Chapter 12 Sequences; Induction; the Binomial Theorem · 2017-12-11 · 102. When the last deposit is made at the …

1272

Copyright © 2017 Pearson Education, Inc.

Chapter 12 Sequences; Induction; the Binomial Theorem

Section 12.1

1. 2 1 12

2 2f

; 3 1 23

3 3f

2. True

3.

2 2

4

1

0.041000 1

2

1000 1.02

1082.43

n tr

A Pn

After two years, the account will contain $1082.43.

4.

12 1

1

0.0510,000 1

12

n tr

A Pn

P

120.05

10,000 112

10,000 1.051162

10,000

1.0511629513.28

P

P

P

P

To have $10,000 at the end of one year, you need to invest $9513.28 now.

5. sequence

6. True

7. ( 1) 3 2 1n n

8. b

9. summation

10. b

11. 10! 10 9 8 7 6 5 4 3 2 1 3,628,800

12. 9! 9 8 7 6 5 4 3 2 1 362,880

13. 9! 9 8 7 6!

9 8 7 5046! 6!

14. 12! 12 11 10!

12 11 13210! 10!

15. 3! 7! 3 2 1 7 6 5 4!

4! 4!3 2 1 7 6 5 1, 260

16. 5! 8! 5 4 3! 8!

3! 3!5 4 8 7 6 5 4 3 2 1

806,400

17. 1 2 3 4 51, 2, 3, 4, 5s s s s s

18. 2 2 21 2 3

2 24 5

1 1 2, 2 1 5, 3 1 10,

4 1 17, 5 1 26

s s s

s s

19. 1 2

3 4

1 1 2 2 1, ,

1 2 3 2 2 4 23 3 4 4 2

, ,3 2 5 4 2 6 3

a a

a a

55 5

5 2 7a

20. 1 2

3 4

2 1 1 3 2 2 1 5, ,

2 1 2 2 2 42 3 1 7 2 4 1 9

, ,2 3 6 2 4 8

b b

b b

52 5 1 11

2 5 10b

21. 1 1 2 2 1 21 2

3 1 2 4 1 23 4

( 1) (1 ) 1, ( 1) (2 ) 4,

( 1) (3 ) 9, ( 1) (4 ) 16,

c c

c c

5 1 25 ( 1) (5 ) 25c

Section 12.1: Sequences

1273


22. 1 11

2 12

3 13

1( 1) 1,

2 1 1

2 2( 1) ,

2 2 1 3

3 3( 1) ,

2 3 1 5

d

d

d

4 14

5 15

4 4( 1) ,

2 4 1 7

5 5( 1)

2 5 1 9

d

d

23. 1 2

1 21 2

3 4

3 43 4

5

5 5

2 2 1 2 4 2, ,

4 2 10 53 1 3 1

2 8 2 2 16 8, ,

28 7 82 413 1 3 1

2 32 8

244 613 1

s s

s s

s

24. 1 2

1 2

3 4

3 4

5

5

4 4 4 16, ,

3 3 3 9

4 64 4 256, ,

3 27 3 81

4 1024

3 243

s s

s s

s

25. 1

1( 1) 1 1

,(1 1)(1 2) 2 3 6

t

2

2

3

3

4

4

5

5

( 1) 1 1,

(2 1)(2 2) 3 4 12

( 1) 1 1,

(3 1)(3 2) 4 5 20

( 1) 1 1,

(4 1)(4 2) 5 6 30

( 1) 1 1

(5 1)(5 2) 6 7 42

t

t

t

t

26. 1 2 3

1 2 3

4 5

4 5

3 3 3 9 3 273, , 9,

1 1 2 2 3 3

3 81 3 243,

4 4 5 5

a a a

a a

27. 1 2 31 2 3

4 54 5

1 1 2 3, , ,

4 5,

b b bee e e

b be e

28. 2 2 2

1 2 31 2 3

2 2

4 54 5

1 1 2 3 9, 1, ,

2 82 2 2

4 16 5 251,

16 322 2

c c c

c c

29. Each term is a fraction with the numerator equal to the term number and the denominator equal to one more than the term number.

1nn

an

30. Each term is a fraction with the numerator equal to 1 and the denominator equal to the product of the term number and one more than the term number.

1

1nan n

31. Each term is a fraction with the numerator equal to 1 and the denominator equal to a power of 2. The power is equal to one less than the term number.

1

1

2n n

a

32. Each term is equal to a fraction with the numerator equal to a power of 2 and the denominator equal to a power of 3. Both powers are equal to the term number. Since the powers are the same, we can use rules for exponents to

write each term as a power of 23 .

2

3

n

na

33. The terms form an alternating sequence. Ignoring the sign, each term always contains a 1. The sign alternates by raising 1 to a power. Since the first term is positive, we use 1n as the power.

11

nna

Chapter 12: Sequences; Induction; the Binomial Theorem

1274


34. The terms appear to alternate between whole numbers and fractions. If we write the whole

numbers as fractions (e.g. 11 1 , 33 1 , etc.), we

see that each term consists of a 1 and the term number. When n is odd, the numerator is n and the denominator is 1. When n is even, the numerator is 1 and the denominator is n. This alternating behavior occurs if we have a power that alternates sign. The alternating sign is

obtained by using 11

n . Thus, we get

11

n

na n

35. The terms (ignoring the sign) are equal to the term number. The alternating sign is obtained by

using 11

n .

11

nna n

36. Here again we have alternating signs which will

be taken care of by using 11

n . The rest of the

term is twice the term number.

11 2

nna n

37. 1 2 3

4 5

2, 3 2 5, 3 5 8,

3 8 11, 3 11 14

a a a

a a

38. 1 2 3

4 5

3, 4 3 1, 4 1 3,

4 3 1, 4 1 3

a a a

a a

39. 1 2 3

4 5

2, 2 ( 2) 0, 3 0 3,

4 3 7, 5 7 12

a a a

a a

40. 1 2 3

4 5

1, 2 1 1, 3 1 2,

4 2 2, 5 2 3

a a a

a a

41. 1 2 3

4 5

5, 2 5 10, 2 10 20,

2 20 40, 2 40 80

a a a

a a

42. 1 2 3

4 5

2, 2, ( 2) 2,

2, ( 2) 2

a a a

a a

43. 1 2 3

4 5

33 123, , ,2 3 2

111 182 ,

4 8 5 40

a a a

a a

44. 1 2

3 4

2, 2 3( 2) 4,

3 3( 4) 9, 4 3( 9) 23,

a a

a a

5 5 3( 23) 64a

45. 1 2 3 4

5

1, 2, 2 1 2, 2 2 4,

4 2 8

a a a a

a

46. 1 2 3

4 5

1, 1, 1 3 1 2,

1 4 2 9, 2 5 9 47

a a a

a a

47. 1 2 3

4

5

, , ( ) 2 ,

( 2 ) 3 ,

( 3 ) 4

a A a A d a A d d A d

a A d d A d

a A d d A d

48.

2

1 2 3

2 3 3 44 5

, , ( ) ,

,

a A a rA a r rA r A

a r r A r A a r r A r A

49. 1 2 3

4

2, 2 2 , 2 2 2 ,

2 2 2 2 ,

a a a

a

5 2 2 2 2 2a

50. 1 2 3

22 22, , ,

2 2a a a

4 5

22 2

2 22 2,

2 2a a

51. 1( 2) 3 4 5 6 7 2

n

kk n

52. 1(2 1) 3 5 7 9 2 1

n

kk n

53. 2 2

1

1 9 25 492 8 18 32

2 2 2 22 2

n

k

k n

54. 2 2

11 4 9 16 25 36 1

n

kk n


1275


55. 0

1 1 1 1 11

3 9 273 3

n

k nk

56. 0

3 3 9 31

2 2 4 2

k nn

k

57. 1

10

1 1 1 1 1

3 9 273 3

n

k nk

58. 1

0(2 1) 1 3 5 7 2( 1) 1

1 3 5 7 (2 1)

n

kk n

n

59. 2( 1) ln ln 2 ln 3 ln 4 ( 1) ln

nk n

kk n

60. 1

3

4 3 5 4 6 5 1

( 1) 2

( 1) 2 ( 1) 2 ( 1) 2 ( 1) 2

nk k

k

n n

3 4 5 6 1

1

2 2 2 2 ( 1) 2

8 16 32 64 ... ( 1) 2

n n

n n

61. 20

11 2 3 20

kk

62. 8

3 3 3 3 3

11 2 3 8

kk

63. 13

1

1 2 3 13

2 3 4 13 1 1k

k

k

64. 12

11 3 5 7 2(12) 1 (2 1)

kk

65. 6

66

0

1 1 1 1 11 ( 1) ( 1)

3 9 27 3 3k

kk

66. 11 11

11 1 1

1

2 4 8

3 9 27

2 2

3 3( 1) ( 1)

kk

k

67. 2 3

1

3 3 3 33

2 3

n kn

kn k

68. 2 3

1

1 2 3 n

n kk

n k

e e e e e

69.

0

1

( ) ( 2 ) ( ) ( )

or ( 1 )

n

k

n

k

a a d a d a nd a kd

a k d

70. 2 1 1

1

nn k

ka a r a r a r a r

71. 40

1 40 times

5 5 5 5 5 40 5 200k

72. 50

1 50 times

8 8 8 8 8 50(8) 400k

73.

40

1

40 40 120 41 820

2kk

74. 24 24

1 1

24 24 1( ) 300

2k kk k

75.

20 20 20 20 20

1 1 1 1 1(5 3) (5 ) 3 5 3

20 20 15 3 20

2

1050 60 1110

k k k k kk k k

76.

26 26 26 26 26

1 1 1 1 13 7 3 7 3 7

26 26 13 7 26

2

1053 182 871

k k k k kk k k

77.

