �

0.01

�0.

001

�…

4.4

�2

�0

�2

�…

conv

erg

ent;

1.11

div

erg

ent

5.2

�4

�8

�16

�…

6.1

��1 5�

�� 21 5�

�� 11 25�

�…

div

erg

ent

conv

erg

ent;

�5 6�

7.4

�2.

4 �

1.44

�0.

864

�…

8.�1 8�

��1 4�

��1 2�

�1

�…

conv

erg

ent;

10d

iver

gen

t

9.�

�5 3��

�1 90 ��

�2 20 7��

�4 80 1��

…10

.48

�12

�3

��3 4�

�…

conv

erg

ent;

�1

conv

erg

ent;

64

Bo

nu

s:Is

1 �

�1 2��

�1 3��

�1 4��

�1 5��

… c

onve

rgen

t or

div

erge

nt?

div

erg

ent

En

rich

men

t

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

11-5

11-5

Exam

ple

Exam

ple


An

swer

s


Stu

dy G

uid

e a

nd I

nte

rven

tion

Rec

urs

ion

an

d S

pec

ial S

equ

ence

s

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

11-6

11-6

©G

lenc

oe/M

cGra

w-H

ill66

1G

lenc

oe A

lgeb

ra 2

Lesson 11-6

Spec

ial S

equ

ence

sIn

a r

ecu

rsiv

e fo

rmu

la,e

ach

su

ccee

din

g te

rm i

s fo

rmu

late

d fr

omon

e or

mor

e pr

evio

us

term

s.A

rec

urs

ive

form

ula

for

a s

equ

ence

has

tw

o pa

rts:

1.th

e va

lue(

s) o

f th

e fi

rst

term

(s),

and

2.an

equ

atio

n t

hat

sh

ows

how

to

fin

d ea

ch t

erm

fro

m t

he

term

(s)

befo

re i

t.

Fin

d t

he

firs

t fi

ve t

erm

s of

th

e se

qu

ence

in

wh

ich

a1

�6,

a2

�10

,an

d a

n�

2an

�2

for

n

3.a 1

�6

a 2�

10a 3

�2a

1�2(

6) �

12a 4

�2a

2�

2(10

) �

20a 5

�2a

3�

2(12

) �

24

Th

e fi

rst

five

ter

ms

of t

he

sequ

ence

are

6,1

0,12

,20,

24.

Fin

d t

he

firs

t fi

ve t

erm

s of

eac

h s

equ

ence

.

1.a 1

�1,

a 2�

1,a n

�2(

a n�

1�

a n�

2),n

3

1,1,

4,10

,28

2.a 1

�1,

a n�

,n

2

1,,

,,

3.a 1

�3,

a n�

a n�

1�

2(n

�2)

,n

23,

3,5,

9,15

4.a 1

�5,

a n�

a n�

1�

2,n

2

5,7,

9,11

,13

5.a 1

�1,

a n�

(n�

1)a n

�1,

n

21,

1,2,

6,24

6.a 1

�7,

a n�

4an

�1

�1,

n

27,

27,1

07,4

27,1

707

7.a 1

�3,

a 2�

4,a n

�2a

n�

2�

3an

�1,

n

33,

4,18

,62,

222

8.a 1

�0.

5,a n

�a n

�1

�2n

,n

20.

5,4.

5,10

.5,1

8.5,

28.5

9.a 1

�8,

a 2�

10,a

n�

,n

38,

10,0

.8,1

2.5,

0.06

4

10.a

1�

100,

a n�

,n

210

0,50

,,

,50 � 60

50 � 1250 � 3

a n�

1�

n

a n�

2� a n

�1

5 � 83 � 5

2 � 31 � 2

1�

�1

�a n

�1

Exam

ple

Exam

ple

Exer

cises

Exer

cises

©G

lenc

oe/M

cGra

w-H

ill66

2G

lenc

oe A

lgeb

ra 2

Iter

atio

nC

ombi

nin

g co

mpo

siti

on o

f fu

nct

ion

s w

ith

th

e co

nce

pt o

f re

curs

ion

lea

ds t

o th

epr

oces

s of

ite

rati

on.I

tera

tion

is

the

proc

ess

of c

ompo

sin

g a

fun

ctio

n w

ith

its

elf

repe

ated

ly.

Fin

d t

he

firs

t th

ree

iter

ates

of

f(x)

�4x

�5

for

an

init

ial

valu

e of

x0

�2.

To

fin

d th

e fi

rst

iter

ate,

fin

d th

e va

lue

of t

he

fun

ctio

n f

or x

0�

2x 1

�f(

x 0)

Itera

te t

he f

unct

ion.

�f(

2)x 0

�2

�4(

2) �

5 or

3S

impl

ify.

To

fin

d th

e se

con

d it

erat

ion

,fin

d th

e va

lue

of t

he

fun

ctio

n f

or x

1�

3.x 2

�f(

x 1)

Itera

te t

he f

unct

ion.

�f(

3)x 1

�3

�4(

3) �

5 or

7S

impl

ify.

To

fin

d th

e th

ird

iter

atio

n,f

ind

the

valu

e of

th

e fu

nct

ion

for

x2

�7.

x 3�

f(x 2

)Ite

rate

the

fun

ctio

n.

�f(

7)x 2

�7

�4(

7) �

5 or

23

Sim

plify

.

Th

e fi

rst

thre

e it

erat

es a

re 3

,7,a

nd

23.

Fin

d t

he

firs

t th

ree

iter

ates

of

each

fu

nct

ion

for

th

e gi

ven

in

itia

l va

lue.

1.f(

x) �

x�

1;x 0

�4

2.f(

x) �

x2�

3x;x

0�

13.

f(x)

�x2

�2x

�1;

x 0�

�2

3,2,

1�

2,10

,70

1,4,

25

4.f(

x) �

4x�

6;x 0

��

55.

f(x)

�6x

�2;

x 0�

36.

f(x)

�10

0 �

4x;x

0�

�5

�26

,�11

0,�

446

16,9

4,56

212

0,�

380,

1620

7.f(

x) �

3x�

1;x 0

�47

8.f(

x) �

x3�

5x2 ;

x 0�

19.

f(x)

�10

x�

25;x

0�

2

140,

419,

1256

�4,

�14

4,�

3,08

9,66

4�

5,�

75,�

775

10.f

(x)

�4x

2�

9;x 0

��

111

.f(x

) �

2x2

�5;

x 0�

�4

12.f

(x)

�;x

0�

1

�5,

91,3

3,11

537

,274

3,15

,048

,103

0,�

,�1

13.f

(x)

�(x

�11

);x 0

�3

14.f

(x)

�

;x0

�9

15.f

(x)

�x

�4x

2 ;x 0

�1

7,9,

10,9

,�

3,�

39,�

6123

16.f

(x)

�x

�;x

0�

217

.f(x

) �

x3�

5x2

�8x

�10

;18

.f(x

) �

x3�

x2;x

0�

�2

2.5,

2.9,

abo

ut

3.24

5x 0

�1

�12

,�18

72,

�6,

�45

4,�

94,6

10,8

86�

6,56

3,71

1,23

2

1 � x

1 � 31 � 3

3 � x1 � 2

1 � 2

x�

1� x

�2

Stu

dy G

uid

e a

nd I

nte

rven

tion

(c

onti

nued

)

Rec

urs

ion

an

d S

pec

ial S

equ

ence

s

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

11-6

11-6

Exam

ple

Exam

ple

Exer

cises

Exer

cises



Skil

ls P

ract

ice

Rec

urs

ion

an

d S

pec

ial S

equ

ence

s

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

11-6

11-6

©G

lenc

oe/M

cGra

w-H

ill66

3G

lenc

oe A

lgeb

ra 2

Lesson 11-6

Fin

d t

he

firs

t fi

ve t

erm

s of

eac

h s

equ

ence

.

1.a 1

�4,

a n�

1�

a n�

72.

a 1�

�2,

a n�

1�

a n�

3

4,11

,18,

25,3

2�

2,1,

4,7,

10

3.a 1

�5,

a n�

1�

2an

4.a 1

��

4,a n

�1

�6

�a n

5,10

,20,

40,8

0�

4,10

,�4,

10,�

4

5.a 1

�1,

a n�

1�

a n�

n6.

a 1�

�1,

a n�

1�

n�

a n1,

2,4,

7,11

�1,

2,0,

3,1

7.a 1

��

6,a n

�1

�a n

�n

�1

8.a 1

�8,

a n�

1�

a n�

n�

2

�6,

�4,

�1,

3,8

8,5,

1,�

4,�

10

9.a 1

��

3,a n

�1

�2a

n�

710

.a1

�4,

a n�

1�

�2a

n�

5

�3,

1,9,

25,5

74,

�13

,21,

�47

,89

11.a

1�

0,a 2

�1,

a n�

1�

a n�

a n�

112

.a1

��

1,a 2

��

1,a n

�1

�a n

�a n

�1

0,1,

1,2,

3�

1,�

1,0,

1,1

13.a

1�

3,a 2

��

5,a n

�1

��

4an

�a n

�1

14.a

1�

�3,

a 2�

2,a n

�1

�a n

�1

�a n

3,�

5,23

,�97

,411

�3,

2,�

5,7,

�12

Fin

d t

he

firs

t th

ree

iter

ates

of

each

fu

nct

ion

for

th

e gi

ven

in

itia

l va

lue.

