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Analysis Packet #2 – Sequences, Series Limits (1) Assignment 1 – Exponents, Factorials (2) Assignment 2 – Binomial Expansion (3) Assignment 3 – Recursion, Explicit Formulas (4) Assignment 4 – Method of Finite Differences (5) Assignment 5 – Review Worksheet (6) TEST (7) Assignment 6 – Arithmetic Progressions (8) Assignment 7 – Geometric Progressions (9) Assignment 8 – AP/GP Combined (10) Assignment 9 – Review Worksheet (11) TEST (12) Assignment 10 – Infinite Limits (13) Assignment 11 – Infinite Geometric Progressions (14) Assignment 12 – Review Worksheet (15) TEST

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Page 1: Analysis Packet #2 – Sequences, Series Limitsfaculty.pingry.org/bpoprik/documents/Packet2SeqSerLim.pdfAnalysis Packet #2 – Sequences, Series Limits Assignment #2 (1) Write the

Analysis Packet #2 – Sequences, Series Limits

(1) Assignment 1 – Exponents, Factorials

(2) Assignment 2 – Binomial Expansion

(3) Assignment 3 – Recursion, Explicit Formulas

(4) Assignment 4 – Method of Finite Differences

(5) Assignment 5 – Review Worksheet

(6) TEST

(7) Assignment 6 – Arithmetic Progressions

(8) Assignment 7 – Geometric Progressions

(9) Assignment 8 – AP/GP Combined

(10) Assignment 9 – Review Worksheet

(11) TEST

(12) Assignment 10 – Infinite Limits

(13) Assignment 11 – Infinite Geometric Progressions

(14) Assignment 12 – Review Worksheet

(15) TEST

Page 2: Analysis Packet #2 – Sequences, Series Limitsfaculty.pingry.org/bpoprik/documents/Packet2SeqSerLim.pdfAnalysis Packet #2 – Sequences, Series Limits Assignment #2 (1) Write the

Analysis Packet #2 – Sequences, Series Limits Combinations 1. Papa Domino’s Hut has four different toppings they can put on their pizzas. They have a strict, no-half toppings rule. If you want a topping, it has to be put on the entire pie. They also only have one size pizza and one crust style. Papa refuses to put more than three toppings on any pizza, claiming it gets too crowded for the pizza and your taste buds. You friend wants to try every possible pizza combination that Papa Domino’s Hut makes. How many different pizzas is he going to wind up ordering?

How many pies with no toppings are there? How many pies with one topping are there? How many pies with two toppings are there? How many pies with three toppings are there?

2. A Cluck U stores opens next to Papa Domino’s Hut and Papa decides to start offering their chicken as a new 5th topping. How many pies is your friend going to have to order now in order to try every combination?

Page 3: Analysis Packet #2 – Sequences, Series Limitsfaculty.pingry.org/bpoprik/documents/Packet2SeqSerLim.pdfAnalysis Packet #2 – Sequences, Series Limits Assignment #2 (1) Write the

Analysis Packet #2 – Sequences, Series Limits

Assignment #1

(1) Simplify each of the following please.

(a) ( ) ( ) 3

2 33 2

2x y

(c) x x + 1 x2 + 2 + 2

(b) ( ) ( )2

2 a

3

2x x

4x

− (d)

x 2 x 2 x 2

x 3

3 + 3 + 3

3

− − −

(2) Evaluate each of the following please.

(a) 8

6 (c)

8

5

(b) 9

5 4 (d)

9 9 +

3 4

(3) Simplify each of the following please.

(a) ( )( )n + 2 n + 1 (b) ( )( )n + 3

n + 1

(4) Solve for n please.

(a) n = 45

2

(b) ( )( )n + 1

= 110n 1 −

!

! !

! ! !

!

!

!

!

Page 4: Analysis Packet #2 – Sequences, Series Limitsfaculty.pingry.org/bpoprik/documents/Packet2SeqSerLim.pdfAnalysis Packet #2 – Sequences, Series Limits Assignment #2 (1) Write the

Analysis Packet #2 – Sequences, Series Limits

Answers

(1) (a) 18 18

64x y− (c) x + 22

(b) a + 1x (d) 9

(2) (a) 56 (c) 56

(b) 126 (d) 210

(3) (a) ( )n + 2 (b) ( )( )n + 3 n + 2

(4) (a) n = 10 (b) n = 10

!