16 16 162 2

1 1 14 4

16 16 1 2 16 14 16

61496 64 1560

k k kk k

78.

14 142 2 2

0 1

14 142

1 1

4 0 4 4

4

14 14 1 2 14 14 4 14

64 1015 64 955

k k

k k

k k

k


1276


79.

60 60 60 9

10 10 1 12 2 2 2

60 60 1 9 9 12

2 2

2 1830 45 3570

k k k kk k k k

80.

40 40 40 7

8 8 1 13 3 3

40 40 1 7 7 13

2 2

3 820 28 2376

k k k kk k k k

81.

20 20 43 3 3

5 1 1

2 2

2 2

20 20 1 4 4 1

2 2

210 10 44,000

k k kk k k

82.

24 24 33 3 3

4 1 1

2 2

2 2

24 24 1 3 3 1

2 2

300 6 89,964

k k kk k k

83. a. 1

2

1.03(2000) 20 2080;

1.03(2080) 20 2162.4

p

p

There are approximately 2162 trout in the pond at the end of the second month.

b. Scrolling down the table, we find the trout population exceeds 5000 at the end of the 26th month when the population is 5084.

84. a. 1

2

0.9(250) 15 240;

0.9(240) 15 231

p

p

There are 240 tons of pollutants at the end of the first year, and 231 tons of pollutants at the end of the second year.

b. Scrolling down the table, we display the pollutant levels for the next 20 years.

c. The equilibrium level of pollution occurs when 0.9 15x x . That is, when 150x tons.

0.9 15

0.1 15

150

x x

x

x

85. a.

0 1

0.050.0125

41500, (1.0125) 750n n

r

A A A

b. Scrolling down the table, we find the account exceeds $150,000 at the end of 99 quarters.


1277


c. Scrolling down the table, we find the account value in 30 years (120 months) is $213,073.

86. a.

0 1

0.0150.00125

124000, (1.00125) 75n n

r

A A A

b. Scrolling down the table, we find the account exceeds $10,000 at the end of 72 months.

c. Scrolling down the table, we find the account value in 16 years (192 months) is $21,348.

87. a. 1 1.01(3000) 100 $2930B

b. Put the graphing utility in SEQuence mode. Enter Y= as follows, then examine the TABLE:


1278


From the table we see that the balance is below $2000 after 14 payments have been made. The balance then is $1953.70.

c. Scrolling down the table, we find that balance is paid off in the 36th month. The last payment is $83.78. There are 35 payments of $100 and the last payment of $83.78. The total amount paid is: 35(100) + 83.78(1.01) = $3584.62. (we have to add the interest for the last month).

d. The interest expense is: 3584.62 – 3000.00 = $584.62

88. a. 1 1.005(18500) 534.47 $18,058.03B

b. Put the graphing utility in SEQuence mode. Enter Y= as follows, then examine the TABLE:

From the table we see that the balance is below $10,000 after 19 payments have been made. The balance then is $9713.76.

c. Scrolling down the table, we find that balance is paid off in the 39th month. The last payment is $54.18. There are 38 payments of $534.47 and the last payment of $54.18 plus interest. The total amount paid is: 38(534.47) + 54.18(1.005) =

$20,364.31.

d. The interest expense is: 20,364.31 – 18,500.00 = $1864.31

89. a. Since the interest rate is 6% per annum compounded monthly, this is equivalent to a rate of 0.5% each month. Defining a recursive sequence, we have:

0 1150,000, 1.005 899.33n na a a

b. 1.005(150,000) 899.33 $149,850.67

c. Enter the recursive formula in Y= and create the table:

d. Scroll through the table:

After 58 payments have been made, the balance is below $140,000. The balance is about $139,981.

e. Scroll through the table:

The loan will be paid off at the end of 360 months or 30 years.


1279


Total amount paid = (359)($899.33) + $890.65(1.005) = $323,754.57.

f. The total interest expense is the difference of the total of the payments and the original loan: 323,754.57 150,000 $173,754.57

g. (a) Since the interest rate is 6% per annum compounded monthly, this is equivalent to a rate of 0.5% each month. Defining a recursive sequence, we have:

0 1150,000, 1.005 999.33n na a a

(b) 1.005(150,000) 999.33 $149,750.67

(c) Enter the recursive formula in Y= and create the table:

(d) Scroll through the table:

After 37 payments have been made, the balance is below $140,000. The balance is $139,894.

(e) Scroll through the table:

The loan will be paid off at the end of 279 months or 23 years and 3 months. Total amount paid = (278)($999.33) + 353.69(1.005) = $278,169.20

(f) The total interest expense is the difference of the total of the payments

and the original loan: 278,169.20 150,000 $128,169.20

h. Yes, if they can afford the additional monthly payment. They would save $44,586.07 in interest payments by paying the loan off sooner.

90. a. Since the interest rate is 6.5% per annum compounded at a rate of (6.5/12)% each month. Defining a recursive sequence, we have:

0

1

120,000,

0.0651 758.48

12n n

a

a a

b. 0.065

1 (120,000) 758.4812

$119,891.52

c. Enter the recursive formula in Y= and create the table:

d. Scroll through the table:

After 129 payments have been made, the balance is below $100,000. The balance is about $99,824.

e. Scroll through the table:

The loan will be paid in the 360th month i.e. after 30 years.


1280


Total amount paid = (359)(758.48) + 756.19(1+0.065/12) = $273,054.60.

f. The total interest expense is the difference of the total of the payments and the original loan: 273,054.60 120,000 $153,054.60

g. (a) Since the interest rate is 6.5% per annum compounded monthly, this is equivalent to a rate of (6.5/12)% each month. Defining a recursive sequence, we have:

0

1

120,000,

0.0651 858.48

12n n

a

a a

(b) 0.065

1 (120, 000) 858.4812

$119,791.52

(c) Enter the recursive formula in Y= and create the table:

(d) Scroll through the table:

After 78 payments have been made, the balance is below $100,000. The balance is $99,831.

(e) Scroll through the table:

The loan will be paid off at the end of 262 months or 21 years and 10 months.

Total amount paid = (261)(858.48) + (851.23)(1+0.065/12) = $224,919.12.

(f) The total interest expense is the difference of the total of the payments and the original loan: 224,919.12 120,000 $104,919.12

h. Yes, if they can afford the additional monthly payment. They would save $48,238.49 in interest payments by paying the loan off sooner.

91. 1 2 3 4 5

6 7 8 1 2

1, 1, 2, 3, 5,

8, 13, 21, n n n

a a a a a

a a a a a a

8 7 6 13 8 21a a a

After 7 months there are 21 mature pairs of rabbits.

92. a. 1 1

1 1

1 5 1 5

2 5

1 5 1 5 2 51

2 5 2 5

u

2 2

2 2

1 5 1 5

2 5

1 2 5 5 1 2 5 5 4 51

4 5 4 5

u


1281


b.

1 1

1

1

1 5 1 5 1 5 1 5

2 5 2 5

n n n n

n n

nnu u

1 1

1

1

1

2 2

2 2

1

2 2

1

2 2

1

2 2

2

1 5 1 5 2 1 5 2 1 5

2 5

1 5 1 5 2 1 5 1 5 2

2 5

1 5 3 5 1 5 3 5

2 5

3 5 3 51 5 1 5

1 5 1 5

2 5

3 5 3 51 5 1 5

6 2 5 6 2 5

2 51 1

1 5 1 52 2

2 5

1 5 1 5

2

n n n n

n

n n

n

n n

n

n n

n

n n

n

n n

n

n n

n

2

5

nu

c. Since 1 2 2 11, 1, ,n n n nu u u u u u

is the Fibonacci sequence.

93. 1, 1, 2, 3, 5, 8, 13 This is the Fibonacci sequence.

94. a. 1 2 3 4 5

6 7 8 9 10

11

1, 1, 2, 3, 5,

8, 13, 21, 34, 55,

89

u u u u u

u u u u u

u

b. 32 4

1 2 3

5 6

4 5

1 2 31, 2, 1.5,

1 1 2

5 81.67, 1.6,

3 5

uu u

u u u

u u

u u

7 8

6 7

9 10

8 9

11

10

13 211.625, 1.615,

8 13

34 551.619, 1.618,

21 34

891.618

55

u u

u u

u u

u u

u

u

c. 1.618 1 5

The exact value is 2

d. 31 2

2 3 4

1 1 21, 0.5, 0.667,

1 2 3

uu u

u u u

54

5 6

6 7

7 8

8 9

9 10

10

11

3 50.6, 0.625,

5 8

8 130.615, 0.619,

13 21

21 340.618, 0.618,

34 55

550.618

89

uu

u u

u u

u u

u u

u u

u

u

e. 0.618 2

The exact value is 1 5

95. a. 0 1 4

41.3 1.3

!0

1.3 1.3 1.3

0! 1! 4!

1.3

...

3.630170833

k

kk

f e

b. 7

1.3 1.3!

0

0 1 7

1.3

1.3 1.3 1.3...

0! 1! 7!3.669060828

k

kk

f e

c. 1.31.3 3.669296668f e

d. It will take 12n to approximate

1.31.3f e correct to 8 decimal places.


1282


96. a.

0 1 2 3

2.43 2.4

!0

2.4 2.4 2.4 2.4

0! 1! 2! 3!

2.4

0.824

k

kk

f e

b.

0 1 6

2.46 2.4

!0

2.4 2.4 2.40! 1! 6!

2.4

0.1602688

...

k

kk

f e

c. 2.42.4 0.0907179533f e

d. It will take 17n to approximate

2.42.4f e correct to 8 decimal places.