15.f

(x)

�2x

�1,

x 0�

35,

9,17

16.f

(x)

�5x

�3,

x 0�

27,

32,1

57

17.f

(x)

�3x

�4,

x 0�

�1

1,7,

2518

.f(x

) �

4x�

7,x 0

��

5�

13,�

45,�

173

19.f

(x)

��

x�

3,x 0

�10

�13

,10,

�13

20.f

(x)

��

3x�

6,x 0

�6

�12

,42,

�12

0

21.f

(x)

��

3x�

4,x 0

�2

�2,

10,�

2622

.f(x

) �

6x�

5,x 0

�1

1,1,

1

23.f

(x)

�7x

�1,

x 0�

�4

24.f

(x)

�x2

�3x

,x0

�5

�27

,�18

8,�

1315

10,7

0,46

90

©G

lenc

oe/M

cGra

w-H

ill66

4G

lenc

oe A

lgeb

ra 2

Fin

d t

he

firs

t fi

ve t

erm

s of

eac

h s

equ

ence

.

1.a 1

�3,

a n�

1�

a n�

52.

a 1�

�7,

a n�

1�

a n�

83,

8,13

,18,

23�

7,1,

9,17

,25

3.a 1

��

3,a n

�1

�3a

n�

24.

a 1�

�8,

a n�

1�

10 �

a n�

3,�

7,�

19,�

55,�

163

�8,

18,�

8,18

,�8

5.a 1

�4,

a n�

1�

n�

a n6.

a 1�

�3,

a n�

1�

3an

4,�

3,5,

�2,

6�

3,�

9,�

27,�

81,�

243

7.a 1

�4,

a n�

1�

�3a

n�

48.

a 1�

2,a n

�1

��

4an

�5

4,�

8,28

,�80

,244

2,�

13,4

7,�

193,

767

9.a 1

�3,

a 2�

1,a n

�1

�a n

�a n

�1

10.a

1�

�1,

a 2�

5,a n

�1

�4a

n�

1�

a n3,

1,�

2,�

3,�

1�

1,5,

�9,

29,�

65

11.a

1�

2,a 2

��

3,a n

�1

�5a

n�

8an

�1

12.a

1�

�2,

a 2�

1,a n

�1

��

2an

�6a

n�

12,

�3,

�31

,�13

1,�

407

�2,

1,�

14,3

4,�

152

Fin

d t

he

firs

t th

ree

iter

ates

of

each

fu

nct

ion

for

th

e gi

ven

in

itia

l va

lue.

13.f

(x)

�3x

�4,

x 0�

�1

1,7,

2514

.f(x

) �

10x

�2,

x 0�

�1

�8,

�78

,�77

8

15.f

(x)

�8

�3x

,x0

�1

11,4

1,13

116

.f(x

) �

8 �

x,x 0

��

311

,�3,

11

17.f

(x)

�4x

�5,

x 0�

�1

1,9,

4118

.f(x

) �

5(x

�3)

,x0

��

25,

40,2

15

19.f

(x)

��

8x�

9,x 0

�1

1,1,

120

.f(x

) �

�4x

2 ,x 0

��

1�

4;�

64;�

16,3

84

21.f

(x)

�x2

�1,

x 0�

38,

63,3

968

22.f

(x)

�2x

2 ;x 0

�5

50;

5000

;50

,000

,000

23.I

NFL

ATI

ON

Iter

atin

g th

e fu

nct

ion

c(x

) �

1.05

xgi

ves

the

futu

re c

ost

of a

n i

tem

at

aco

nst

ant

5% i

nfl

atio

n r

ate.

Fin

d th

e co

st o

f a

$200

0 ri

ng

in f

ive

year

s at

5%

in

flat

ion

.$2

552.

56

FRA

CTA

LSF

or E

xerc

ises

24–

27,u

se t

he

fo

llow

ing

info

rmat

ion

.R

epla

cin

g ea

ch s

ide

of t

he

squ

are

show

n w

ith

th

eco

mbi

nat

ion

of

segm

ents

bel

ow i

t gi

ves

the

figu

re

to i

ts r

igh

t.

24.W

hat

is

the

peri

met

er o

f th

e or

igin

al s

quar

e?12

in.

25.W

hat

is

the

peri

met

er o

f th

e n

ew s

hap

e?20

in.

26.I

f yo

u r

epea

t th

e pr

oces

s by

rep

laci

ng

each

sid

e of

th

e n

ew s

hap

e by

a p

ropo

rtio

nal

com

bin

atio

n o

f 5

segm

ents

,wh

at w

ill

the

peri

met

er o

f th

e th

ird

shap

e be

?33

in.

27.W

hat

fu

nct

ion

f(x

) ca

n y

ou i

tera

te t

o fi

nd

the

peri

met

er o

f ea

ch s

ucc

essi

ve s

hap

e if

you

con

tin

ue

this

pro

cess

?f(

x)

�x

5 � 3

1 � 3

1 in

.1

in.

1 in

.

3 in

.

1 in

.1

in.

Pra

ctic

e (

Ave

rag

e)

Rec

urs

ion

an

d S

pec

ial S

equ

ence

s

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

11-6

11-6


An

swer

s


Readin

g t

o L

earn

Math

em

ati

csR

ecu

rsio

n a

nd

Sp

ecia

l Seq

uen

ces

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

11-6

11-6

©G

lenc

oe/M

cGra

w-H

ill66

5G

lenc

oe A

lgeb

ra 2

Lesson 11-6

Pre-

Act

ivit

yH

ow i

s th

e F

ibon

acci

seq

uen

ce i

llu

stra

ted

in

nat

ure

?

Rea

d th

e in

trod

ucti

on t

o L

esso

n 11

-6 a

t th

e to

p of

pag

e 60

6 in

you

r te

xtbo

ok.

Wh

at a

re t

he

nex

t th

ree

nu

mbe

rs i

n t

he

sequ

ence

th

at g

ives

th

e n

um

ber

ofsh

oots

cor

resp

ondi

ng

to e

ach

mon

th?

8,13

,21

Rea

din

g t

he

Less

on

1.C

onsi

der

the

sequ

ence

in

wh

ich

a1

�4

and

a n�

2an

�1

�5.

a.E

xpla

in w

hy

this

is

a re

curs

ive

form

ula

.S

amp

le a

nsw

er:

Eac

h t

erm

is f

ou

nd

fro

m t

he

valu

e o

f th

e p

revi

ou

s te

rm.

b.

Exp

lain

in

you

r ow

n w

ords

how

to

fin

d th

e fi

rst

fou

r te

rms

of t

his

seq

uen

ce.(

Do

not

actu

ally

fin

d an

y te

rms

afte

r th

e fi

rst.

)S

amp

le a

nsw

er:T

he

firs

t te

rm is

4.T

ofi

nd

th

e se

con

d t

erm

,do

ub

le t

he

firs

t te

rm a

nd

ad

d 5

.To

fin

d t

he

thir

dte

rm,d

ou

ble

th

e se

con

d t

erm

an

d a

dd

5.T

o f

ind

th

e fo

urt

h t

erm

,d

ou

ble

th

e th

ird

ter

m a

nd

ad

d 5

.

c.W

hat

hap

pen

s to

th

e te

rms

of t

his

seq

uen

ce a

s n

incr

ease

s?S

amp

le a

nsw

er:

Th

ey k

eep

get

tin

g la

rger

an

d la

rger

.

2.C

onsi

der

the

fun

ctio

n f

(x)

�3x

�1

wit

h a

n i

nit

ial

valu

e of

x0

�2.

a.W

hat

doe

s it

mea

n t

o it

erat

eth

is f

un

ctio

n?

to c

om

po

se t

he

fun

ctio

n w

ith

itse

lf r

epea

ted

ly

b.

Fil

l in

th

e bl

anks

to

fin

d th

e fi

rst

thre

e it

erat

es.T

he

blan

ks t

hat

fol

low

th

e le

tter

xar

e fo

r su

bscr

ipts

.

x 1�

f(x

) �

f()

�3(

) �

1 �

�1

�

x 2�

f(x

) �

f()

�3(

) �

1 �

x 3�

f(x

) �

f()

�3(

) �

1 �

c.A

s th

is p

roce

ss c

onti

nu

es,w

hat

hap

pen

s to

th

e va

lues

of

the

iter

ates

?S

amp

le a

nsw

er:T

hey

kee

p g

etti

ng

larg

er a

nd

larg

er.

Hel

pin

g Y

ou

Rem

emb

er

3.U

se a

dic

tion

ary

to f

ind

the

mea

nin

gs o

f th

e w

ords

rec

urr

ent

and

iter

ate.

How

can

th

em

ean

ings

of

thes

e w

ords

hel

p yo

u t

o re

mem

ber

the

mea

nin

g of

th

e m

ath

emat

ical

ter

ms

recu

rsiv

ean

d it

erat

ion

? H

ow a

re t

hes

e id

eas

rela

ted?

Sam

ple

an

swer

:R

ecu

rren

tm

ean

s h

app

enin

g r

epea

ted

ly,w

hile

iter

ate

mea

ns

to r

epea

t a

pro

cess

or

op

erat

ion

.A r

ecu

rsiv

efo

rmu

la is

use

d r

epea

ted

ly t

o f

ind

th

e va

lue

of

on

ete

rm o

f a

seq

uen

ce b

ased

on

th

e p

revi

ou

s te

rm.I

tera

tio

nm

ean

s to

com

po

se a

fu

nct

ion

wit

h it

sel

f re

pea

ted

ly.B

oth

idea

s h

ave

to d

o w

ith

rep

etit

ion

—d

oin

g t

he

sam

e th

ing

ove

r an

d o

ver

agai

n.

4114

142

145

51

56

22

0

©G

lenc

oe/M

cGra

w-H

ill66

6G

lenc

oe A

lgeb

ra 2

Co

nti

nu

ed F

ract

ion

sT

he

frac

tion

bel

ow i

s an

exa

mpl

e of

a c

onti

nu

ed f

ract

ion

.Not

e th

at e

ach

frac

tion

in

th

e co

nti

nu

ed f

ract

ion

has

a n

um

erat

or o

f 1.