Page 5: Analysis Packet #2 – Sequences, Series Limitsfaculty.pingry.org/bpoprik/documents/Packet2SeqSerLim.pdfAnalysis Packet #2 – Sequences, Series Limits Assignment #2 (1) Write the

Analysis Packet #2 – Sequences, Series Limits Binomial Theorem Expand:

1. 2(x 1)+ 2.

3(x 1)+ 3. 4(x 1)+

4. 4(x 2)+ 5.

3(2x 3)+

Page 6: Analysis Packet #2 – Sequences, Series Limitsfaculty.pingry.org/bpoprik/documents/Packet2SeqSerLim.pdfAnalysis Packet #2 – Sequences, Series Limits Assignment #2 (1) Write the

Analysis Packet #2 – Sequences, Series Limits

Assignment #2

(1) Write the first three terms in the binomial expansion of

72 2x +

x

(2) Find the indicated term on the binomial expansion of each of the following please.

(a) ( )73 th2x y ; 6 term− (c)

62

4 x + ; middle term

x 4

(b) ( )93 2 rdx + 3y ; 3 term (d) ( )n + 1 th

x + y ; n term

(3) Write the term in the expansion of ( )82x 2y− in which the exponent of y is 3.

(4) If the coefficients of the third and ninth terms of ( )nx + y are equal, find the middle term.

(5) Find the term(s) in the expansion of ( )11x + y having the coefficient 11

8

.

(6) Find the first four terms in the expansion of ( )51 + .2

Page 7: Analysis Packet #2 – Sequences, Series Limitsfaculty.pingry.org/bpoprik/documents/Packet2SeqSerLim.pdfAnalysis Packet #2 – Sequences, Series Limits Assignment #2 (1) Write the

Analysis Packet #2 – Sequences, Series Limits

Answers

(1) 14 11 8x + 14x + 84x

(2) (a) 6 5

84x y− (c) 3

20x

(b) 21 4

324x y (d) ( )n n + 1 2 n 1

2x y

(3) 10 3

448x y−

(4) 5 5

252x y

(5) 8 3 3 8

165x y or 165x y

(6) 1 + 1 + .4 + .08

Page 8: Analysis Packet #2 – Sequences, Series Limitsfaculty.pingry.org/bpoprik/documents/Packet2SeqSerLim.pdfAnalysis Packet #2 – Sequences, Series Limits Assignment #2 (1) Write the

Analysis Packet #2 – Sequences, Series Limits

Assignment #3

(1) Write the first six terms of each of the following sequences given each recursion formula.

(a) 1

n+1 n

a = 5

a = a 3− (c)

( )

11 2

n 1n 1 n

a =

a = 1 a+

+ − i

(b) 1

n+1 n

a = 2

a = 2a + 1

(2) Write the first five terms of each of the following sequences given each explicit formula.

(a) 2

na = 2n 3− (b) ( )n

n

1a =

n

(3) Write recursion and explicit formulas for each of the following sequences. (a) 1 , 5 , 25 , 125 , 625 , 3125 ...

(b) 11 , 14 , 17 , 20 , 23 , 26 ...

(c) 1 1 1 1 1 12 4 8 16 32 64 , , , , , ...

(d) 4 , 16 , 36 , 64 , 100 , 144 ...

(e) 1 1 1 1 11 2 1 2 3 1 2 3 4 1 2 3 4 5 1 2 3 4 5 6

1 , , , , , ... i i i i i i i i i i i i i i i

(f) 5 10 17 26 3722 2 2 2 2 2 , , , , , ...

(4) Evaluate each of the following series.

(a) ( )5

j = 2

2j + 1∑ (b) ( )3

j = 0

j ∑

!

Page 9: Analysis Packet #2 – Sequences, Series Limitsfaculty.pingry.org/bpoprik/documents/Packet2SeqSerLim.pdfAnalysis Packet #2 – Sequences, Series Limits Assignment #2 (1) Write the

Analysis Packet #2 – Sequences, Series Limits

Answers

(1) (a) 5 , 2 , 1, 4 , 7 , 10− − − − (c) 1 1 1 1 1 12 2 2 2 2 2 , , , , , − −

(b) 2 , 5 , 11 , 23 , 47 , 95

(2) (a) –1 , 5 , 15 , 29 , 47 (b) 1 1 1 12 3 4 5

1 , , , , − − −

(3) (a) recursion : 1

n + 1 n

a = 1

a = 5 a⋅

explicit: n 1

na = 5−

(b) recursion: 1

n + 1 n

a = 11

a = a + 3

explicit: na = 3n + 8

(c) recursion : 1

1 2

1n + 1 n2

a =

a = a

explicit: n n

1a =

2

(d) recursion:

( )1

2

n + 1 n

a = 4

a = a + 2

explicit: ( )2na = 2n

(e) recursion : 1

1n + 1 nn + 1

a = 1

a = a⋅

explicit: n1

a = n

(f) recursion:

( )

21 2

2

n

n + 1

a =

2 a 1 + 1 + 1a =

2

⋅ −

explicit: 2

nn + 1

a = 2

(4) (a) 32 (b) 10

!

Page 10: Analysis Packet #2 – Sequences, Series Limitsfaculty.pingry.org/bpoprik/documents/Packet2SeqSerLim.pdfAnalysis Packet #2 – Sequences, Series Limits Assignment #2 (1) Write the

Analysis Packet #2 – Sequences, Series Limits

Assignment #4

Use the method of finite differences to generate na for each of the following sequences.

(1) −3 , 18 , 53 , 102 , 165 , 242 , 333 , 438 , 557 , 690 ... (2) 6 , 11 , 20 , 33 , 50 , 71 , 96 , 125 , 158 , 195 …

(3) −7 , −16 , −35 , −64 , −103 , −152 , −211 , −280 , −359 , −448 …

(4) −2 , 12 , 62 , 166 , 342 , 608 , 982 , 1482 , 2126 , 2932 …

(5) 4 , 16 , 30 , 40 , 40 , 24 , −14 , −80 , −180 , −320 … (6) 0 , 13 , 76 , 249 , 616 , 1285 , 2388 , 4081 , 6544 , 9981 …

Page 11: Analysis Packet #2 – Sequences, Series Limitsfaculty.pingry.org/bpoprik/documents/Packet2SeqSerLim.pdfAnalysis Packet #2 – Sequences, Series Limits Assignment #2 (1) Write the

Analysis Packet #2 – Sequences, Series Limits Answers

(1) 10 7n a 2n −=

(2) 5 n 2n a 2n +−=

(3) 8 6n n5 a 2n −+−=

(4) 2 7n 3n a 3n +−=

(5) 2n 7n n a 23n −+−=

(6) 1 2n n a 4n +−=

Page 12: Analysis Packet #2 – Sequences, Series Limitsfaculty.pingry.org/bpoprik/documents/Packet2SeqSerLim.pdfAnalysis Packet #2 – Sequences, Series Limits Assignment #2 (1) Write the

Analysis Packet #2 – Sequences, Series Limits

Assignment #5 – Review Worksheet (1) Simplify each of the following please.

(a) ( ) ( ) 3 55 2 4 72x y x y− − (c)

x + 1 x 2 x + 12 2 2

++ +

(b) ( ) ( ) 3 2

2 3 4

8 15

2x y 3x y

4x y

− − (d)

( )( )x 1

x 1

+

(2) Evaluate each of the following numbers please.

(a)

7

10 (b)

3

20 (c) ( ) ( )( )∑

=

−−7

3 i

i 5 i21

(3) Find the indicated term in each binomial expansion please.

(a) ( )12y x + ; 4th (c)

8

3

3

x

1 x

− ; middle

(b) ( )73 4 x 3y− ; 2nd (d) ( )ny x + ; (n−1)st

(4) Write a recursion formula and explicit formula to generate the terms for each of the

following sequences.

(a) 2 , 5 , 8 , 11 , 14 , 17 … (b) 5 25 125 625 3125, , , , ...9 27 81 243 729

− −

(5) Write the first 5 terms for the sequence whose recursion formula is;

( )

1

nn 1 n

a 4

2n 1a 1 a

n 2+

=

+= −

+i i

(6) Use the method of finite differences to write an explicit formula for the following

sequence. 4 , 13 , 40 , 97 , 196 , 349 , 568 , 865 , 1252 , 1741 …

! !