97. a. 1 0.4a , 2 2

2 0.4 0.3 2 0.4 0.3 0.7a

3 23 0.4 0.3 2 0.4 0.3 2 1.0a

4 24 0.4 0.3 2 0.4 0.3 4 1.6a

5 25 0.4 0.3 2 0.4 0.3 8 2.8a

6 26 0.4 0.3 2 0.4 0.3 16 5.2a

7 27 0.4 0.3 2 0.4 0.3 32 10.0a

8 28 0.4 0.3 2 0.4 0.3 64 19.6a

The first eight terms of the sequence are 0.4, 0.7, 1.0, 1.6, 2.8, 5.2, 10.0, and 19.6.

b. Except for term 5, which has no match, Bode’s formula provides excellent approximations for the mean distances of the planets from the Sun.

c. The mean distance of Ceres from the Sun is approximated by 5 2.8a , and that of

Uranus is 8 19.6a .

d. 9 29 0.4 0.3 2 0.4 0.3 128 38.8a

10 210 0.4 0.3 2 0.4 0.3 256 77.2a

e. Pluto’s distance is approximated by 9a , but

no term approximates Neptune’s mean distance from the Sun.

f. 11 211 0.4 0.3 2 0.4 0.3 512 154a

According to Bode’s Law, the mean orbital distance of 2003 UB 313 will be 154 AU from the Sun.

98. To show that 11 2 3 ... 1

2

n nn n

Let

1 2 3 ... 1 , we can reverse the order to get

1 2 +...+ 2 + 1, now add these two lines to get

2 1 2 1 3 2 ...... 1 2 1

S n n

S n n n

S n n n n n

So we have 2 1 1 1 .... 1 1 1S n n n n n n n

2 1

1

2

S n n

n nS


1283


99. 5 We begin with an initial guess of 0 2a .

1

2

3

1 52 2.25

2 2

1 52.25 2.236111111

2 2.25

1 52.236111111

2 2.2361111112.236067978

a

a

a

4

5

1 52.236067978

2 2.2360679782.236067977

1 52.236067977

2 2.2360679772.236067977

a

a

For both 5a and the calculator approximation,

we obtain 5 2.236067977 .


1 00

2 11

3 22

4 33

1 8 1 83 2.833333333

2 2 3

1 8

2

1 82.833333333

2 2.2.8333333332.828431373

1 8

2

1 82.828431373

2 2.8284313732.828427125

1 8

2

12.828427125

2

a aa

a aa

a aa

a aa

5 44

8

2.8284271252.828427125

1 8

2

1 82.828427125

2 2.8284271252.828427125

a aa


we obtain 8 2.828427125 .


1284



1

2

3

1 215 4.625

2 5

1 214.625 4.58277027

2 4.625

1 214.58277027

2 4.582770274.582575699

a

a

a

4

1 214.582575699

2 4.5825756994.582575695

a

5

1 214.582575695

2 4.5825756954.582575695

a


we obtain 21 4.582575695 .


1

2

3

4

5

1 895 9.444444444

2 5

1 899.444444444

2 9.4444444449.433986928

1 899.433986928

2 9.4339869289.433981132

1 899.433981132

2 9.4339811329.433981132

1 899.433981132

2 9.4339811

a

a

a

a

a

32

9.433981132


we obtain 89 9.433981132 .

103. 1 1u and 1 ( 1)n nu u n : So

1

2 1

3 2

4 3

5 4

6 5

7 6

1

(1 1) 1 2 3

(2 1) 3 3 6

(3 1) 6 4 10

(4 1) 10 5 15

(5 1) 15 6 21

(6 1) 21 7 28

u

u u

u u

u u

u u

u u

u u

104. Note that: 1 1 2 3 ( 2) ( 1) ( 1)nu n n n n

and rewriting 1 ( 1) ( 1) ( 2) 3 2 1nu n n n n .

So adding these together we have

1

1

1

1 2 3 ( 1) ( 1)

( 1) ( 1) 3 2 1

2( ) ( 2) ( 2) ( 2) ( 2) ( 2) ( 2)

n

n

n

u n n n

u n n n

u n n n n n n

1

1

2( ) ( 1)( 2)

( 1)( 2)

2

n

n

u n n

n nu

105. We know from number 97 and 98 that:

1( 1)( 2) ( )( 1)

and 2 2n n

n n n nu u

Adding these together we get

1

2 2

2

22 2

( 1)( 2) ( )( 1)+

2 2( 1)( 2) ( )( 1)

2

3 2

2

2 4 2

2

2( 2 1)2 1 ( 1)

2

n nn n n n

u u

n n n n

n n n n

n n

n nn n n

106 - 107. Answers will vary.

Section 12.2: Arithmetic Sequences

1285


108.

12(2)

1

0.032500 1

12

$2654.39

ntr

A Pn

109. 2 2 2 2( 1) ( 1) 2r x y

1tan 1

1225º

yx

The polar form of 1z i is

cos sin 2 cos 225º sin 225ºz r i i .

110. (2)(1) ( 1)(2)

2 ( 2)

0

v w

111. The vertex is (–3, 4) and the focus is (1, 4). Both lie on the horizontal line 4y . 3 1 4a

and since (1, 4) is to the right of (–3, 4), the parabola opens to the right. The equation of the parabola is:

2

2

2

4

4 4 4 ( 3)

4 16 3

y k a x h

y x

y x

Section 12.2

1. arithmetic

2. False; the sum of the first and last terms equals twice the sum of all the terms divided by the number of terms.

3. 1 1

6

12 (5 1)5 8

So

8 (6 1)5 17

a a

a

4. True

5. d

6. c

7. 1

( 4) ( 1 4) ( 4) ( 3)

4 3 1

n nd s s

n n n n

n n

The difference between consecutive terms is constant, therefore the sequence is arithmetic.

1 2 3

4

1 4 5, 2 4 6, 3 4 7,

4 4 8

s s s

s

8.

1

( 5) ( 1 5) 5 6

5 6 1

n nd s s

n n n n

n n


1 2 3

4

1 5 4, 2 5 3, 3 5 2,

4 5 1

s s s

s

9.

1

2 5 (2( 1) 5)

2 5 2 2 5

2 5 2 7 2

n nd a a

n n

n n

n n


1 2

3 4

2 1 5 3, 2 2 5 1,

2 3 5 1, 2 4 5 3

a a

a a

10.

1

3 1 (3( 1) 1)

3 1 3 3 1

3 1 3 2 3

n nd b b

n n

n n

n n


1 2

3 4

3 1 1 4, 3 2 1 7,

3 3 1 10, 3 4 1 13

b b

b b

11.

1

6 2 (6 2( 1))

6 2 6 2 2

6 2 6 2 2 2

n nd c c

n n

n n

n n


1 2

3 4

6 2 1 4, 6 2 2 2,

6 2 3 0, 6 2 4 2

c c

c c


1286


12.

1

4 2 (4 2( 1))

4 2 4 2 2

4 2 4 2 2 2

n nd d d

n n

n n

n n


1 2

3 4

4 2 1 2, 4 2 2 0,

4 2 3 2, 4 2 4 4

d d

d d

13. 1

1 1 1 1( 1)

2 3 2 3

1 1 1 1 1

2 3 2 3 3

1 1 1 1 1 1

2 3 2 3 3 3

n nd t t

n n

n n

n n


1 2

3 4

1 1 1 1 1 11 , 2 ,

2 3 6 2 3 61 1 1 1 1 5

3 , 42 3 2 2 3 6

t t

t t

14. 1

2 1 2 1( 1)

3 4 3 4

2 1 2 1 1

3 4 3 4 4

2 1 2 1 1 1

3 4 3 4 4 4

n nd t t

n n

n n

n n


1 2

3 4

2 1 11 2 1 71 , 2 ,

3 4 12 3 4 62 1 17 2 1 5

3 , 43 4 12 3 4 3

t t

t t

15.

1

1ln 3 ln 3

ln 3 1 ln 3

ln 3 ( 1 ) ln 3 1

ln 3

n n

n n

d s s

n n

n n n n

The difference between consecutive terms is

constant, therefore the sequence is arithmetic.

1 21 2

3 43 4

ln 3 ln 3 , ln 3 2ln 3 ,

ln 3 3ln 3 , ln 3 4ln 3

s s

s s

16. ln ln( 1)1 1 1n n

n nd s s e e n n


ln1 ln 2 ln31 2 3

ln 44

1, 2, 3,

4

s e s e s e

s e

17. 1 ( 1)

2 ( 1)3

2 3 3

3 1

na a n d

n

n

n

51 3 51 1 152a

18. 1 ( 1)

2 ( 1)4

2 4 4

4 6

na a n d

n

n

n

51 4 51 6 198a

19. 1 ( 1)

5 ( 1)( 3)

5 3 3

8 3

na a n d

n

n

n

51 8 3 51 145a

20. 1 ( 1)

6 ( 1)( 2)

6 2 2

8 2

na a n d

n

n

n

51 8 2 51 94a

21.