2 �

1�

�3

�� 4

�11 – 5

�En

rich

men

t

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

11-6

11-6

Eva

luat

e th

e co

nti

nu

edfr

acti

on a

bov

e.S

tart

at

the

bot

tom

an

dw

ork

you

r w

ay u

p.

Ste

p 1

:4

��1 5�

��2 50 �

��1 5�

��2 51 �

Ste

p 2

:�

� 25 1�

Ste

p 3

:3

�� 25 1�

��6 23 1�

�� 25 1�

��6 28 1�

Ste

p 4

:�

�2 61 8�

Ste

p 5

:2

��2 61 8�

�2�

2 61 8�

1 � �6 28 1�1 � �2 51 �

Ch

ange

�2 15 1�in

to a

con

tin

ued

fra

ctio

n.

Fol

low

th

e st

eps.

Ste

p 1

:�2 15 1�

��2 12 1�

�� 13 1�

�2

�� 13 1�

Ste

p 2

:� 13 1�

�

Ste

p 3

:�1 31 �

��9 3�

��2 3�

�3

��2 3�

Ste

p 4

:�2 3�

�

Ste

p 5

:�3 2�

��2 2�

��1 2�

�1

��1 2�

Sto

p, b

ecau

se t

he n

umer

ator

is 1

.

Th

us,

�2 15 1�ca

n b

e w

ritt

en a

s 2

�1

��

3 �

� 1 �1

1 – 2

�

1 � �3 2�

1 � �1 31 �

Exam

ple1

Exam

ple1

Exam

ple2

Exam

ple2

Eva

luat

e ea

ch c

onti

nu

ed f

ract

ion

.

1.1

�1

2.0

�� 59 6�

1 �

1�1 27 4�

3.2

�1

4.5

�5

�1 70 10 1�4

�2

� 24 09 66 5�

Ch

ange

eac

h f

ract

ion

in

to a

con

tin

ued

fra

ctio

n.

5.�7 35 1�

6.�2 89 �

7.�1 13 9�

1�

�6

�� 8

�11 — 10�

1�

�7

�� 9

�11 — 11�

1

2 �

� 3�1

1 – 3

�

1�

�6

�� 4

�11 – 2

�

2 �

1

2 �

1

2 �

1

1 �

1� 1

��1 2�

3 �

1

1 �

1

1 �

1� 1

��1 2�

0 �

1

1 �

1� 2

��1 6�



Stu

dy G

uid

e a

nd I

nte

rven

tion

Th

e B

ino

mia

l Th

eore

m

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

11-7

11-7

©G

lenc

oe/M

cGra

w-H

ill66

7G

lenc

oe A

lgeb

ra 2

Lesson 11-7

Pasc

al’s

Tri

ang

leP

asca

l’s t

rian

gle

is t

he

patt

ern

of

coef

fici

ents

of

pow

ers

of b

inom

ials

disp

laye

d in

tri

angu

lar

form

.Eac

h r

ow b

egin

s an

d en

ds w

ith

1 a

nd

each

coe

ffic

ien

t is

th

esu

m o

f th

e tw

o co

effi

cien

ts a

bove

it

in t

he

prev

iou

s ro

w.

(a�

b)0

1(a

�b)

11

1

Pas

cal’s

Tri

ang

le(a

�b)

21

21

(a�

b)3

13

31

(a�

b)4

14

64

1(a

�b)

51

510

105

1

Use

Pas

cal’s

tri

angl

e to

fin

d t

he

nu

mb

er o

f p

ossi

ble

seq

uen

ces

con

sist

ing

of 3

as

and

2 b

s.T

he

coef

fici

ent

10 o

f th

e a3

b2-t

erm

in

th

e ex

pan

sion

of

(a�

b)5

give

s th

e n

um

ber

ofse

quen

ces

that

res

ult

in

th

ree

as a

nd

two

bs.

Exp

and

eac

h p

ower

usi

ng

Pas

cal’s

tri

angl

e.

1.(a

�5)

4a4

�20

a3

�15

0a2

�50

0a�

625

2.(x

�2y

)6x

6�

12x

5 y�

60x

4 y2

�16

0x3 y

3�

240x

2 y4

�19

2xy

5�

64y

6

3.(j

�3k

)5j5

�15

j4k

�90

j3k

2�

270j

2 k3

�40

5jk

4�

243k

5

4.(2

s�

t)7

128s

7�

448s

6 t�

672s

5 t2

�56

0s4 t

3�

280s

3 t4

�84

s2 t

5�

14st

6�

t7

5.(2

p�

3q)6

64p

6�

576p

5 q�

2160

p4 q2

�43

20p

3 q3

�48

60p

2 q4

�29

16pq

5�

729q

6

6.�a

��4

a4�

2a3 b

�a2

b2

�ab

3�

b4

7.R

ay t

osse

s a

coin

15

tim

es.H

ow m

any

diff

eren

t se

quen

ces

of t

osse

s co

uld

res

ult

in

4h

eads

an

d 11

tai

ls?

1365

8.T

her

e ar

e 9

tru

e/fa

lse

ques

tion

s on

a q

uiz

.If

twic

e as

man

y of

th

e st

atem

ents

are

tru

e as

fals

e,h

ow m

any

diff

eren

t se

quen

ces

of t

rue/

fals

e an

swer

s ar

e po

ssib

le?

84

1 � 161 � 2

3 � 2b � 2

Exam

ple

Exam

ple

Exer

cises

Exer

cises

©G

lenc

oe/M

cGra

w-H

ill66

8G

lenc

oe A

lgeb

ra 2

The

Bin

om

ial T

heo

rem

Bin

om

ial

If n

is a

non

nega

tive

inte

ger,

then

Th

eore

m(a

�b)

n�

1an b

0�

an�

1 b1

�an

�2 b

2�

an�

3 b3

�…

�1a

0 bn

An

oth

er u

sefu

l fo

rm o

f th

e B

inom

ial T

heo

rem

use

s fa

ctor

ial

not

atio

n a

nd

sigm

a n

otat

ion

.

Fact

ori

alIf

nis

a p

ositi

ve in

tege

r, th

en n

! �

n(n

�1)

(n�

2) �

… �

2 �

1.

Bin

om

ial

Th

eore

m,

(a�

b)n

�an

b0�

an�

1 b1

�an

�2 b

2�

… �

a0bn

Fact

ori

al

Fo

rm�

�n

k�0

an�

k bk

Eva

luat

e .

� �11

�10

�9

�99

0

Exp

and

(a

�3b

)4.

(a�

3b)4

��4

k�

0a4

�k (

�3b

)k

�a4

�a3

(�3b

)1�

a2(�

3b)2

�a(

�3b

)3�

(�3b

)4

�a4

�12

a3b

�54

a2b2

�10

8ab3

�81

b4

Eva

luat

e ea

ch e

xpre

ssio

n.

1.5!

120

2.36

3.21

0

Exp

and

eac

h p

ower

.

4.(a

�3)

6a6

�18

a5�

135a

4�

540a

3�

1215

a2

�14

58a

�72

9

5.(r

�2s

)7r7

�14

r6s

�84

r5s

2�

280r

4 s3

�56

0r3 s

4�

672r

2 s5

�44

8rs

6�

128s

7

6.(4

x�

y)4

256x

4�

256x

3 y�

96x

2 y2

�16

xy3

�y

4

7.�2

��5

32 �

40m

�20

m2

�5m

3�

m4

�m

5

Fin

d t

he

ind

icat

ed t

erm

of

each

exp

ansi

on.

8.th

ird

term

of

(3x

�y)

527

0x3 y

29.

fift

h t

erm

of

(a�

1)7

35a

3

10.f

ourt

h t

erm

of

(j�

2k)8

448j

5 k3

11.s

ixth

ter

m o

f (1

0 �

3t)7

�51

0,30

0t5

12.s

econ

d te

rm o

f �m

��9

6m8

13.s

even

th t

erm

of

(5x

�2)

1192

,400

,000

x5

2 � 3

1 � 325 � 8

m � 2

10!

� 6!4!

9!� 7!

2!

4!� 0!

4!4!

� 1!3!

4!� 2!

2!4!

� 3!1!

4!� 4!

0!

4!�

�(4

�k)

!k!

11 �

10 �

9 �

8 �

7 �

6 �

5 �

4 �

3 �

2 �

1�

��

��

8 �

7 �

6 �

5 �

4 �

3 �

2 �

111

!� 8!

11!

� 8!

n!

��

(n�

k)!k

!

n!

� 0!n

!n

!�

�(n

�2)

!2!

n!

��

(n�

1)!1

!n

!� n

!0!

n(n

�1)

(n�

2)�

�1

�2

�3

n(n

�1)

��

1 �

2n � 1

Stu

dy G

uid

e a

nd I

nte

rven

tion

(c

onti

nued

)

Th

e B

ino

mia

l Th

eore

m

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

11-7

11-7

Exam

ple1

Exam

ple1

Exam

ple2

Exam

ple2

Exer

cises

Exer

cises


An

swer

s


Skil

ls P

ract

ice

Th

e B

ino

mia

l Th

eore

m

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

11-7

11-7

©G

lenc

oe/M

cGra

w-H

ill66

9G

lenc

oe A

lgeb

ra 2

Lesson 11-7

Eva

luat

e ea

ch e

xpre

ssio

n.

1.8!

40,3

202.

10!

3,62

8,80

0

3.12

!47

9,00

1,60

04.

210

5.12

06.

45

7.84

8.15

,504

Exp

and

eac

h p

ower

.