Page 13: Analysis Packet #2 – Sequences, Series Limitsfaculty.pingry.org/bpoprik/documents/Packet2SeqSerLim.pdfAnalysis Packet #2 – Sequences, Series Limits Assignment #2 (1) Write the

Analysis Packet #2 – Sequences, Series Limits Answers

(1) (a) 4135yx8 (c) x 32

+

(b) 4

4

y

x18− (d) x(x + 1)

(2) (a) 120 (b) 1140 (c) −5

(3) (a) 39yx220 (c) 70

(b) 418yx21− (d) ( ) 2 n2yx2

1 nn −−

(4) (a) recursion formula (b) recursion formula

3 a a

2 a

n1n

1

+=

=

+

1

n 1 n

5 a

9

5a a

3+

=

= −

explicit formula explicit formula

1 3n a n −= ( )n

n 1n n 1

5a 1

3

++

= − i

(5) 212

4 , 4 , 5 , 7 , − −

(6) 3 2na 2n 3n 4n 1= − + +

Page 14: Analysis Packet #2 – Sequences, Series Limitsfaculty.pingry.org/bpoprik/documents/Packet2SeqSerLim.pdfAnalysis Packet #2 – Sequences, Series Limits Assignment #2 (1) Write the

Analysis Packet #2 – Sequences, Series Limits

Assignment #6

(1) Find the 51st term in the A.P. 4 , 9 , 14 , 19 , 24 , 29 ....

(2) Find the first term and common difference for the A.P. nt = 5n 7− .

(3) Fill in the blanks in each of the following A.P.’s (a) 8 , ____ , 14 , ____ , ____ (b) ____ , 9 , ____ , ____ , –1 , ____ (4) 413 is what term of the A.P. 5 , 13 , 21 , 29 , 37 ...? (5) In an A.P. 20 43t = 110 , t = 248. Find a and d.

(6) How many terms of the A.P. –10 , –7 , –4 , –1 , 2 ... must be added to give a sum of 200? (7) Find all values of x so that 2x – 5 , 5x + 4 , 9x – 1 will be consecutive terms of an A.P. (8) Find the required number in each A.P.

(a) 14d = 7 , t = 93 , a = ? (b) 17 17t = 43 , S = 323 , a = ?− −

(9) Find the sum of all multiples of 9 which are between 100 and 1000. (10) Evaluate each of the following sums please.

(a) ( )25

y = 1

5y + 3∑ (b) ( )506

j = 7

20 5j−∑

Page 15: Analysis Packet #2 – Sequences, Series Limitsfaculty.pingry.org/bpoprik/documents/Packet2SeqSerLim.pdfAnalysis Packet #2 – Sequences, Series Limits Assignment #2 (1) Write the

Analysis Packet #2 – Sequences, Series Limits

Answers

(1) 254 (2) a = –2 , d = 5 (3) (a) 8 , ____ , 14 , ____ , ____ (b) ____ , 9 , ____ , ____ , –1 , ____ (4) 52 (5) a = –4 , d = 6 (6) 16 (7) 14 (8) (a) a = 2 (b) a = 5 (9) 55,350 (10) (a) 1700 (b) –631,250

11 17 20 13

12 235 1

32 1

34−

Page 16: Analysis Packet #2 – Sequences, Series Limitsfaculty.pingry.org/bpoprik/documents/Packet2SeqSerLim.pdfAnalysis Packet #2 – Sequences, Series Limits Assignment #2 (1) Write the

Analysis Packet #2 – Sequences, Series Limits

Assignment #7 (1) Find the indicated term for each of the following G.P.’s.

(a) 16a = 3 , r = 1 , t = ? − (b) 10a = 5 , r = 3 , t = ?

(2) Find the 7th term for each of the following G.P.’s.

(a) 14 , 1 , 4 , 16 ...− − (b) 1 1 2 4

2 7 49 343 , , , ...

(3) Find the value of the specified variable for each of the following G.P.’s. (a) na = 2 , r = 2 , t = 64 , n = ? − − (b) 7t = 243 , r = 3 , a = ?

(4) Fill in the blanks in each of the following G.P.’s. (a) ____ , 3 , ____ , ____ , ____ , 27 (b) ____ , ____ , 5 , ____ , ____ , 625

(5) Evaluate each of the following sums.

(a) ( )6

z 1

z = 1

6 3−∑ (b) ( )

7

12

= 2

ϕ

ϕ

−∑

(6) Find all values of x if the terms 2x – 3 , 4x + 5 , 8x + 1 form a G.P.

(7) In a G.P. 3 7 10t = 15 and t = 240. Find t .