1 ( 1)

10 ( 1)

21 1

2 21

12

na a n d

n

n

n

511

51 1 252

a


1287


22. 1 ( 1)

11 ( 1)

3

1 11

3 34 1

3 3

na a n d

n

n

n

514 1 4 51 47

513 3 3 3 3

a

23. 1 ( 1)

2 ( 1) 2

2 2 2 2

na a n d

n

n n

51 51 2a

24. 1 ( 1) 0 ( 1) 1na a n d n n

51 51 50a

25. 1 12, 2, ( 1)na d a a n d

100 2 (100 1)2 2 99(2) 2 198 200a

26. 1 11, 2, ( 1)na d a a n d

80 1 (80 1)2 1 79(2) 1 158 157a

27. 1 11, 2 1 3, ( 1)na d a a n d

90 1 (90 1)( 3) 1 89( 3)

1 267 266

a

28. 1 15, 0 5 5, ( 1)na d a a n d

80 5 (80 1)( 5) 5 79( 5)

5 395 390

a

29. 1 15 1

2, 2 , ( 1)2 2 na d a a n d

801 83

2 (80 1)2 2

a

30. 1

1

2 5, 4 5 2 5 2 5,

( 1)n

a d

a a n d

70 2 5 (70 1)2 5

2 5 69 2 5

2 5 138 5

140 5

a

31. 8 1 20 17 8 19 44a a d a a d

Solve the system of equations by subtracting the first equation from the second:

1

12 36 3

8 7(3) 8 21 13

d d

a

Recursive formula: 1 113 3n na a a

nth term:

1 1

13 1 3

13 3 3

3 16

na a n d

n

n

n

32. 4 1 20 13 3 19 35a a d a a d


1

16 32 2

3 3(2) 3 6 3

d d

a


nth term:

1 1

3 1 2

3 2 2

2 5

na a n d

n

n

n

33. 9 1 15 18 5 14 31a a d a a d


1

6 36 6

5 8(6) 5 48 53

d d

a


nth term:

1 1

53 1 6

53 6 6

6 59

na a n d

n

n

n

34. 8 1 18 17 4 17 96a a d a a d


1

10 100 10

4 7( 10) 4 70 74

d d

a


nth term:

1 1

74 1 10

74 10 10

84 10

na a n d

n

n

n


1288


35. 15 1 40 114 0 39 50a a d a a d


1

25 50 2

14( 2) 28

d d

a


nth term:

1 1

28 1 2

28 2 2

30 2

na a n d

n

n

n

36. 5 1 13 14 2 12 30a a d a a d


1

8 32 4

2 4(4) 18

d d

a


nth term:

1 1

18 1 4

18 4 4

4 22

na a n d

n

n

n

37. 14 1 18 113 1 17 9a a d a a d


1

4 8 2

1 13( 2) 1 26 25

d d

a


nth term:

1 1

25 1 2

25 2 2

27 2

na a n d

n

n

n

38. 12 1 18 111 4 17 28a a d a a d


1

6 24 4

4 11(4) 4 44 40

d d

a


nth term:

1 1

40 1 4

40 4 4

4 44

na a n d

n

n

n

39. 21 1 2 1 2

2 2 2n nn n n

S a a n n n

40. 21 2 2 1

2 2n nn n

S a a n n n n n

41. 1 7 2 5 9 52 2 2n nn n n

S a a n n

42.

1

2

1 4 52 2

4 6 2 32

2 3

n nn n

S a a n

nn n n

n n

43. 1 12, 4 2 2, ( 1)na d a a n d

70 2 ( 1)2

70 2 2 2

70 2

35

n

n

n

n

135

2 702 235

72 35 362

1260

n nn

S a a

44. 1 11, 3 1 2, ( 1)na d a a n d

59 1 ( 1)2

59 1 2 2

60 2

30

n

n

n

n

130

1 59 15 60 9002 2n nn

S a a

45. 1 15, 9 5 4, ( 1)na d a a n d

49 5 1 4

49 5 4 4

48 4

12

n

n

n

n

112

5 49 6 54 3242 2n nn

S a a


1289


46. 1 12, 5 2 3, ( 1)na d a a n d

41 2 1 3

41 2 3 3

42 3

14

n

n

n

n

114

2 41 7 43 3012 2n nn

S a a

47. 1 73a , 78 73 5d , 1 1na a n d

558 73 1 5

485 5 1

97 1

98

n

n

n

n

198

73 5582 2

49 631 30,919

n nn

S a a

48. 1 7a , 1 7 6d , 1 1na a n d

299 7 1 6

306 6 1

51 1

52

n

n

n

n

1

527 299

2 2

26 292 7592

n n

nS a a

49. 1 4a , 4.5 4 0.5d , 1 1na a n d

100 4 1 0.5

96 0.5 1

192 1

193

n

n

n

n

1

1934 100

2 2193

104 10,0362

n n

nS a a

50. 1 8a , 1 1

8 84 4

d , 1 1na a n d

150 8 1

4

142 1

4168 1

169

n

n

n

n

1

1698 50

2 2169

58 49012

n n

nS a a

51. 1 2 1 5 3a , 80 2 80 5 155a

80

803 155 40 152 6080

2S

52. 1 3 2 1 1a , 90 3 2 90 177a

90

901 177 45 176 7920

2S

53. 1

1 116 1

2 2a , 100

16 100 44

2a

100100 11

442 2

7750 1925

2

S

54. 11 1 5

13 2 6

a , 801 1 163

803 2 6

a

8080 5 163

40 28 11202 6 6

S

55. 1 14a , 16 14 2d , 1 1na a n d

120 14 120 1 2 14 119 2 252a

120120

14 252 60 266 15,9602

S

56. 1 2a , 1 2 3d , 1 1na a n d

46 2 46 1 3 2 45 3 133a

4646

2 133 23 131 30132

S


1290


57. Find the common difference of the terms and solve the system of equations:

(2 1) ( 3) 2

(5 2) (2 1) 3 1

3 1 2

2 3

3

2

x x d x d

x x d x d

x x

x

x

58. Find the common difference of the terms and solve the system of equations:

(3 2) (2 ) 2

(5 3) (3 2) 2 1

2 1 2

1

x x d x d

x x d x d

x x

x

59. 13, 11, and 1092d a S

2

2

1092 2(11) ( 1)(3)2

1092 22 3 32

2194 19 3

2194 19 3

3 19 2184 0

(3 91)( 24) 0

nn

nn

n n

n n

n n

n n

So 24n .

60. 14, 78, and 702d a S

2

2

2

702 2(78) ( 1)( 4)2

702 156 4 42

1404 160 4

1404 160 4

4 160 1404 0

40 351 0

( 13)( 27) 0

nn

nn

n n

n n

n n

n n

n n

So 13n or 27n .

61. The total number of seats is:

25 26 27 25 29 1S

This is the sum of an arithmetic sequence with

11, 25, and 30d a n .

Find the sum of the sequence:

3030

2(25) (30 1)(1)2

15(50 29) 15(79)

1185

S

There are 1185 seats in the theater.

62. The total number of seats is:

15 17 19 15 39 2S

This is the sum of an arithmetic sequence with

12, 15, and 40d a n .

Find the sum of the sequence:

4040

2(15) (40 1)(2)2

20(30 78) 20(108)

2160

S

The corner section has 2160 seats.

63. The lighter colored tiles have 20 tiles in the bottom row and 1 tile in the top row. The number decreases by 1 as we move up the triangle. This is an arithmetic sequence with

1 20, 1, and 20a d n . Find the sum:

202(20) (20 1)( 1)

210(40 19) 10(21)

210

S

There are 210 lighter tiles.

The darker colored tiles have 19 tiles in the bottom row and 1 tile in the top row. The number decreases by 1 as we move up the triangle. This is an arithmetic sequence with

1 19, 1, and 19a d n . Find the sum:

192(19) (19 1)( 1)

219 19

(38 18) (20) 1902 2

S

There are 190 darker tiles.

64. The number of bricks required decreases by 2 on each successive step. This is an arithmetic sequence with 1 100, 2, and 30a d n .

a. The number of bricks for the top step is:

30 1 ( 1) 100 (30 1)( 2)

100 29( 2) 100 58

42

a a n d

42 bricks are required for the top step.


1291


b. The total number of bricks required is the sum of the sequence:

30100 42 15(142) 2130

2S

2130 bricks are required to build the staircase.

65. The air cools at the rate of 5.5 F per 1000 feet. Since n represents thousands of feet, we have

5.5d . The ground temperature is 67 F so we have 1 67 5.5 61.5T . Therefore,

61.5 1 5.5

5.5 67 or 67 5.5

nT n

n n

After the parcel of air has risen 5000 feet, we

have 5 61.5 5 1 5.5 39.5T .

The parcel of air will be 39.5 F after it has risen 5000 feet.

66. If we treat the length of each rung as the term of an arithmetic sequence, we have 1 49a ,

2.5d , and 24na .

1 1na a n d

24 49 1 2.5

25 2.5 1

10 1

11

n

n

n

n

Therefore, the ladder contains 11 rungs.

To find the total material required for the rungs, we need the sum of their lengths. Since there are 11 rungs, we have

1111 11

49 24 73 401.52 2

S

It would require 401.5 feet of material to construct the rungs for the ladder.

67. 1 35a , 37 35 2d , 1 1na a n d

27 35 27 1 2 35 26 2 87a

2727 27

35 87 122 16472 2

S

The amphitheater has 1647 seats.

68. Find n in an arithmetic sequence with

1 10, 4, 2040na d S .

1

2

2

2 ( 1)2

2040 2(10) ( 1)42

4080 20 4 4

4080 (4 16)

4080 4 16

1020 4

nn

S a n d

nn

n n

n n

n n

n n

2 4 1020 0

( 34)( 30) 0 34 or 30

n n

n n n n

There are 30 rows in the corner section of the stadium.

69. The yearly salaries form an arithmetic sequence with 1 35,000, 1400, 280,000na d S .

Find the number of years for the aggregate salary to equal $280,000.

1

2

2

2 ( 1)2

280,000 2(35,000) ( 1)14002

280,000 35,000 700 700

280,000 (700 34,300)

280,000 700 34,300

400 49

nn

S a n d

nn

n n

n n

n n

n n

2 49 400 0n n

249 49 4(1)( 400)

2(1)

49 4001 49 63.25

2 27.13 or 56.13

n

n n

It takes about 8 years to have an aggregate salary of at least $280,000. The aggregate salary after 8 years will be $319,200.

70. Answers will vary.

71. Answers will vary. Both increase (or decrease) at a constant rate, but the domain of an arithmetic sequence is the set of natural numbers while the domain of a linear function is the set of all real numbers.


1292


72. 12

0.1531 1 0.1642

1216.42%

er

73. 2 1 2 1( ) ( )

(3 ( 1)) ( 4 2)

4 6

x x y y

v i j

i j

i j

74. 2 2

2 2

25 4 100

14 25

x y

x y

The center of the ellipse is at the origin. 2, 5a b . The vertices are (0, 5) and (0, –5).