9.(x

�y)

310

.(a

�b)

5

x3

�3x

2 y�

3xy

2�

y3

a5�

5a4 b

�10

a3b

2�

10a2

b3

�5a

b4

�b

5

12.(

m�

1)4

11.(

g�

h)4

m4

�4m

3�

6m2

�4m

�1

g4

�4g

3 h�

6g2 h

2�

4gh

3�

h4

13.(

r�

4)3

14.(

a�

5)4

r3�

12r2

�48

r�

64a

4�

20a

3�

150a

2�

500a

�62

5

15.(

y�

7)3

16.(

d�

2)5

y3

�21

y2

�14

7y�

343

d5

�10

d4

�40

d3

�80

d2

�80

d�

32

17.(

x�

1)4

18.(

2a�

b)4

x4

�4x

3�

6x2

�4x

�1

16a

4�

32a

3 b�

24a

2 b2

�8a

b3

�b

4

19.(

c�

4d)3

20.(

2a�

3)3

c3

�12

c2 d

�48

cd2

�64

d3

8a3

�36

a2

�54

a�

27

Fin

d t

he

ind

icat

ed t

erm

of

each

exp

ansi

on.

21.f

ourt

h t

erm

of

(m�

n)1

012

0m7 n

322

.sev

enth

ter

m o

f (x

�y)

828

x2 y

6

23.t

hir

d te

rm o

f (b

�6)

536

0b3

24.s

ixth

ter

m o

f (s

�2)

9�

4032

s4

25.f

ifth

ter

m o

f (2

a�

3)6

4860

a2

26.s

econ

d te

rm o

f (3

x�

y)7

�51

03x

6 y

20!

� 15!5

!9!

� 3!6!

10!

� 2!8!

6! � 3!

15!

� 13!

©G

lenc

oe/M

cGra

w-H

ill67

0G

lenc

oe A

lgeb

ra 2

Eva

luat

e ea

ch e

xpre

ssio

n.

1.7!

5040

2.11

!39

,916

,800

3.30

244.

380

5.28

6.56

7.92

48.

10,6

60

Exp

and

eac

h p

ower

.

9.(n

�v)

5n

5�

5n4 v

�10

n3 v

2�

10n

2 v3

�5n

v4

�v

5

10.(

x�

y)4

x4

�4x

3 y�

6x2 y

2�

4xy

3�

y4

11.(

x�

y)6

x6

�6x

5 y�

15x

4 y2

�20

x3 y

3�

15x

2 y4

�6x

y5

�y

6

12.(

r�

3)5

r5�

15r4

�90

r3�

270

r2�

405r

�24

3

13.(

m�

5)5

m5

�25

m4

�25

0m3

�12

50m

2�

3125

m�

3125

14.(

x�

4)4

x4

�16

x3

�96

x2

�25

6x�

256

15.(

3x�

y)4

81x

4�

108x

3 y�

54x

2 y2

�12

xy3

�y

4

16.(

2m�

y)4

16m

4�

32m

3 y�

24m

2 y2

�8m

y3

�y

4

17.(

w�

3z)3

w3

�9w

2 z�

27w

z2

�27

z3

18.(

2d�

3)6

64d

6�

576d

5�

2160

d4

�43

20d

3�

4860

d2

�29

16d

�72

9

19.(

x�

2y)5

x5

�10

x4 y

�40

x3 y

2�

80x

2 y3

�80

xy4

�32

y5

20.(

2x�

y)5

32x

5�

80x

4 y�

80x

3 y2

�40

x2 y

3�

10xy

4�

y5

21.(

a�

3b)4

a4

�12

a3 b

�54

a2 b

2�

108a

b3

�81

b4

22.(

3 �

2z)4

16z4

�96

z3�

216z

2�

216z

�81

23.(

3m�

4n)3

27m

3�

108m

2 n�

144m

n2

�64

n3

24.(

5x�

2y)4

625x

4�

1000

x3 y

�60

0x2 y

2�

160x

y3

�16

y4

Fin

d t

he

ind

icat

ed t

erm

of

each

exp

ansi

on.

25.s

even

th t

erm

of

(a�

b)10

210a

4 b6

26.s

ixth

ter

m o

f (m

�n

)10

�25

2m5 n

5

27.n

inth

ter

m o

f (r

�s)

1430

03r6

s8

28.t

enth

ter

m o

f (2

x�

y)12

1760

x3 y

9

29.f

ourt

h t

erm

of

(x�

3y)6

�54

0x3 y

330

.fif

th t

erm

of

(2x

�1)

940

32x

5

31.G

EOM

ETRY

How

man

y li

ne

segm

ents

can

be

draw

n b

etw

een

ten

poi

nts

,no

thre

e of

wh

ich

are

col

lin

ear,

if y

ou u

se e

xact

ly t

wo

of t

he

ten

poi

nts

to

draw

eac

h s

egm

ent?

45

32.P

RO

BA

BIL

ITY

If y

ou t

oss

a co

in 4

tim

es,h

ow m

any

diff

eren

t se

quen

ces

of t

osse

s w

ill

give

exa

ctly

3 h

eads

an

d 1

tail

or

exac

tly

1 h

ead

and

3 ta

ils?

8

41!

� 3!38

!12

!� 6!

6!8!

� 5!3!

8!� 6!

2!

20!

� 18!

9! � 5!

Pra

ctic

e (

Ave

rag

e)

Th

e B

ino

mia

l Th

eore

m

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

11-7

11-7



Readin

g t

o L

earn

Math

em

ati

csT

he

Bin

om

ial T

heo

rem

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

11-7

11-7

©G

lenc

oe/M

cGra

w-H

ill67

1G

lenc

oe A

lgeb

ra 2

Lesson 11-7

Pre-

Act

ivit

yH

ow d

oes

a p

ower

of

a b

inom

ial

des

crib

e th

e n

um

ber

s of

boy

s an

dgi

rls

in a

fam

ily?

Rea

d th

e in

trod

ucti

on t

o L

esso

n 11

-7 a

t th

e to

p of

pag

e 61

2 in

you

r te

xtbo

ok.

•If

a f

amil

y h

as f

our

chil

dren

,lis

t th

e se

quen

ces

of b

irth

s of

gir

ls a

nd

boys

that

res

ult

in

th

ree

girl

s an

d on

e bo

y.B

GG

GG

BG

GG

GB

GG

GG

B•

Des

crib

e a

way

to

figu

re o

ut

how

man

y su

ch s

equ

ence

s th

ere

are

wit

hou

tli

stin

g th

em.

Sam

ple

an

swer

:Th

e b

oy c

ou

ld b

e th

e fi

rst,

seco

nd

,th

ird

,or

fou

rth

ch

ild,s

o t

her

e ar

e fo

ur

seq

uen

ces

wit

h t

hre

e g

irls

an

d o

ne

boy

.

Rea

din

g t

he

Less

on

1.C

onsi

der

the

expa

nsi

on o

f (w

�z)

5 .

a.H

ow m

any

term

s do

es t

his

exp

ansi

on h

ave?

6

b.

In t

he

seco

nd

term

of

the

expa

nsi

on,w

hat

is

the

expo

nen

t of

w?

4

Wh

at i

s th

e ex

pon

ent

of z

?1

Wh

at i

s th

e co

effi

cien

t of

th

e se

con

d te

rm?

5

c.In

th

e fo

urt

h t

erm

of

the

expa

nsi

on,w

hat

is

the

expo

nen

t of

w?

2

Wh

at i

s th

e ex

pon

ent

of z

?3

Wh

at i

s th

e co

effi

cien

t of

th

e fo

urt

h t

erm

?10

d.

Wh

at i

s th

e la

st t

erm

of

this

exp

ansi

on?

z5

2.a.

Sta

te t

he

defi

nit

ion

of

a fa

ctor

ial

in y

our

own

wor

ds.(

Do

not

use

mat

hem

atic

alsy

mbo

ls i

n y

our

defi

nit

ion

.)S

amp

le a

nsw

er:T

he

fact

ori

al o

f an

y p

osi

tive

inte

ger

is t

he

pro

du

ct o

f th

at in

teg

er a

nd

all

the

smal

ler

inte

ger

s d

ow

nto

on

e.T

he

fact

ori

al o

f ze

ro is

on

e.

b.

Wri

te o

ut

the

prod

uct

th

at y

ou w

ould

use

to

calc

ula

te 1

0!.(

Do

not

act

ual

ly c

alcu

late

the

prod

uct

.)10

�9

�8

�7

�6

�5

�4

�3

�2

�1

c.W

rite

an

expr

essi

on i

nvol

ving

fac

tori

als

that

cou

ld b

e us

ed t

o fi

nd t

he c

oeff

icie

nt o

f th

e

thir

d te

rm o

f th

e ex

pans

ion

of (

m�

n)6 .

(Do

not

actu

ally

cal

cula

te t

he c

oeff

icie

nt.)

Hel

pin

g Y

ou

Rem

emb

er

3.W

ith

out

usi

ng

Pas

cal’s

tri

angl

e or

fac

tori

als,

wh

at i

s an

eas

y w

ay t

o re

mem

ber

the

firs

ttw

o an

d la

st t

wo

coef

fici

ents

for

th

e te

rms

of t

he

bin

omia

l ex

pan

sion

of

(a�

b)n?

Sam

ple

an

swer

:Th

e fi

rst

and

last

co

effi

cien

ts a

re a

lway

s 1.

Th

e se

con

dan

d n

ext-

to-l

ast

coef

fici

ents

are

alw

ays

n,t

he

po

wer

to

wh

ich

th

eb

ino

mia

l is

bei

ng

rai

sed

.

6!� 4!

2!

©G

lenc

oe/M

cGra

w-H

ill67

2G

lenc

oe A

lgeb

ra 2

Pat

tern

s in

Pas

cal’s

Tri

ang

leYo

u h

ave

lear

ned

th

at t

he

coef

fici

ents

in

th

e ex

pan

sion

of

(x�

y)n

yiel

d a

nu

mbe

r py

ram

id c

alle

d P

asca

l’s t

rian

gle.