(8) Show that three consecutive integers can never form three consecutive terms of a G.P.

Page 17: Analysis Packet #2 – Sequences, Series Limitsfaculty.pingry.org/bpoprik/documents/Packet2SeqSerLim.pdfAnalysis Packet #2 – Sequences, Series Limits Assignment #2 (1) Write the

Analysis Packet #2 – Sequences, Series Limits

Answers

(1) (a) 16t = 3− (b) 10t = 98,415

(2) (a) 1024 (b) 32117,649

(3) (a) n = 6 (b) a = 13

(4) (a) ____ , 3 , ____ , ____ , _____ , 27

(b) ____ , ____ , 5 , ____ , ____ , 625

(5) (a) 2184 (b) 21128

(6) 1431

(7) 1920±

3± 3 3± 9 9 3±

15

1 25 125

Page 18: Analysis Packet #2 – Sequences, Series Limitsfaculty.pingry.org/bpoprik/documents/Packet2SeqSerLim.pdfAnalysis Packet #2 – Sequences, Series Limits Assignment #2 (1) Write the

Analysis Packet #2 – Sequences, Series Limits

Assignment #8

(1) The 12th term of an A.P. is 49. The 41st term is 165. Find t3 and S50. (2) Insert two numbers between 18 and –2 so that the first three numbers form a G.P. while the

last three numbers form an A.P. (3) In a set of four numbers, the first three are in a G.P., while the last three form an A.P. whose

difference is 12. Find the numbers if the first number is the same as the fourth number.

(4) Solve for x. ( ) 532 5 k3

x

2 k

=−∑=

(5) The sum of the first two terms of a G.P. is 18. The sum of the first three terms is 21. Find the

fourth term of the G.P. (6) The 5th term of an A.P. is 9. The 1st, 3rd, and 7th terms of the A.P. form the first three terms of a

G.P. Find the first term of the A.P. (7) Find three numbers in an A.P. whose sum is 12 and whose product is 55.

(8) Find all values of x for which x + 2 , 3x + 1 , 7x − 1 form; (a) an A.P. (b) a G.P.

(9) A student receives 5¢ on Nov. 1st , 10¢ on Nov 2nd, 20¢ on Nov 3rd and so on, each day

receiving twice as much as the previous day. How much money will this student have after receiving the final payment on Nov. 30th?

Page 19: Analysis Packet #2 – Sequences, Series Limitsfaculty.pingry.org/bpoprik/documents/Packet2SeqSerLim.pdfAnalysis Packet #2 – Sequences, Series Limits Assignment #2 (1) Write the

Analysis Packet #2 – Sequences, Series Limits

Answers

(1) t3 = 13 , S50 = 5,150

(2) 12

6 & 2 or 3 &

(3) 16 , −8 , 4 , 16

(4) 20

(5) t4 = −1 or t4 = 23

(6) 3 or 9

(7) 25 , 4 ,

211 or

211 , 4 ,

25

(8) (a) 21 (b)

21 or 3

(9) $53,687,091.15

Page 20: Analysis Packet #2 – Sequences, Series Limitsfaculty.pingry.org/bpoprik/documents/Packet2SeqSerLim.pdfAnalysis Packet #2 – Sequences, Series Limits Assignment #2 (1) Write the

Analysis Packet #2 – Sequences, Series Limits

Assignment #9 – Review Worksheet (1) Which term of the sequence 5 , 14 , 23 , 32 , 41 … is 239? (2) Find three numbers in an A.P. whose sum is 15 and whose product is 45. (3) Find the sum of all multiples of 7 that are between 60 and 600. (4) How long will it take to repay a debt of $880 if $25 is paid the first month, $27 the second

month, $29 the third month, continuing this pattern until the loan is repaid? (5) The first term of a G.P. is 375, and the fourth term is 192. Find the common ratio and the sum

of the first four terms. (6) Insert two numbers between 4 and 18 so that the first three numbers form an A.P. while the

last three numbers form a G.P. (7) The 4th term of an A.P. is 12. The 1st, 3rd, and 9th terms of the A.P. form the first three terms

of a G.P. Find the 5th term of the A.P. (8) Insert 5 arithmetic means between 8 and 26.

(9) The first term of a G.P. is 160 and the common ratio is 23 . How many terms must be added

so that the sum will be 2110 ?