Find the value of c: 2 2 2 25 4 21 21c b a c

The foci are 0, 21 and 0, 21 .

75. 2 0

3 1A

Augment the matrix with the identity and use row operations to find the inverse:

( )

( )

( )

112

1 12

12

2 1 232

12

2 232

2 0 1 0

3 1 0 1

1 0 0

3 1 0 1

1 0 0 3

0 1 1

1 0 0 1

0 1 1

R r

R r r

R r

é ùê úê ú-ë ûé ùê ú =ê ú-ë ûé ùê ú =- +ê ú- -ê úë ûé ùê ú =-ê ú-ê úë û

Thus, 1

1 232

0

1A-

é ùê ú= ê ú-ê úë û

.

Section 12.3

1. 11

nna a r

2. 8 3 24 2

9 4 36 3

3. c

4. 1

a

r

5. b

6. True

7. False; the common ratio can be positive or negative (or 0, but this results in a sequence of only 0s).

8. True

9. 1

133 3

3

nn n

nr

The ratio of consecutive terms is constant, therefore the sequence is geometric.

1 21 2

3 43 4

3 3, 3 9,

3 27, 3 81

s s

s s

10. 1

1( 5)( 5) 5

( 5)

nn n

nr


1 21 2

3 43 4

( 5) 5, ( 5) 25,

( 5) 125, ( 5) 625

s s

s s

11.

1

11

31 12

2 213

2

n

n n

nr


1 2

1 2

3 4

3 4

1 3 1 33 , 3 ,

2 2 2 4

1 3 1 33 , 3

2 8 2 16

a a

a a

Section 12.3: Geometric Sequences; Geometric Series

1293


12.

1

15

5 522 25

2

n

n n

nr


1 2

1 2

3 4

3 4

5 5 5 25, ,

2 2 2 4

5 125 5 625,

2 8 2 16

b b

b b

13.

1 1

( 1)11

2

4 22 2

22

4

n

nn n

nnr


1 1 02

1 2

2 1 11

2 2

3 1 2

3 2

4 1 3

4 2

2 2 12 ,

4 42

2 2 12 ,

4 22

2 21,

4 2

2 22

4 2

c

c

c

c

14.

1

11

3

9 33 3

33

9

n

nn n

nnr


1 2

1 2

3 4

3 4

3 1 3 9, 1,

9 3 9 9

3 27 3 813, 9

9 9 9 9

d d

d d

15.

113

1/33 3

3

22 2

2

nn n

nr


1/3 2 /3 3/3 4/ 31 2 3 42 , 2 , 2 2, 2e e e e

16. 2( 1)

2 2 2 22

33 3 9

3

nn n

nr


2 1 2 2 41 2

2 3 6 2 4 83 4

3 9, 3 3 81,

3 3 729, 3 3 6561

f f

f f

17.

1 1

1

1 11

( 1) ( 1) 1

3

2 3 2

3 23

2

33 2 3 2

2

n

n n n

n nn

n

n n n n

r


1 1 0 2 1 1

1 21 2 2

3 1 2 4 1 3

3 43 3 4 4

3 3 1 3 3 3, ,

2 2 42 2 2

3 3 9 3 3 27,

8 162 2 2 2

t t

t t

18.

1

1 1 1 1

1

1 1 1

2

3 3 2

3 22

3

23 2 3 2

3

n

n n n

n nn

n

n n n n

r


1 2

1 21 1 0 2 1

3 4

3 43 1 2 4 1 3

2 2 2 2 42, ,

1 33 3 3

2 8 8 2 16 16,

9 273 3 3 3

u u

u u

19. 5 1 45

1

2 3 2 3 2 81 162

2 3nn

a

a

20. 5 1 45

1

2 4 2 4 2 256 512

2 4nn

a

a

21. 5 1 45

1

5( 1) 5( 1) 5 1 5

5 ( 1)nn

a

a

22. 5 1 45

1

6( 2) 6( 2) 6 16 96

6 ( 2)nn

a

a


1294


23. 5 1 4

5

1

1 10 0 0

2 2

10 0

2

n

n

a

a

24. 5 1 4

5

1 1

1 1 11 1

3 3 81

1 11

3 3

n n

n

a

a

25.

5 1 4

5

1

2 2 2 2 2 4 4 2

2 2 2n n

n

a

a

26. 5 1 4

5

1

1 10 0 0

10 0

n

n

a

a

27. 1

7 1 6

7

11, , 7

2

1 1 11

2 2 64

a r n

a

28. 1

8 1 78

1, 3, 8

1 3 3 2187

a r n

a

29.

1

9 1 89

1, 1, 9

1 1 1 1

a r n

a

30.

1

10 1 910

1, 2, 10

1 2 1 2 1( 512) 512

a r n

a

31.

1

8 1 78

0.4, 0.1, 8

0.4 0.1 0.4 0.1 0.00000004

a r n

a

32.

1

67 17

0.1, 10, 7

0.1 10 0.1 10 100,000

a r n

a

33. 1 7a , 14

27

r , 11

nna a r

17 2nna

34. 1 5a , 10

25

r , 11

nna a r

15 2nna

35. 1 3a , 1 1

3 3r

, 1

1n

na a r

1 21 1

33 3

n n

na

36. 1 4a , 1

4r , 1

1n

na a r

1 21 1

44 4

n n

na

37. 11

nna a r

6 1

1

5

1

1

1

243 3

243 3

243 243

1

a

a

a

a

Therefore, 1( 3)nna

38. 11

nna a r

2 1

1

1

1

17

3

17

321

a

a

a

Therefore, 1

121

3

n

na

.

39. 4 1 3

24 12 1

2 1

a a r rr

a ra r

2 1575225

7

225 15

r

r

11

2 11

1

1

7 15

7 15

7

15

nna a r

a

a

a

Therefore, 1 2715 7 15

15n n

na .


1295


40. 6 1 5

36 13 1 2

3 1

a a r rr

a a r r

13 81

13

3

1 13

81 27

1 1

27 3

r

r

11

3 1

1

1

1

1 1

3 3

1 1

3 93

nna a r

a

a

a

Therefore, 1 2

1 13

3 3

n n

na

.

41.

1

1

1, 2

4

1 1 1 2 11 2

1 4 1 2 4

12 1

4

n nn

n

n

a r

rS a

r

42.

1

1

3 1, 3

9 3

1 1 1 3 1 1 3

1 3 1 3 3 2

1 11 3 3 1

6 6

n n n

n

n n

a r

rS a

r

43. 12 2

,3 3

a r

1

21

1 2 321 3 13

21

2 232 1

13 33

n

n

n

n

n

rS a

r

44.

1

1

4, 3

1 1 3 1 34 4

1 1 3 2

2 1 3 2 3 1

n n n

n

n n

a r

rS a

r

45. 1

1

1, 2

1 1 21 1 2

1 1 2

n nn

n

a r

rS a

r

46. 13

2,5

a r

1

3 31 1

1 5 52 2

3 21 15 5

35 1

5

n n

n

n

n

rS a

r

47. Using the sum of the sequence feature:





1296




53. 1

1

11,

3Since 1, the series converges.

1 1 31 21 2

13 3

a r

r

aS

r

54. 1

1

22,


2 26

2 111

3 3

a r

r

aS

r

55. 1

1

18,


8 816

1 111

2 2

a r

r

aS

r

56. 1

1

16,


6 69

1 211

3 3

a r

r

aS

r

57. 1

1

12,


2 2 851 51

144

a r

r

aS

r

58. 1

1

31,


1 1 471 73

144

a r

r

aS

r

59. 1 8a , 3

2r

Since 1r , the series diverges.

60. 1 9a , 4

3r


61. 1

1

15,


5 5 20

1 31 31

4 4

a r

r

aS

r

62. 1

1

18,


8 812

1 211

3 3

a r

r

aS

r

63. 11

2a , 3r


64. 1 3a , 3

2r



1297


65. 1

1

26,


6 6 18

51 521

33

a r

r

aS

r

66. 1

1

14,


4 4 831 31

122

a r

r

aS

r

67. 1 1

1 1 1

2 2 2 23 3 2

3 3 3 3

k k k

k k k

1 2a , 2

3r

12 13 3

Since 1, the series converges.

2 26

1 1

r

aS

r

68. 1 1

1 1 1

3 3 3 3 32 2

4 4 4 2 4

k k k

k k k

13

2a ,

3

4r

13214


332 4 6

31 21 4

r

aS

r

69. 2n

( 1 2) ( 2) 3 2 1d n n n n

The difference between consecutive terms is constant. Therefore the sequence is arithmetic.

50 50 50

501 1 1

( 2) 2

50(50 1)2(50) 1275 100 1375

2

k k k

S k k

70. 2 5n

2( 1) 5 (2 5)

2 2 5 2 5 2

d n n

n n


50 50 50

501 1 1

(2 5) 2 5

50(50 1)2 5(50) 2550 250 2300

2

k k k

S k k

71. 24n Examine the terms of the sequence: 4,

16, 36, 64, 100, ... There is no common difference and there is no common ratio. Therefore the sequence is neither arithmetic nor geometric.

72. 25 1n Examine the terms of the sequence:

6, 21, 46, 81, 126, ... There is no common difference and there is no common ratio. Therefore the sequence is neither arithmetic nor geometric.

73. 2

33

n

2 23 ( 1) 3

3 3

2 2 2 23 3

3 3 3 3

d n n

n n


50 50 50

501 1 1

2 23 3

3 3

2 50(50 1)3(50) 150 850 700

3 2

k k k

S k k

74. 3

84

n

3 38 ( 1) 8

4 4

3 3 3 38 8

4 4 4 4

d n n

n n


50 50 50

501 1 1

3 38 8

4 4

3 50(50 1)8(50)

4 2

400 956.25 556.25

k k k

S k k


1298


75. 1, 3, 6, 10, ... There is no common difference and there is no common ratio. Therefore the sequence is neither arithmetic nor geometric.