As

man

y ro

ws

can

be

adde

d to

th

e bo

ttom

of

the

pyra

mid

as

you

ple

ase.

Th

is a

ctiv

ity

expl

ores

som

e of

th

e in

tere

stin

g pr

oper

ties

of

this

fam

ous

nu

mbe

r py

ram

id.

1.P

ick

a ro

w o

f P

asca

l’s t

rian

gle.

a.W

hat

is

the

sum

of

all

the

nu

mbe

rs i

n a

ll t

he

row

s ab

ove

the

row

yo

u p

icke

d?S

ee s

tud

ents

’wo

rk.

b.

Wh

at i

s th

e su

m o

f al

l th

e n

um

bers

in

th

e ro

w y

ou p

icke

d?S

ee s

tud

ents

’wo

rk.

c.H

ow a

re y

our

answ

ers

for

part

s a

and

bre

late

d?T

he

answ

er f

or

Par

t b

is 1

mo

re t

han

th

e an

swer

fo

r P

art

a.

d.

Rep

eat

part

s a

thro

ugh

cfo

r at

lea

st t

hre

e m

ore

row

s of

Pas

cal’s

tr

ian

gle.

Wh

at g

ener

aliz

atio

n s

eem

s to

be

tru

e?It

ap

pea

rs t

hat

th

e su

m o

f th

e n

um

ber

s in

any

ro

w is

1 m

ore

th

an t

he

sum

of

the

nu

mb

ers

in a

ll o

f th

e ro

ws

abov

e it

.

e.S

ee if

you

can

pro

ve y

our

gene

rali

zati

on.

Su

m o

f nu

mb

ers

in r

ow n

�2n

�1;

20�

21�

22�

… �

2n�

2 ,w

hic

h,

by t

he

form

ula

fo

r th

e su

m o

f a

geo

met

ric

seri

es,i

s 2n

�1

�1.

2.P

ick

any

row

of

Pas

cal’s

tri

angl

e th

at c

omes

aft

er t

he

firs

t.

a.S

tart

ing

at t

he

left

en

d of

th

e ro

w,a

dd t

he

firs

t n

um

ber,

the

thir

d n

um

ber,

the

fift

h n

um

ber,

and

so o

n.S

tate

th

e su

m.

See

stu

den

ts’w

ork

.

b.

In t

he

sam

e ro

w,a

dd t

he

seco

nd

nu

mbe

r,th

e fo

urt

h n

um

ber,

and

so o

n.

Sta

te t

he

sum

.S

ee s

tud

ents

’wo

rk.

c.H

ow d

o th

e su

ms

in p

arts

aan

d b

com

pare

?T

he

sum

s ar

e eq

ual

.

d.

Rep

eat

part

s a

thro

ugh

cfo

r at

lea

st t

hre

e ot

her

row

s of

Pas

cal’s

tr

ian

gle.

Wh

at g

ener

aliz

atio

n s

eem

s to

be

tru

e?In

any

ro

w o

f P

asca

l’s t

rian

gle

aft

er t

he

firs

t,th

e su

m o

f th

e o

dd

n

um

ber

ed t

erm

s is

eq

ual

to

th

e su

m o

f th

e ev

en n

um

ber

ed t

erm

s.

Row

1Ro

w 2

Row

3Ro

w 4

Row

5Ro

w 6

Row

71

615

2015

61

15

1010

51

46

41

34

12

11

11

11

1

En

rich

men

t

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

11-7

11-7


An

swer

s


Stu

dy G

uid

e a

nd I

nte

rven

tion

Pro

of

and

Mat

hem

atic

al In

du

ctio

n

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

11-8

11-8

©G

lenc

oe/M

cGra

w-H

ill67

3G

lenc

oe A

lgeb

ra 2

Lesson 11-8

Mat

hem

atic

al In

du

ctio

nM

ath

emat

ical

in

duct

ion

is

a m

eth

od o

f pr

oof

use

d to

pro

vest

atem

ents

abo

ut

posi

tive

in

tege

rs.

Ste

p 1

Sho

w t

hat

the

stat

emen

t is

tru

e fo

r so

me

inte

ger

n.M

ath

emat

ical

Ste

p 2

Ass

ume

that

the

sta

tem

ent

is t

rue

for

som

e po

sitiv

e in

tege

r k

whe

re k

n.

In

du

ctio

n P

roo

fT

his

assu

mpt

ion

is c

alle

d th

e in

du

ctiv

e hy

po

thes

is.

Ste

p 3

Sho

w t

hat

the

stat

emen

t is

tru

e fo

r th

e ne

xt in

tege

r k

�1.

Pro

ve t

hat

5 �

11 �

17 �

… �

(6n

�1)

�3n

2�

2n.

Ste

p 1

Wh

en n

�1,

the

left

sid

e of

th

e gi

ven

equ

atio

n i

s 6(

1) �

1 �

5.T

he

righ

t si

de i

s3(

1)2

�2(

1) �

5.T

hu

s th

e eq

uat

ion

is

tru

e fo

r n

�1.

Ste

p 2

Ass

um

e th

at 5

�11

�17

�…

�(6

k�

1) �

3k2

�2k

for

som

e po

siti

ve i

nte

ger

k.

Ste

p 3

Sh

ow t

hat

th

e eq

uat

ion

is

tru

e fo

r n

�k

�1.

Fir

st,a

dd [

6(k

�1)

�1]

to

each

sid

e.5

�11

�17

�…

�(6

k�

1) �

[6(k

�1)

�1]

�3k

2�

2k�

[6(k

�1)

�1]

�3k

2�

2k�

6k�

5A

dd.

�3k

2�

6k�

3 �

2k�

2R

ewrit

e.

�3(

k2�

2k�

1) �

2(k

�1)

Fac

tor.

�3(

k�

1)2

�2(

k�

1)F

acto

r.

Th

e la

st e

xpre

ssio

n a

bove

is

the

righ

t si

de o

f th

e eq

uat

ion

to

be p

rove

d,w

her

e n

has

bee

nre

plac

ed b

y k

�1.

Th

us

the

equ

atio

n i

s tr

ue

for

n�

k�

1.T

his

pro

ves

that

5 �

11 �

17 �

… �

(6n

�1)

�3n

2�

2nfo

r al

l po

siti

ve i

nte

gers

n.

Pro

ve t

hat

eac

h s

tate

men

t is

tru

e fo

r al

l p

osit

ive

inte

gers

.

1.3

�7

�11

�…

�(4

n�

1) �

2n2

�n

.S

tep

1T

he

stat

emen

t is

tru

e fo

r n

�1

sin

ce 4

(1)

�1

�3

and

2(1

)2�

1 �

3.S

tep

2A

ssu

me

that

3 �

7 �

11 �

… �

(4k

�1)

�2k

2�

kfo

r so

me

po

siti

ve in

teg

er k

.S

tep

3A

dd

ing

th

e (k

�1)

st t

erm

to

eac

h s

ide

fro

m s

tep

2,w

e g

et

3 �

7 �

11 �

… �

(4k

�1)

�[4

(k�

1) �

1] �

2k2

�k

�[4

(k�

1) �

1].

Sim

plif

yin

g t

he

rig

ht

sid

e o

f th

e eq

uat

ion

giv

es 2

(k�

1)2

�(k

�1)

,wh

ich

isth

e st

atem

ent

to b

e p

rove

d.

2.50

0 �

100

�20

�…

�4

�54

� n

�62

5 �1 �

�.S

tep

1T

he

stat

emen

t is

tru

e fo

r n

�1,

sin

ce 4

�54

�1

�4

�53

�50

0 an

d

625�1

��

(625

) �

500.

Ste

p 2

Ass

um

e th

at 5

00 �

100

�20

�…

�4

�54

�k

�62

5 �1 �

�for

som

e p

osi

tive

inte

ger

k.

Ste

p 3

Ad

din

g t

he

(k�

1)st

ter

m t

o e

ach

sid

e fr

om

ste

p 2

an

d s

imp

lifyi

ng

g

ives

500

�10

0 �

20 �

… �

4 �

54 �

k�

4 �

53 �

k�

625�1

��

4 �

53 �

k�

625 �1

��,w

hic

h is

th

e st

atem

ent

to b

e p

rove

d.

1� 5k

�1

1 � 5k

1 � 5k

4 � 51 � 51

1 � 5n

Exam

ple

Exam

ple

Exer

cises

Exer

cises

©G

lenc

oe/M

cGra

w-H

ill67

4G

lenc

oe A

lgeb

ra 2

Co

un

tere

xam

ple

sT

o sh

ow t

hat

a f

orm

ula

or

oth

er g

ener

aliz

atio

n i

s n

ottr

ue,

fin

d a

cou

nte

rexa

mp

le.O

ften

th

is i

s do

ne

by s

ubs

titu

tin

g va

lues

for

a v

aria

ble.

Fin

d a

cou

nte

rexa

mp

le f

or t

he

form

ula

2n

2�

2n�

3 �

2n�

2�

1.C

hec

k th

e fi

rst

few

pos

itiv

e in

tege

rs.

nL

eft

Sid

e o

f F

orm

ula

Rig

ht

Sid

e o

f F

orm

ula

12(

1)2

�2(

1) �

3 �

2 �

2 �

3 or

721

�2

�1

�23

�1

or 7

true

22(

2)2

�2(

2) �

3 �

8 �

4 �

3 or

15

22 �

2�

1 �

24�

1 or

15

true

32(

3)2

�2(

3) �

3 �

18 �

6 �

3 or

27

23 �

2�

1 �

25�

1 or

31

fals

e

Th

e va

lue

n�

3 pr

ovid

es a

cou

nte

rexa

mpl

e fo

r th

e fo

rmu

la.