(10) Find all values of x if the terms x − 4 , 3x + 2 , 9x + 5 form an (a) A.P. (b) G.P. (11) Evaluate each of the following sums.

(a) ( )∑=

−193

8 j

1 2j (b) ∑=

5

0 j

j

2

1 8 (c) ∑

=

305

6 j

11

Page 21: Analysis Packet #2 – Sequences, Series Limitsfaculty.pingry.org/bpoprik/documents/Packet2SeqSerLim.pdfAnalysis Packet #2 – Sequences, Series Limits Assignment #2 (1) Write the

Analysis Packet #2 – Sequences, Series Limits Answers

(1) n = 27

(2) 1 , 5 , 9 or 9 , 5 , 1

(3) 25,333

(4) 20 months

(5) 1107 S , r 454 ==

(6) 8 and 12 , or 21 and −3

(7) 12 or 15

(8) 11 , 14 , 17 , 20 , 23 (9) 5

(10) (a) 43 (b)

4324−

(11) (a) 37,200 (b) 463 (c) 3,300

Page 22: Analysis Packet #2 – Sequences, Series Limitsfaculty.pingry.org/bpoprik/documents/Packet2SeqSerLim.pdfAnalysis Packet #2 – Sequences, Series Limits Assignment #2 (1) Write the

Analysis Packet #2 – Sequences, Series Limits

Assignment #10

(1)

n

7n + 3

3n 2lim→ ∞

(11) ( )

( )( )3

2n

1 3n

5n + 7 4n + 3lim→ ∞

(2) 2

2n

2n + 5n 2

3n + 4n + 5lim→ ∞

− −

(12) ( )

( )

3

22n

5n 3

n + 7lim→ ∞

(3) 2

3n

9n 5n + 3

2n + 7lim→ ∞

(13) ( )nn

2n + 31

3n + 5lim→ ∞

i

(4) 4

3n

5n 2n

16n 9nlim→ ∞

+ −

(14) ( )n2

n

2n + 31

5n + 7lim→ ∞

i

(5) 4

4 2n

5n 7n

4n + 3n 2lim→ ∞

− −

(15) 2 2

n

2n + 1 2n + 3n

n 2 n + 1lim→ ∞

− −

(6)

n

2n 3 3n 2 +

5n + 4 2n + 1lim→ ∞

− −

(16) 2 3

n

3 5 4 6n +

n n nlim→ ∞

(7)

n

5n 1 n + 6

3n + 2 2n + 7lim→ ∞

− −

(17)

n

n

2

3lim→ ∞

(8) 3

2 2n

n + 1 2n + 3 +

n 5 n + 7lim→ ∞

(18)

n

n

4

3lim→ ∞

(9) 3

2 2n

n + 1 2n + 3

n 5 n + 7lim→ ∞

i (19) 10

n

n

nlim→ ∞

(10) ( )( )

2

2n

2n 76 +

3n + 5lim→ ∞

(20) n + 1

n 2n

2 5

2 9lim −→ ∞

+ −

!

Page 23: Analysis Packet #2 – Sequences, Series Limitsfaculty.pingry.org/bpoprik/documents/Packet2SeqSerLim.pdfAnalysis Packet #2 – Sequences, Series Limits Assignment #2 (1) Write the

Analysis Packet #2 – Sequences, Series Limits

Answers

(1) 73 (11) 27

20−

(2) 23

− (12) 0

(3) 0 (13) No Limit

(4) ∞ (14) 0

(5) 74

− (15) 3

(6) 1910 (16) 30−

(7) 76 (17) 0

(8) ∞ (18) ∞

(9) 2 (19) ∞

(10) 589 (20) 8

Page 24: Analysis Packet #2 – Sequences, Series Limitsfaculty.pingry.org/bpoprik/documents/Packet2SeqSerLim.pdfAnalysis Packet #2 – Sequences, Series Limits Assignment #2 (1) Write the

Analysis Packet #2 – Sequences, Series Limits

Assignment #11 (1) Find the sum of the infinite geometric series for each of the following.

(a) 13

a = 7 , r = (b) 36

a = 11 , r =

(2) Write the first three terms of an infinite G.P. if 9 152 2

a = and S = .

(3) Find the range of values for which each of the following infinite geometric series converges.

(a) ( ) ( ) ( )2 31 + 3 x + 1 + 9 x + 1 + 27 x + 1 + ...