76. 2, 4, 6, 8, ... The common difference is 2. The difference between consecutive terms is constant. Therefore the sequence is arithmetic.

50 50

501 1

50(50 1)2 2 2 2550

2k k

S k k

77. 2

3

n

1

12

2 233 32

3

n

n n

nr

The ratio of consecutive terms is constant. Therefore the sequence is geometric.

50

50

501

21

2 2 31.999999997

23 3 13

k

k

S

78. 5

4

n

1

15

5 54

4 45

4

n

n n

nr


50

50

501

51

5 5 4350,319.6161

54 4 14

k

k

S

79. –1, 2, –4, 8, ... 2 4 8

21 2 4

r


50501

501

14

1 ( 2)1 ( 2) 1

1 ( 2)

3.752999689 10

k

k

S

80. 1, 1, 2, 3, 5, 8, ... There is no common difference and there is no common ratio. Therefore the sequence is neither arithmetic nor geometric.

81. / 23n

112

1/ 22 2

2

33 3

3

nn n

nr


501/ 250/ 2 1/ 2

50 1/ 21

12

1 33 3

1 3

2.004706374 10

k

k

S

82. ( 1)n

11( 1)

( 1) 1( 1)

nn n

nr


5050

501

1 ( 1)( 1) ( 1) 0

1 ( 1)k

k

S

83. Find the common ratio of the terms and solve the system of equations:

2 2

2 3;

22 3

4 4 3 42

x xr r

x xx x

x x x x xx x

84. Find the common ratio of the terms and solve the system of equations:

2 2

2;

12

2 21

x xr r

x xx x

x x x xx x

85. This is a geometric series with

1 $42,000, 1.03, 5a r n . Find the 5th

term:

5 1 45 42000 1.03 42000 1.03 $47, 271.37a

86. This is a geometric series with

1 $15,000, 0.85, 6a r n . Find the 6th

term:

6 1 56 15000 0.85 15000 0.85 $6655.58a


1299


87. a. Find the 10th term of the geometric sequence:

1

10 1 910

2, 0.9, 10

2(0.9) 2(0.9) 0.775 feet

a r n

a

b. Find when 1nn a :

1

1

2(0.9) 1

0.9 0.5

( 1) log 0.9 log 0.5

log 0.51

log 0.9

log 0.51 7.58

log 0.9

n

n

n

n

n

On the 8th swing the arc is less than 1 foot.

c. Find the sum of the first 15 swings:

1515

15

15

1 0.91 (0.9)2 2

1 0.9 0.1

20 1 0.9 15.88 feet

S

d. Find the infinite sum of the geometric series: 2 2

20 feet1 0.9 0.1

S

88. a. Find the 3rd term of the geometric sequence:

1

3 1 23

24, 0.8, 3

24(0.8) 24(0.8) 15.36 feet

a r n

a

b. The height after the n th bounce is:

1124(0.8) 24 0.8 0.8

30 0.8 ft

nnn

n

a

c. Find when 0.5nn a :

1

1

24(0.8) 0.5

0.8 0.020833

( 1) log 0.8 log 0.020833

log 0.0208331

log 0.8

log 0.0208331 18.35

log 0.8

n

n

n

n

n

On the 19th bounce the height is less than 0.5 feet.

d. Find the infinite sum of the geometric series:

24 24120 feet

1 0.8 0.2S

on the upward

bounce.

For the downward motion of the ball: 30 30

150 feet1 0.8 0.2

S

The total distance the ball travels is 120 + 150 = 270 feet.

89. This is an ordinary annuity with $100P and

12 30 360n payment periods. The

interest rate per period is 0.12

0.0112

. Thus,

3601 0.01 1

100 $349,496.410.01

A


12 3 36n payment periods. The interest

rate per period is 0.10

12. Thus,

360.10

1 112

400 $16,712.730.10

12

A




0.024

. Thus,

801 0.02 1

500 $96,885.980.02

A




0.052

. Thus,

301 0.05 1

1000 $66,438.850.05

A


1300


93. This is an ordinary annuity with $50,000A

and 12 10 120n payment periods. The


0.00512

. Thus,

120

120

1 0.005 150,000

0.005

0.00550,000 $305.10

1 0.005 1

P

P

94. This is an ordinary annuity with $160,000A

and 12 18 216n payment periods. The


12. Thus,

216

216

0.081 1

12160,000

0.08

12

0.08

12160,000 $333.270.08

1 112

P

P

95. This is a geometric sequence with

1 1, 2, 64a r n .

Find the sum of the geometric series: 64 64

6464

19

1 2 1 21 2 1

1 2 1

1.845 10 grains

S

96. This is an infinite geometric series with

11 1,4 4a r .

Find the sum of the infinite geometric series:

1 114 4

1 3 31 4 4

S

1

3 of the square is eventually shaded.

97. The common ratio, 0.90 1r . The sum is: 1 1

101 0.9 0.10

S

.

The multiplier is 10.

98. The common ratio, 0.95 1r . The sum is: 1 1

201 0.95 0.05

S

.

The multiplier is 20.

99. This is an infinite geometric series with 1.03

4, and 1.09

a r .

Find the sum: 4

Price $72.671.03

11.09

.

100. This is an infinite geometric series with

11.04

2.5, and 1.11

a r .

Find the sum: 2.5

Price $39.641.04

11.11

.

101. Given: 1 1000, 0.9a r

Find when 0.01nn a :

1

1

1000(0.9) 0.01

0.9 0.00001

( 1) log 0.9 log 0.00001

log 0.000011

log 0.9

log 0.000011 110.27

log 0.9

n

n

n

n

n

On the 111th day or December 20, 2015, the amount will be less than $0.01.

Find the sum of the geometric series:

111

111 1

111

1 0.911000

1 1 0.9

1 0.91000 $9999.92

0.1

nrS a

r


1301


102. When the last deposit is made at the nth payment period, the first deposit of $P has earned interest compounded for n – 1 payment periods, the second deposit of $P has earned interest compounded for n – 2 payment periods, and so on. The following table shows the value of each deposit after n deposits have been made.

Deposit 1 2 3 … n – 1 n

Value 1(1 )nP i 2(1 )nP i 3(1 )nP i … (1 )P i P

The amount A of the annuity is the sum of the values shown in the table.

1 2 3

2 1

(1 ) (1 ) (1 ) (1 )

1 (1 ) (1 ) (1 )

n n n

n n

A P i P i P i P i P

P i i i

The expression in brackets is the sum of a geometric sequence with n terms and a common ratio of (1 )i . Thus,

1 (1 ) 1 (1 ) (1 ) 1

.1 (1 )

n n ni i iA P P P

i i i

103. First, determine the number of seats in the

section:

12 ( 1)214

2 2 13 2 210 seats2

nn

S a n d

Now, find the sum of a geometric sequence with 1 0.01 and 1.05 and 210.a r n

210

210

1 1.050.01 $5633.36

1 1.05S

104. Both options are geometric sequences: Option A: 1 $40,000; 1.06; 5a r n

5 1 45

5

5

40,000(1.06) 40,000(1.06) $50,499

1 1.0640000 $225, 484

1 1.06

a

S

Option B: 1 $44,000; 1.03; 5a r n

5 1 45

5

5

44,000(1.03) 44,000(1.03) $49,522

1 1.0344000 $233,602

1 1.03

a

S

Option A provides more money in the 5th year, while Option B provides the greatest total amount of money over the 5 year period.

105. Find the sum of each sequence: A: Arithmetic series with:

1 $1000, 1, 1000a d n

Find the sum of the arithmetic series:

10001000

1000 1 500(1001) $500,5002

S

B: This is a geometric sequence with 1 1, 2, 19a r n .

Find the sum of the geometric series: 19 19

1919

1 2 1 21 2 1 $524,287

1 2 1S

B results in more money.

106. Option 1: Total Salary $2,000,000(7) $100,000(7)

$14,700,000

Option 2: Geometric series with:

1 $2,000,000, 1.045, 7a r n

Find the sum of the geometric series:

71 1.0452,000,000 $16,038,304

1 1.045S

Option 3: Arithmetic series with:

1 $2,000,000, $95,000, 7a d n

Find the sum of the arithmetic series:

77

2(2,000,000) (7 1)(95,000)2$15,995,000

S

Option 2 provides the most money; Option 1 provides the least money.

107. The amount paid each day forms a geometric sequence with 1 0.01a and 2r .

22 22

22 11 1 2

0.01 41,943.031 1 2

rS a

r

The total payment would be $41,943.03 if you worked all 22 days.

2122 122 1 0.01 2 20,971.52a a r

The payment on the 22nd day is $20,971.52.


1302


Answers will vary. With this payment plan, the bulk of the payment is at the end so missing even one day can dramatically reduce the overall payment. Notice that with one sick day you would lose the amount paid on the 22nd day which is about half the total payment for the 22 days.

108. Yes, a sequence can be both arithmetic and geometric. For example, the constant sequence 3,3,3,3,..... can be viewed as an arithmetic

sequence with 1 3a and 0.d Alternatively,

the same sequence can be viewed as a geometric sequence with 1 3a and 1.r



111. Answers will vary. Both increase (or decrease) exponentially, but the domain of a geometric sequence is the set of natural numbers while the domain of an exponential function is the set of all real numbers.

112. 7log 62

log 62 2.121log 7

113. 2 2

8 15 8 158 15 8 ( 15)

8 15 8 1517 17 17

i j i jvu

v i j

i ji j

114. Hyperbola: Vertices: (–2, 0), (2, 0); Focus: (4, 0); Center: (0, 0); Transverse axis is the x-axis; 2; 4a c .

Find b: 2 2 2 16 4 12

12 2 3

b c a

b

Write the equation: 2 2

14 12

x y

115.