Fin

d a

cou

nte

rexa

mp

le f

or t

he

stat

emen

t x2

�4

is e

ith

er p

rim

e or

div

isib

le b

y 4.

nx

2�

4Tr

ue?

nx

2�

4Tr

ue?

11

�4

or 5

Prim

e6

36 �

4 or

40

Div

. by

4

24

�4

or 8

Div

. by

47

49 �

4 or

53

Prim

e

39

�4

or 1

3P

rime

864

�4

or 6

8D

iv.

by4

416

�4

or 2

0D

iv.

by4

981

�4

or 8

5N

eith

er

525

�4

or 2

9P

rime

Th

e va

lue

n�

9 pr

ovid

es a

cou

nte

rexa

mpl

e.

Fin

d a

cou

nte

rexa

mp

le f

or e

ach

sta

tem

ent.

Sam

ple

an

swer

s ar

e g

iven

.1.

1 �

5 �

9 �

… �

(4n

�3)

�4n

�3

n�

2

2.10

0 �

110

�12

0 �

… �

(10n

�90

) �

5n2

�95

n�

2

3.90

0 �

300

�10

0 �

… �

100(

33 �

n)

�90

0 �

n�

3

4.x2

�x

�1

is p

rim

e.n

�4

5.2n

�1

is a

pri

me

nu

mbe

r.n

�4

6.7n

�5

is a

pri

me

nu

mbe

r.n

�2

7.�

1 �

�…

��

n�

n�

3

8.5n

2�

1 is

div

isib

le b

y 3.

n�

3

9.n

2�

3n�

1 is

pri

me

for

n�

2.n

�9

10.4

n2

�1

is d

ivis

ible

by

eith

er 3

or

5.n

�6

1 � 2n � 2

3 � 21 � 2

2n� n

�1

Stu

dy G

uid

e a

nd I

nte

rven

tion

(c

onti

nued

)

Pro

of

by M

ath

emat

ical

Ind

uct

ion

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

11-8

11-8

Exam

ple1

Exam

ple1

Exam

ple2

Exam

ple2

Exer

cises

Exer

cises



Skil

ls P

ract

ice

Pro

of

and

Mat

hem

atic

al In

du

ctio

n

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

11-8

11-8

©G

lenc

oe/M

cGra

w-H

ill67

5G

lenc

oe A

lgeb

ra 2

Lesson 11-8

Pro

ve t

hat

eac

h s

tate

men

t is

tru

e fo

r al

l p

osit

ive

inte

gers

.

1.1

�3

�5

�…

�(2

n�

1) �

n2

Ste

p 1

:W

hen

n�

1,2n

�1

�2(

1) �

1 �

1 �

12.S

o,t

he

equ

atio

n is

tru

efo

r n

�1.

Ste

p 2

:A

ssu

me

that

1 �

3 �

5 �

… �

(2k

�1)

�k

2fo

r so

me

po

siti

vein

teg

er k

.S

tep

3:

Sh

ow

th

at t

he

giv

en e

qu

atio

n is

tru

e fo

r n

�k

�1.

1 �

3 �

5 �

… �

(2k

�1)

�[2

(k�

1) �

1]�

k2

�[2

(k�

1) �

1]

�k

2�

2k�

1 �

(k�

1)2

So

,1 �

3 �

5 �

… �

(2n

�1)

�n

2fo

r al

l po

siti

ve in

teg

ers

n.

2.2

�4

�6

�…

�2n

�n

2�

n

Ste

p 1

:W

hen

n�

1,2n

�2(

1) �

2 �

12�

1.S

o,t

he

equ

atio

n is

tru

e fo

r n

�1.

Ste

p 2

:A

ssu

me

that

2 �

4 �

6 �

… �

2k�

k2

�k

for

som

e p

osi

tive

inte

ger

k.

Ste

p 3

:S

ho

w t

hat

th

e g

iven

eq

uat

ion

is t

rue

for

n�

k�

1.2

�4

�6

�…

.�2k

�2(

k�

1) �

k2

�k

�2(

k�

1)�

(k2

�2k

�1)

�(k

�1)

�(k

�1)

2�

(k�

1)S

o,2

�4

�6

�…

�2n

�n

2�

nfo

r al

l po

siti

ve in

teg

ers

n.

3.6n

�1

is d

ivis

ible

by

5.

Ste

p 1

:W

hen

n�

1,6n

�1

�61

�1

�5.

So

,th

e st

atem

ent

is t

rue

for

n�

1.S

tep

2:

Ass

um

e th

at 6

k�

1 is

div

isib

le b

y 5

for

som

e p

osi

tive

inte

ger

k.

Th

en t

her

e is

a w

ho

le n

um

ber

rsu

ch t

hat

6k

�1

�5r

.S

tep

3:

Sh

ow

th

at t

he

stat

emen

t is

tru

e fo

r n

�k

�1.

6k�

1 �

5r6k

�5r

�1

6(6k

) �

6(5r

�1)

6k�

1�

30r

�6

6k�

1�

1 �

30r

�5

6k�

1�

1 �

5(6r

�1)

Sin

ce r

is a

wh

ole

nu

mb

er,6

r�

1 is

a w

ho

le n

um

ber

,an

d 6

k�

1�

1 is

div

isib

le b

y 5.

Th

e st

atem

ent

is t

rue

for

n�

k�

1.S

o,6

n�

1 is

div

isib

leby

5 f

or

all p

osi

tive

inte

ger

s n

.

Fin

d a

cou

nte

rexa

mp

le f

or e

ach

sta

tem

ent.

4.3n

�3n

is d

ivis

ible

by

6.5.

1 �

4 �

8 �

… �

2n�

Sam

ple

an

swer

:n

�2

Sam

ple

an

swer

:n

�3

n(n

�1)

(2n

�1)

�� 6

©G

lenc

oe/M

cGra

w-H

ill67

6G

lenc

oe A

lgeb

ra 2

Pro

ve t

hat

eac

h s

tate

men

t is

tru

e fo

r al

l p

osit

ive

inte

gers

.

1.1

�2

�4

�8

�…

�2n

�1

�2n

�1

Ste

p 1

:W

hen

n�

1,th

en 2

n�

1�

21 �

1�

20�

1 �

21�

1.S

o,t

he

equ

atio

n is

tru

e fo

r n

�1.

Ste

p 2

:A

ssu

me

that

1 �

2 �

4 �

8 �

… �

2k�

1�

2k�

1 fo

r so

me

po

siti

vein

teg

er k

.S

tep

3:

Sh

ow

th

at t

he

giv

en e

qu

atio

n is

tru

e fo

r n

�k

�1.

1 �

2 �

4 �

8 �

… �

2k�

1�

2(k

�1)

�1

�(2

k�

1) �

2(k

�1)

�1

�2k

�1

�2k

�2

�2k

�1

�2k

�1

�1

So

,1 �

2 �

4 �

8 �

… �

2n�

1�

2n�

1 fo

r al

l po

siti

ve in

teg

ers

n.

2.1

�4

�9

�…

�n

2�

Ste

p 1

:W

hen

n�

1,n

2�

12�

1 �

;tr

ue

for

n�

1.

Ste

p 2

:A

ssu

me

that

1 �

4 �

9 �

… �

k2

��k

(k�

1) 6(2k

�1)

�fo

r so

me

po

siti

vein

teg

er k

.S

tep

3:

Sh

ow

th

at t

he

giv

en e

qu

atio

n is

tru

e fo

r n

�k

�1.

1 �

4 �

9 �

… �

k2

�(k

�1)

2�

�(k

�1)

2

��

�

��

�

So

,1 �

4 �

9 �

… �

n2

�fo

r al

l po

siti

ve in

teg

ers

n.

3.18

n�

1 is

a m

ult

iple

of

17.

Ste

p 1

:W

hen

n�

1,18

n�

1 �

18 �

1 o

r 17

;tr

ue

for

n�

1.S

tep

2:

Ass

um

e th

at 1

8k�

1 is

div

isib

le b

y 17

fo

r so

me

po

sitiv

e in

teg

er k

.Th

ism

ean

s th

at t

her

e is

a w

ho

le n

um

ber

rsu

ch t

hat

18k

�1

�17

r.S

tep

3:

Sh

ow

th

at t

he

stat

emen

t is

tru

e fo

r n

�k

�1.

18k

�1

�17

r,so

18k

�17

r�

1,an

d 1

8(18

k)

�18

(17r

�1)

.Th

is is

equ

ival

ent

to 1

8k�

1�

306r

�18

,so

18k

�1

�1

�30

6r�

17,a

nd

18

k�

1�

1 �

17(1

8r�

1).

Sin

ce r

is a

wh

ole

nu

mb

er,1

8r�

1 is

a w

ho

le n

um

ber

,an

d 1

8k�

1�

1 is

div

isib

le b

y 17

.Th

e st

atem

ent

is t

rue

for

n�

k�

1.S

o,1

8n�

1 is

div

isib

le b

y17

fo

r al

l po

siti

ve in

teg

ers

n.

Fin

d a

cou

nte

rexa

mp

le f

or e

ach

sta

tem

ent.

4.1

�4

�7

�…

�(3

n�

2) �

n3

�n

2�

15.

5n�

2n�

3 is

div

isib

le b

y 3.