(b) ( ) ( ) ( )3 6 91 1 18 64 512

1 + 2x + 3 + 2x + 3 + 2x + 3 + ...

(4) Find r for the infinite geometric series in which 4 + 3 22

S = and a = 1 + 2 .

(5) Find all values of x for which each of the following infinite geometric series has the given

sum.

(a) 2 32

3 = 1 + 3x + 9x + 27x + ... (b)

2 32x + 1 = 1 + x + x + x + ...

2x

(6) Find the sum of an infinite geometric series if the sum of the first two terms is 36, and the third term is 3.

(7) Evaluate each of the following sums please.

(a) ( )5

d13

d = 1

12∑ (c) ( )d + 112

d = 1

16

−∑

(b) ( )d13d = 1

12

∑ (d) ( )d 152

d = 1

12

∞ −

(8) Each side of a square measures 16. The midpoints of each side are joined to form an inscribed

square. The midpoints of each side of this second square are joined to form a third square. If this process is continued endlessly, find the sum of the areas and perimeters of all the squares, including the original one.

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Analysis Packet #2 – Sequences, Series Limits

Answers

(1) (a) 212

S = (b) S = 12 + 2 3

(2) 9 9 182 5 25 , ,

(3) (a) 4 23 3 < x < − − (b) 5 1

2 2 < x < − −

(4) 2 1−

(5) (a) 16

− (b) 12

(6) 81 1922 5

S = or S =

(7) (a) 48481 (c) 32

3

(b) 6 (d) No Sum

(8) Sum of areas is 512 , Sum of perimeters is 128 + 64 2

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Analysis Packet #2 – Sequences, Series Limits

Assignment #12

(1) Find each of the following limits please.

(a)

∞→ 2 + 5n

3 7nlim

n

(g)

∞→2

3

3n 5n

2 7n

n

5

n

1lim

(b)

∞→ 1 + n

8 + 5n

2nlim (h)

−−

−+

∞→ 3n

2 6n

5 2n

1 4n

22

nlim

(c)

+

∞→ 6 n +2n

1 3n + n

2

23

nlim (i)

+

− −

∞→ 2 2

2 2

n

1n1+n

nlim

(d) ( )

∞→3

23

n 3n 4

7n 5nlim (j)

−+

∞→

n 5n2

nlim

(e) 25

n

n

n lim→∞

(k)

+−+

∞→

109n 7n9n 22

nlim

(f) n

n4

3lim

∞→

(l)

−−+

∞→

2n 1 5n 8n3 23

nlim

(2) Find the first three terms of an infinite G.P. if the ratio is 31 and the sum of the series is 9.

!

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Analysis Packet #2 – Sequences, Series Limits (3) The sum of the first two terms of an infinite G.P. is 12. The third term is 2. Find the sum of

the infinite series if it exists.

(4) Find the ratio of the infinite G.P. whose sum is 223+ and whose first term is 21+ . (5) Find all values of x for which the sum of the following infinite geometric series exists.

(2x − 3) , 5(2x − 3)2 , 25(2x − 3)3 , 125(2x − 3)4 , .....

(6) Each side of an equilateral triangle is 16. The midpoints of each side are connected to form a second equilateral triangle. The midpoints of each side of this triangle are connected to form a third triangle. If this process is continued endlessly, find the sum of the areas of all the triangles.

(7) Evaluate each of the following sums.

(a)

k

k 0

263

=

∑ (c)

k

k 1

543

=

(b) ( )k 1

k 0

.2

∞−

=∑ (d) ( )

109

k 10

2k 3

=

+∑

Page 28: Analysis Packet #2 – Sequences, Series Limitsfaculty.pingry.org/bpoprik/documents/Packet2SeqSerLim.pdfAnalysis Packet #2 – Sequences, Series Limits Assignment #2 (1) Write the

Analysis Packet #2 – Sequences, Series Limits Answers

(1) (a) 5

7 (g) −7

(b) 0 (h) 5

(c) ∞ (i) 2

3

(d) 27

5− (j) 0

(e) 0 (k) 6

7

(f) 0 (l) 12

5

(2) 6 , 2 , 3

2

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Analysis Packet #2 – Sequences, Series Limits Answers

(3) 16 or 2

27

(4) 22 −

(5) 7 8 x 5 5< <

(6) 3

3256

(7) (a) 18 (c) ∞

(b) 4

25 (d) 12,200