3 016 0 6 02 20 6 3 1 02

1 2 4 2 4 14 1 2

3(4 6) 1(0 24) 0

3(10) ( 24) 30 24

54

Section 12.4

1. I: 1: 2 1 2 and 1(1 1) 2n

II: If 2 4 6 2 ( 1)k k k , then

2 4 6 2 2( 1)

2 4 6 2 2( 1)

( 1) 2( 1)

( 1)( 2)

1 1 1

k k

k k

k k k

k k

k k

Conditions I and II are satisfied; the statement is true.

2. I: 1: 4 1 3 1 and 1(2 1 1) 1n

II: If 1 5 9 (4 3) (2 1)k k k , then

2

2

1 5 9 (4 3) (4( 1) 3)

1 5 9 (4 3) 4 4 3

(2 1) 4 1

2 4 1

2 3 1

( 1)(2 1)

1 2 1 1

k k

k k

k k k

k k k

k k

k k

k k


Section 12.4: Mathematical Induction

1303


3. I: 1

1: 1 2 3 and 1(1 5) 32

n

II: If 1

3 4 5 ( 2) ( 5)2

k k k , then

2

2

2

3 4 5 ( 2) [( 1) 2]

3 4 5 ( 2) ( 3)

1( 5) ( 3)

21 5

32 21 7

32 21

7 621

( 1)( 6)21

( 1) 1 52

k k

k k

k k k

k k k

k k

k k

k k

k k


4. I: 1: 2 1 1 3 and 1(1 2) 3n

II: If 3 5 7 (2 1) ( 2)k k k , then

2

2

3 5 7 (2 1) [2( 1) 1]

3 5 7 (2 1) (2 3)

( 2) (2 3)

2 2 3

4 3

( 1)( 3)

( 1) 1 2

k k

k k

k k k

k k k

k k

k k

k k


5. I: 1

1: 3 1 1 2 and 1(3 1 1) 22

n

II: If 1

2 5 8 (3 1) (3 1)2

k k k ,

then

2

2 2

2 5 8 (3 1) [3( 1) 1]

2 5 8 (3 1) (3 2)

1 3 1(3 1) (3 2) 3 2

2 2 23 7 1

2 3 7 42 2 21

( 1)(3 4)21

( 1) 3 1 12

k k

k k

k k k k k k

k k k k

k k

k k

Conditions I and II are satisfied; the statement is

true.

6. I: 1

1: 3 1 2 1 and 1(3 1 1) 12

n

II: If 1

1 4 7 (3 2) (3 1)2

k k k ,

then

2

2 2

1 4 7 (3 2) [3( 1) 2]

1 4 7 (3 2) (3 1)

1 3 1(3 1) (3 1) 3 1

2 2 23 5 1

1 3 5 22 2 21

( 1)(3 2)21

( 1) 3 1 12

k k

k k

k k k k k k

k k k k

k k

k k

Conditions I and II are satisfied; the statement is

true.

7. I: 1 1 11: 2 1 and 2 1 1n

II: If 2 11 2 2 2 2 1k k , then 2 1 1 1

2 1

1

1 2 2 2 2

1 2 2 2 2

2 1 2 2 2 1

2 1

k k

k k

k k k

k



1304


8. I: 1 1 111: 3 1 and (3 1) 1

2n

II: If 2 1 11 3 3 3 (3 1)

2k k , then

2 1 1 1

2 1

1

1 3 3 3 3

1 3 3 3 3

1 1 1(3 1) 3 3 3

2 2 23 1 1

3 3 3 12 2 21

3 12

k k

k k

k k k k

k k

k


9. I: 1 1 111: 4 1 and 4 1 1

3n

II: If 2 1 11 4 4 4 4 1

3k k , then

2 1 1 1

2 1

1

1 4 4 4 4

1 4 4 4 4

1 1 14 1 4 4 4

3 3 34 1 1

4 4 4 13 3 31

4 13

k k

k k

k k k k

k k

k


10. I: 1 1 111: 5 1 and 5 1 1

4n

II: If 2 1 11 5 5 5 5 1

4k k , then

2 1 1 1

2 1

1

1 5 5 5 5

1 5 5 5 5

1 1 15 1 5 5 5

4 4 45 1 1

5 5 5 14 4 41

5 14

k k

k k

k k k k

k k

k


11. I: 1 1 1 1

1: and1(1 1) 2 1 1 2

n

II: If 1 1 1 1

1 2 2 3 3 4 ( 1) 1

k

k k k

, then

2

1 1 1 1 1 1 1 1 1 1

1 2 2 3 3 4 ( 1) ( 1)( 1 1) 1 2 2 3 3 4 ( 1) ( 1)( 2)

1 2 1

1 ( 1)( 2) 1 2 ( 1)( 2)

2 1 ( 1)( 1) 1 1

( 1)( 2) ( 1)( 2) 2 1 1

k k k k k k k k

k k k

k k k k k k k

k k k k k k

k k k k k k



1305


12. I: 1 1 1 1

1: and(2 1 1)(2 1 1) 3 2 1 1 3

n

II: If 1 1 1 1

1 3 3 5 5 7 (2 1)(2 1) 2 1

k

k k k

, then

2

1 1 1 1 1

1 3 3 5 5 7 (2 1)(2 1) (2( 1) 1)(2( 1) 1)

1 1 1 1 1

1 3 3 5 5 7 (2 1)(2 1) (2 1)(2 3)

1 2 3 1

2 1 (2 1)(2 3) 2 1 2 3 (2 1)(2 3)

2 3 1 ( 1)(2 1)

(2 1)(2 3) (2 1)(2 3)

k k k k

k k k k

k k k

k k k k k k k

k k k k k

k k k k

1 1

2 3 2 1 1

k

k k


13. I: 2 11: 1 1 and 1(1 1)(2 1 1) 1

6n

II: If 2 2 2 2 11 2 3 ( 1)(2 1)

6k k k k , then

2 2 2 2 2 2 2 2 2 2 2

2 2 2

11 2 3 ( 1) 1 2 3 ( 1) ( 1)(2 1) ( 1)

61 1 1 1 7 1

( 1) (2 1) 1 ( 1) 1 ( 1) 1 ( 1) 2 7 66 3 6 3 6 6

1( 1)( 2)(2 3)

61

( 1) 1 1 2 1 16

k k k k k k k k

k k k k k k k k k k k k k k

k k k

k k k


14. I: 3 2 211: 1 1 and 1 (1 1) 1

4n

II: If 3 3 3 3 2 211 2 3 ( 1)

4k k k , then

3 3 3 3 3 3 3 3 3 3 2 2 3

2 2 2 2

2 2

2 2

11 2 3 ( 1) 1 2 3 ( 1) ( 1) ( 1)

41 1

( 1) 1 ( 1) 4 44 4

1( 1) ( 2)

41

( 1) (( 1) 1)4

k k k k k k k

k k k k k k

k k

k k



1306


15. I: 1

1: 5 1 4 and 1(9 1) 42

n

II: If 1

4 3 2 (5 ) (9 )2

k k k , then

2 2 2

14 3 2 (5 ) 5 ( 1) 4 3 2 (5 ) (4 ) (9 ) (4 )

29 1 1 7 1

4 4 7 82 2 2 2 2

1 1 1( 1)( 8) ( 1)(8 ) ( 1) 9 ( 1)

2 2 2

k k k k k k k

k k k k k k k

k k k k k k


16. I: 1

1: (1 1) 2 and 1(1 3) 22

n

II: If 1

2 3 4 ( 1) ( 3)2

k k k , then

2 2

2

2 3 4 ( 1) ( 1) 1 2 3 4 ( 1) ( 2)

1 1 3 1 5( 3) ( 2) 2 2

2 2 2 2 21 1

5 4 ( 1)( 4)2 21

( 1)(( 1) 3)2

k k k k

k k k k k k k k

k k k k

k k


17. I: 1

1: 1(1 1) 2 and 1(1 1)(1 2) 23

n

II: If 1

1 2 2 3 3 4 ( 1) ( 1)( 2)3

k k k k k , then

1 2 2 3 3 4 ( 1) ( 1)( 1 1) 1 2 2 3 3 4 ( 1) ( 1)( 2)

1 1( 1)( 2) ( 1)( 2) ( 1)( 2) 1

3 3

1( 1)( 2)( 3)

31

( 1)(( 1) 1)(( 1) 2)3

k k k k k k k k

k k k k k k k k

k k k

k k k



1307


18. I: 1

1: (2 1 1)(2 1) 2 and 1(1 1)(4 1 1) 23

n

II: If 1

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

k k k k k , then

2 2

2

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

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

1 1( 1)(4 1) 2( 1)(2 1) ( 1) (4 1) 2(2 1)

3 3

4 1 1( 1) 4 2 ( 1) 4 12 6

3 3 3

1( 1) 4 11

3

k k k k

k k k k

k k k k k k k k k

k k k k k k k k

k k k

16 ( 1)( 2)(4 3)

31

( 1)(( 1) 1)(4( 1) 1)3

k k k

k k k


19. I: 21: 1 1 2 is divisible by 2n

II: If 2 is divisible by 2k k , then 2 2

2

( 1) ( 1) 2 1 1

( ) (2 2)

k k k k k

k k k

Since 2k k is divisible by 2 and 2 2k is

divisible by 2, then 2( 1) ( 1)k k is

divisible by 2.


20. I: 31: 1 2 1 3 is divisible by 3n

II: If 3 2 is divisible by 3k k , then 3

3 2

3 2

( 1) 2( 1)

3 3 1 2 2

( 2 ) (3 3 3)

k k

k k k k

k k k k

Since 3 2k k is divisible by 3 and 23 3 3k k is divisible by 3, then

3( 1) 2( 1)k k is divisible by 3.