Sam

ple

an

swer

:n

�3

Sam

ple

an

swer

:n

�3

6.1

�3

�5

�…

�(2

n�

1) �

7.13

�23

�33

�…

�n

3�

n4

�n

3�

1

Sam

ple

an

swer

:n

�3

Sam

ple

an

swer

:n

�3

n2

�3n

�2

�� 2

n(n

�1)

(2n

�1)

�� 6

(k�

1)[(

k�

1) �

1][2

(k�

1) �

1]�

��

�6

(k�

1)[(

k�

2)(2

k�

3)]

��

�6

(k�

1)(2

k2�

7k�

6)�

��

6

(k�

1)[k

(2k

�1)

�6(

k�

1)]

��

��

66(

k�

1)2

�� 6

k(k

�1)

(2k

�1)

�� 6

k(k

�1)

(2k

�1)

�� 6

1(1

�1)

(2 �

1 �

1)�

��

6

n(n

�1)

(2n

�1)

�� 6

Pra

ctic

e (

Ave

rag

e)

Pro

of

and

Mat

hem

atic

al In

du

ctio

n

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

11-8

11-8


An

swer

s


Readin

g t

o L

earn

Math

em

ati

csP

roo

f an

d M

ath

emat

ical

Ind

uct

ion

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

11-8

11-8

©G

lenc

oe/M

cGra

w-H

ill67

7G

lenc

oe A

lgeb

ra 2

Lesson 11-8

Pre-

Act

ivit

yH

ow d

oes

the

con

cep

t of

a l

add

er h

elp

you

pro

ve s

tate

men

ts a

bou

tn

um

ber

s?

Rea

d th

e in

trod

ucti

on t

o L

esso

n 11

-8 a

t th

e to

p of

pag

e 61

8 in

you

r te

xtbo

ok.

Wh

at a

re t

wo

way

s in

wh

ich

a l

adde

r co

uld

be

con

stru

cted

so

that

you

cou

ldn

ot r

each

eve

ry s

tep

of t

he

ladd

er?

Sam

ple

an

swer

:1.

Th

e fi

rst

step

co

uld

be

too

far

off

th

eg

rou

nd

fo

r yo

u t

o c

limb

on

it.2

.Th

e st

eps

cou

ld b

e to

o f

arap

art

for

you

to

go

up

fro

m o

ne

step

to

th

e n

ext.

Rea

din

g t

he

Less

on

1.F

ill

in t

he

blan

ks t

o de

scri

be t

he

thre

e st

eps

in a

pro

of b

y m

ath

emat

ical

in

duct

ion

.

Ste

p 1

Sh

ow t

hat

th

e st

atem

ent

is

for

the

nu

mbe

r .

Ste

p 2

Ass

um

e th

at t

he

stat

emen

t is

fo

r so

me

posi

tive

k.

Th

is a

ssu

mpt

ion

is

call

ed t

he

.

Ste

p 3

Sh

ow t

hat

th

e st

atem

ent

is

for

the

nex

t in

tege

r .

2.S

upp

ose

that

you

wan

ted

to p

rove

th

at t

he

foll

owin

g st

atem

ent

is t

rue

for

all

posi

tive

inte

gers

.

3 �

6 �

9 �

… �

3n�

a.W

hic

h o

f th

e fo

llow

ing

stat

emen

ts s

how

s th

at t

he

stat

emen

t is

tru

e fo

r n

�1?

ii

i.3

��3

�2 2

�1

�ii

.3 �

iii.

3 �

b.

Wh

ich

of

the

foll

owin

g is

th

e st

atem

ent

for

n�

k�

1?iv

i.3

�6

�9

�…

�3k

�

ii.

3 �

6 �

9 �

… �

3k�

1�

iii.

3 �

6 �

9 �

… �

3k�

1�

3(k

�1)

(k�

2)

iv.

3 �

6 �

9 �

… �

3(k

�1)

�

Hel

pin

g Y

ou

Rem

emb

er

3.M

any

stud

ents

con

fuse

the

rol

es o

f n

and

kin

a p

roof

by

mat

hem

atic

al in

duct

ion.

Wha

t is

ago

od w

ay t

o re

mem

ber

the

diff

eren

ce in

the

way

s th

ese

vari

able

s ar

e us

ed in

suc

h a

proo

f?S

amp

le a

nsw

er:T

he

lett

er n

stan

ds

for

“nu

mb

er”

and

is u

sed

as

a va

riab

leto

rep

rese

nt

any

nat

ura

l nu

mb

er.T

he

lett

er k

is u

sed

to

rep

rese

nt

ap

arti

cula

r va

lue

of

n.

3(k

�1)

(k�

2)�

� 2

3k(k

�1)

�� 2

3k(k

�1)

�� 2

3 �

1 �

2�

� 23

�1

�2

�2

3n(n

�1)

�� 2

k�

1tr

ue

ind

uct

ive

hyp

oth

esis

inte

ger

tru

e

1tr

ue

©G

lenc

oe/M

cGra

w-H

ill67

8G

lenc

oe A

lgeb

ra 2

Pro

of

by In

du

ctio

nM

ath

emat

ical

in

duct

ion

is

a u

sefu

l to

ol w

hen

you

wan

t to

pro

ve t

hat

ast

atem

ent

is t

rue

for

all

nat

ura

l n

um

bers

.

Th

e th

ree

step

s in

usi

ng

indu

ctio

n a

re:

1.P

rove

th

at t

he

stat

emen

t is

tru

e fo

r n

�1.

2.P

rove

th

at i

f th

e st

atem

ent

is t

rue

for

the

nat

ura

l n

um

ber

n,i

t m

ust

als

obe

tru

e fo

r n

�1.

3.C

oncl

ude

th

at t

he

stat

emen

t is

tru

e fo

r al

l n

atu

ral

nu

mbe

rs.

Fol

low

th

e st

eps

to c

omp

lete

eac

h p

roof

.

Th

eore

m A

:Th

e su

m o

f th

e fi

rst

nod

d n

atu

ral

nu

mbe

rs i

s eq

ual

to

n2 .

1.S

how

th

at t

he

theo

rem

is

tru

e fo

r n

�1.

1 �

(1)2

2.S

upp

ose

1 �

3 �

5 �

… �

(2n

�1)

�n

2 .S

how

th

at

1 �

3 �

5 �

… �

(2n

�1)

�(2

n�

1) �

(n�

1)2 .

Ad

d 2

n�

1 to

eac

h s

ide

of

the

equ

atio

n w

ho

se t

ruth

was

ass

um

ed:

1 �

3 �

5 �

… �

(2n

�1)

�(2

n�

1) �

n2

�(2

n�

1) �

(n�

1)2

3.S

um

mar

ize

the

resu

lts

of p

robl

ems

1 an

d 2.

Th

e th

eore

m is

tru

e fo

r n

�1.

If t

he

sum

of

the

firs

t n

od

d n

um

ber

seq

ual

s n

2 ,th

en it

is t

rue

that

th

e su

m o

f th

e fi

rst

n�

1 o

dd

nu

mb

ers

equ

als

(n�

1)2 .

Th

eref

ore

,th

e th

eore

m is

tru

e fo

r al

l nat

ura

l nu

mb

ers.

Th

eore

m B

:Sh

ow t

hat

an

�bn

is e

xact

ly d

ivis

ible

by

a�

bfo

r n

equ

al t

o 1,

2,3,

and

all

nat

ura

l n

um

bers

.

4.S

how

th

at t

he

theo

rem

is

tru

e fo

r n

�1.

(a1

�b

1 )

(a�

b)

�1

5.T

he

expr

essi

on a

n �

1�

bn �

1ca

n b

e re

wri

tten

as

a(an

�bn

) �

bn(a

�b)

.V

erif

y th

at t

his

is

tru

e.a(

an�

bn)

�b

n(a

�b

) �

an �

1�

abn

�ab

n�

bn

�1

�an

�1

�b

n �

1

6.S

upp

ose

a�

bis

a f

acto

r of

an

�bn

.Use

th

e re

sult

in

pro

blem

5 t

o sh

ow

that

a�

bm

ust

th

en a

lso

be a

fac

tor

of a

n �

1�

bn �

1 .a

n �

1�

bn

�1

�a

(an

�b

n)

�b

n(a

�b

);a

�b

is a

fac

tor

of

bo

th

add

end

s o

n t

he

rig

ht

sid

e.S

o,a

�b

is a

lso

a f

acto

r o

f th

e le

ft s

ide.

7.S

um

mar

ize

the

resu

lts

of p

robl

ems

4 th

rou

gh 6

.T

he

theo

rem

is t

rue

for

n�

1.If

a�

bis

a f

acto

r o

f a

n�

bn,i

t is

als

o a

fact

or

of

an

�1

�b

n �

1 .S

o,t

he

theo

rem

is t

rue

for

all n

atu

ral n

um

ber

s n

.

En

rich

men

t

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

11-8

11-8


1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

B:

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11. B

B

D

A

C

B

C

D

B

B

A

�5

n�13(2)n�1

C

D

B

A

D

C

D

A

C

C

B

D

A

B

C

B

D

A

A

D

Chapter 11 Assessment Answer Key Form 1 Form 2APage 679 Page 680 Page 681

(continued on the next page)


12.

13.

14.

15.

16.

17.

18.

19.

20.

B:

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

B:�5

n�16(�3)n�1

B

C

A

D

C

A

B

B

C

C

B

A

B

D

A

D

A

A

C

B

�5

n�18(�2)n�1

C

D

B

D

C

D

A

B

C

Chapter 11 Assessment Answer KeyForm 2A (continued) Form 2BPage 682 Page 683 Page 684

An

swer

s


1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

21.

22.

23.

24.

25.

B: x � 5y; 4

450 ft

405u4v

6, 41, 1686

4, �3, 1, �2, �1

11, 13, 17, 23, 31

�383�

does not exist

75

2

1820

�1247

�

810, 270, 90, 30

an � 12��14

��n�1

�83

�, �196�

405

30

�175

1853

�3, 2, 7, 12

an � �9n � 26

66

3, �2, �7, �12

Chapter 11 Assessment Answer KeyForm 2CPage 685 Page 686

g 4 � 12g3 � 54g2

� 108g � 81

Sample answer: n � 3

See students’answers.


1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

21.

22.

23.

24.

25.

B: x � 3y; 5

750 ft

240k2m4

1, 4, �11

13, 5, �8, �13, �5

�1343�

20

does not exist

�5

762

�3694

�

14, 28, 56, 112

an � 27��13

��n�1

�1265�, �

13225

�

288

�15

�120

2230

�3, 0, 3, 6

an � �8n � 23

43

9, 5, 1, �3

Chapter 11 Assessment Answer KeyForm 2DPage 687 Page 688

An

swer

s

�4, �19, �93, �462,�2306

c5 � 15c4 � 90c3 �

270c2 � 405c � 243

See students’answers.

Sample answer:n � 3


1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

21.

22.

23.

24.

25.

B:

See students’ answer.

Sample Answer: n � 7

�35

8x4�

�34

�, �74

�, �349�

�141�, �

72

�, 23, 129, 783

�25

�, �125�, �

115�, �

215�, �

725�

�490910

�

�6, �65

�, ��265�, �

1625�

�43

�

��98

�

16

�301669

�

�13,650

�6.4, 3.2, �1.6

an � �6561��29

��n�1

0.01024

��29

�, �247�

80,712

149.6, 149.2, 148.8

�223�

�3101�, �

185�, �

170�

17

an � 1.4n � 7.9

��6170�

Chapter 11 Assessment Answer KeyForm 3Page 689 Page 690

a6 � �12

5a5� � �

125a4� � �

3225a3� �

�4182a5

2� � �

139122a5

� � �15

6,6425

�

x � �12

�, y � �2 or

x � �92

�, y � 6

Chapter 11 Assessment Answer KeyPage 691, Open-Ended Assessment

Scoring Rubric


Score General Description Specific Criteria

• Shows thorough understanding of the concepts ofarithmetic and geometric sequences and series, specialsequences and iteration of functions, binomial expansion,and proof by induction.

• Uses appropriate strategies to solve problems.• Computations are correct.• Written explanations are exemplary.• Goes beyond requirements of some or all problems.

• Shows an understanding of the concepts of arithmetic andgeometric sequences and series, special sequences anditeration of functions, binomial expansion, and proof byinduction.

• Uses appropriate strategies to solve problems.• Computations are mostly correct.• Written explanations are effective.• Satisfies all requirements of problems.

• Shows an understanding of most of the concepts ofarithmetic and geometric sequences and series, specialsequences and iteration of functions, binomial expansion,and proof by induction.

• May not use appropriate strategies to solve problems.• Computations are mostly correct.• Written explanations are satisfactory.• Satisfies the requirements of most of the problems.

• Final computation is correct.• No written explanations or work is shown to substantiate

the final computation.• Satisfies minimal requirements of some of the problems.

• Shows little or no understanding of most of arithmetic andgeometric sequences and series, special sequences anditeration of functions, binomial expansion, and proof byinduction.

• Does not use appropriate strategies to solve problems.• Computations are incorrect.• Written explanations are unsatisfactory.• Does not satisfy requirements of problems.• No answer may be given.

0 UnsatisfactoryAn incorrect solutionindicating no mathematicalunderstanding of theconcept or task, or nosolution is given

1 Nearly Unsatisfactory A correct solution with nosupporting evidence orexplanation

2 Nearly SatisfactoryA partially correctinterpretation and/orsolution to the problem

3 SatisfactoryA generally correct solution,but may contain minor flawsin reasoning or computation

4 SuperiorA correct solution that is supported by well-developed, accurateexplanations

An

swer

s


Chapter 11 Assessment Answer KeyPage 691, Open-Ended Assessment

Sample Answers

1a. Students should choose Section 11-3 on Geometric Sequences because the numbers ofbacteria at the end of each period form a list, or sequence, in which each term is amultiple of the previous term.

1b. an � a1 � rn�1 or, more specifically, an � 1000 � 2n�1

1c. By the end of the 5th day, there will have been 20 six-hour periods. Students shouldindicate that they would use the formula in part b with n � 20.

1d. (Table format may vary, but data will not vary.)

2a. Students should explain that the 12th row of Pascal’s triangle gives a numerical factorfor each of the twelve terms of the expansion; that each term contains a power of 2xbeginning with 11 then decreasing by 1 for each subsequent term; and that each termcontains a power of y beginning with 0 then increasing by 1 for each subsequent term.

2b. Students should demonstrate their knowledge of one of the two forms of the BinomialTheorem, stating the corresponding formula and explaining its use.

2c. Students should choose Pascal’s triangle or the Binomial Theorem, explaining thechoice and showing that the 8th term of the expansion is 5280x4y7.

3. Sample answer: (Format may vary. Column 2 and Answer Key entries will vary.)

Students should have at least one expression in Column 2 in sigma notation.

4. Students should indicate that both expressions have six terms and the same commonratio 3, but that the first expression represents a geometric series with a1 � 2, whilethe second expression represents twice a geometric series with a1 � 1. Students should

show that �6

n�12 � 3n�1 � 2 � �

6

n�11 � 3n�1 � 728, meaning that doubling each term of a

series before adding yields the same result as doubling the sum of the terms.

In addition to the scoring rubric found on page A31, the following sample answers may be used as guidance in evaluating open-ended assessment items.

At the end of Day 1

Number of bacteria 8000

Day 5

524,288,000

Day 2

128,000

Day 3

2,048,000

Day 4

32,768,000

Column 11. arithmetic sequence2. arithmetic series

3. geometric sequence

4. geometric series5. infinite geometric series6. binomial expansion7. counterexample

Column 2a. 3 � 7 � 11 � 15 � 19b. 5 � 10 � 20 � 40 � …

c. �5

n�15 � 2n�1

d. 3, 7, 11, 15, 19e. (x � y)3 � x 3 � 3x2y � 3xy2 � y 3

f. n � 2 for 3n � 1 is prime.g. 5, 10, 20, 40, 80

Answer Key1. d2. a

3. g

4. c5. b6. e7. f


1. false; common ratio

2. false; mathematicalinduction

3. false; series

4. true

5. false; infinitegeometric series

6. false; recursiveformula

7. true

8. false; iteration

9. false; term

10. true

11. Sample answer: The Fibonaccisequence is asequence ofnumbers in whichthe first two termsare each 1, andevery term after thatis found by addingthe two terms thatcome immediatelybefore it.

12. Sample answer: The missing term orterms between twononconsecutiveterms of ageometric sequenceare called geometricmeans.

1.

2.

3.

4.

5.6.

7.8.

9.

10.

Quiz (Lessons 11–3 and 11–4)

Page 693

1.

2.

3.

4.

5.

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

Quiz (Lessons 11–7 and 11–8)

Page 694

1.

2.

3.

4.

5.

6.

7.

8.

9.

10. See students’ answers.

Sample answer: n � 2

2268x3y 6

35a3

366

m3 � 18m2 � 108m � 216

1, �5, 43

�11, �51, �251

�2, 5, �9, 19, �37

3, 5, 9, 15, 23

�151�

�23

�

�158�

15

does not exist81

1875

3906

81, 27, 9, 3

3, �6, 12, �24, 48

C

205

�5

2, �1, �4�36

1281, 6, 11

an � �3n � 7

51

4, 11, 18, 25, 32

19, 22, 25, 28

Chapter 11 Assessment Answer KeyVocabulary Test/Review Quiz (Lessons 11–1 and 11–2) Quiz (Lessons 11–5 and 11–6)

Page 692 Page 693 Page 694

An

swer

s

x5 � 5x4y � 10x3y2 �

10x2y3 � 5xy 4 � y5

81r 4 � 216r 3 �216r2 � 96r � 16x 6 � 12x5y � 60x 4y2 �

160x3y3 � 240x2y 4 �

192xy5 � 64y 6


1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

1.

2.

3.4.

5.

6.

D: x � �4, R: y � 0

7.

8.

9.

10.

11.

12.

13.

14.

15.

270

155

7, 4, 1, �2, �5

about 7.4 days

�0.9345

86�11�

�57

�

y

xO

y

xO

12x2 � 7x � 12

(�1, 1, 3)

�74

�

�2, �5, �8

20, �10, 5, ��52

�, �54

�

3, �1, �5

728

an � 40��12

��n�1

18, 14, 10, 6

an � �2n � 7�27, 81

C

D

A

A

B

C

D

C

Chapter 11 Assessment Answer KeyMid-Chapter Test Cumulative ReviewPage 695 Page 696

y � 4(x � 3)2 � 1; parabola

asymptote: x � 8; hole: x � �1

x4 � 12x3y � 54x2y2 �

108xy3 � 81y 4


1.

2.

3.

4.

5.

6.

7.

8.

9.

10. 11.

12. 13.

14.

15.

16. DCBA

DCBA

DCBA

0 0 0

.. ./ /

.

99 9 987654321

87654321

87654321

87654321

1 2 4/

0 0 0

.. ./ /

.

99 9 987654321

87654321

87654321

87654321

2 6

0 0 0

.. ./ /

.

99 9 987654321

87654321

87654321

87654321

7

0 0 0

.. ./ /

.

99 9 987654321

87654321

87654321

87654321

1 2

DCBA

HGFE

DCBA

HGFE

DCBA

HGFE

DCBA

HGFE

DCBA

Chapter 11 Assessment Answer KeyStandardized Test Practice

Page 697 Page 698

An

swer

s

Documents

Chapter 11 Resource Masters - KTL MATH CLASSESktlmathclass.weebly.com/uploads/2/5/7/6/25760552/alg_2_resource_ws_ch_11.pdf©Glencoe/McGraw-Hill iv Glencoe Algebra 2 Teacher’s Guide