21. I: 21: 1 1 2 2 is divisible by 2n

II: If 2 2 is divisible by 2k k , then 2 2

2

( 1) ( 1) 2 2 1 1 2

( 2) (2 )

k k k k k

k k k

Since 2 2k k is divisible by 2 and 2k is

divisible by 2, then 2( 1) ( 1) 2k k is

divisible by 2.


22. I: 1: 1(1 1)(1 2) 6 is divisible by 6n

II: If ( 1)( 2) is divisible by 6k k k , then

( 1)( 1 1)( 1 2)

( 1)( 2)( 3)

( 1)( 2) 3( 1)( 2).

k k k

k k k

k k k k k

Now, ( 1)( 2) is divisible by 6;

and since either 1 or 2 is even,

k k k

k k

3 1 2k k is divisible by 6.

Thus, 1 2 3

1 2 3 1 2

k k k

k k k k k

is divisible by 6.



1308


23. I: 11: If 1 then 1.n x x x

II: Assume, for some natural number k, that if

1x , then 1kx . 1

1

Then 1, for 1,

1 1

( 1)

k

k k

k

x x

x x x x x

x


24. I: 11: If 0 1 then 0 1.n x x

II: Assume, for some natural number k, that if

0 1x , then 0 1kx .

1

1

Then, for 0 1,

0 1 1

Thus, 0 1.

k k

k

x

x x x x x

x


25. I: 1 11: is a factor of .n a b a b a b

II: If is a factor of k ka b a b , show that

a b is a factor of 1 1k ka b .

1 1

( )

k k k k

k k k k

k k k

a b a a b b

a a a b a b b b

a a b b a b

Since a b is a factor of k ka b and a b is a factor of a b , then a b is a factor of

1 1k ka b .


26. I:

2 1 1 2 1 1 3 3

3 3 2 2

1:

is a factor of .

n

a b a b a b

a b a b a ab b

II: 2 1 2 1If is a factor of k ka b a b , show that a b is a factor of

2 1 1 2 1 1k ka b .

2( 1) 1 2( 1) 1 2 3 2 3

2 2 1 2 2 1 2 2 1 2 2 1

2 2 1 2 1 2 1 2 2( )

k k k k

k k k k

k k k

a b a b

a a a b a b b b

a a b b a b

Since a b is a factor of 2 1 2 1k ka b and

a b is a factor of 2 2a b a b a b ,

then a b is a factor of 2 3 2 3k ka b .


27. I: 1n : 11 1 1 1a a a

II: Assume that there is an integer k for which the inequality holds. We need to show that if

1 1k

a ka then

11 1 1

ka k a

.

1

2

2

1 1 1

1 1

1

1 1

1 1

k ka a a

ka a

ka a ka

k a ka

k a

Conditions I and II are satisfied, the statement is true.

28. 2

1:

1 1 41 41 is a prime number.

n

2 2

41:

41 41 41 41 is not a prime number.

n

29. II: If 22 4 6 2 2k k k , then

2

2

2

2 4 6 2 2( 1)

2 4 6 2 2 2

2 2 2

( 2 1) ( 1) 2

( 1) ( 1) 2

k k

k k

k k k

k k k

k k

I: 21: 2 1 2 and 1 1 2 4 2n


1309


30. I: 1

1 1 11: and

1

rn a r a a a

r

II: If 2 1 1

1

kk r

a a r a r a r ar

,

then 2 1 1 1

2 1

1

1

1

1

(1 ) (1 )

1

1

1

1

k k

k k

kk

k k

k k k

k

a a r a r a r a r

a a r a r a r a r

ra a r

r

a r a r r

r

a a r a r a r

r

ra

r


31. I: 1:

1(1 1)(1 1) and 1

2

n

a d a a d a

II: If ( ) ( 2 ) ( 1)

( 1)

2

a a d a d a k d

k kka d

then

2

2

( ) ( 2 ) ( 1)

( ) ( 2 ) ( 1) ( )

( 1)( )

2( 1)

( 1)2

2( 1)

2

( 1)2

( 1)( 1)

2

1 1 11

2

a a d a d a k d a kd

a a d a d a k d a kd

k kka d a kd

k kk a d k

k k kk a d

k kk a d

k kk a d

k kk a d


32. I: 4 :n The number of diagonals of a

quadrilateral is 1

4(4 3) 22 .

II: Assume that for any integer k, the number of diagonals of a convex polygon with k sides

(k vertices) is 1

( 3)2

k k . A convex

polygon with 1k sides ( 1k vertices) consists of a convex polygon with k sides (k vertices) plus a triangle, for a total of ( 1k ) vertices. The diagonals of this

1k -sided convex polygon consist of the diagonals of the k-sided polygon plus 1k additional diagonals. For example, consider the following diagrams.

k = 5 sides

k 1 = 4 new diagonalsk + 1 = 6 sides

Thus, we have the equation:

2

2

2

1 1 3( 3) ( 1) 1

2 2 21 1

12 21

221

( 1)( 2)21

( 1)(( 1) 3)2

k k k k k k

k k

k k

k k

k k



1310


33. I: 3 :n (3 2) 180 180 which is the

sum of the angles of a triangle.

II: Assume that for any integer k, the sum of the angles of a convex polygon with k sides is ( 2) 180k . A convex polygon with

1k sides consists of a convex polygon with k sides plus a triangle. Thus, the sum of the angles is ( 2) 180 180 (( 1) 2) 180 .k k



35. 2

4

log 5 4

2 5

16 5

256 5

251

x

x

x

x

x

36. Let 1F = the tension on the left cable, 2F = the

tension on the right cable, and 3F = the force of

the weight of the box.

2 2

2 2

cos 135º sin 135º

1 1

1

F F i j

F i j

2 2

3 12 2 2

3

cos 30º sin 30º

500

F F i j

F i j

F j

For equilibrium, the sum of the force vectors must be zero.

2 22 2

3 12 22 2

3222 2

2 122 2

500

500

0

1 2 3

1 1

1

1

F F F

F i F j

F i F j j

F F i

F F j

Set the and i j components equal to zero and

solve: 32

22 2

2 122 2

0

500 0

1

1

F F

F F

Solve the first equation for 2F and substitute

the result into the second equation to solve the

system: 2

222 33

2

1 1F F F

2 212 2 3

500 0 1 1F F

2 22 2 3

22 3

500

448.3 lb

(448.3) 366.0 lb

1

1

F

F

F

The tension in the left cable is about 448.3 kg and the tension in the right cable is about 366.0 kg.

37. 4 3 7

2 5 16

x y

x y

Multiply each side of the second equation by –2 and add to eliminate x:

4 3 7

4 10 32

13 39

3

x y

x y

y

y

Substitute and solve for x: 4 3( 3) 7

4 9 7

4 2

1

2

x

x

x

x

The solution is 1

, 32

x y or 1

, 32

.

38.

3 11 2 1

1 00 1 4

2 2

1(3) 2(1) 1( 2) 1( 1) 2(0) 1(2)

0(3) 1(1) 4( 2) 0( 1) 1(0) 4(2)

7 3

7 8

AB

é ù-ê úé ù- ê úê ú= ⋅ ê úê úë û ê ú-ë û

é ù+ - - - + -ê ú= ê ú+ + - - + +ë ûé ù-ê ú= ê ú-ë û

Section 12.5

1. Pascal Triangle

Section 12.5: The Binomial Theorem

1311


2. ! !

10 0! ( 0)! 1 !

n n n

n n

! ( 1)!

1 1! ( 1)! 1 ( 1)!

n n n nn

n n

3. False;

!

! !

n n

j j n j

4. Binomial Theorem

5. 5 5! 5 4 3 2 1 5 4

103 3! 2! 3 2 1 2 1 2 1

6. 7 7! 7 6 5 4 3 2 1 7 6 5

353 3! 4! 3 2 1 4 3 2 1 3 2 1

7. 7 7! 7 6 5 4 3 2 1 7 6

215 5! 2! 5 4 3 2 1 2 1 2 1

8. 9 9! 9 8 7 6 5 4 3 2 1 9 8

367 7! 2! 7 6 5 4 3 2 1 2 1 2 1

9. 50 50! 50 49! 50

5049 49!1! 49! 1 1

10. 100 100! 100 99 98! 100 99

495098 98! 2! 98! 2 1 2 1

11. 1000 1000! 1

11000 1000!0! 1

12. 1000 1000! 1

10 0!1000! 1

13. 1555 55!1.8664 10

23 23!32!

14. 1560 60!4.1918 10

20 20! 40!

15. 1347 47!1.4834 10

25 25! 22!

16. 1037 37!1.7673 10

19 19!18!

17. 5 5 4 3 2 1 05 5 5 5 5 5( 1)

0 1 2 3 4 5x x x x x x x

5 4 3 25 10 10 5 1x x x x x

18. 5 5 4 2 3 3 2 4 1 5 0

5 4 3 2

5 5 5 5 5 5( 1) ( 1) ( 1) ( 1) ( 1) ( 1)

0 1 2 3 4 5

5 10 10 5 1

x x x x x x x

x x x x x

19. 6 6 5 4 2 3 3 2 4 5 0 6

6 5 4 3 2

6 5 4 3 2

6 6 6 6 6 6 6( 2) ( 2) ( 2) ( 2) ( 2) ( 2) ( 2)

0 1 2 3 4 5 6

6 ( 2) 15 4 20 ( 8) 15 16 6 ( 32) 64

12 60 160 240 192 64

x x x x x x x x

x x x x x x

x x x x x x

20. 5 5 4 3 2 2 3 1 4 0 5

5 4 3 2

5 4 3 2

5 5 5 5 5 5( 3) (3) (3) (3) (3) (3)

0 1 2 3 4 5

5 (3) 10 9 10 (27) 5 81 243

15 90 270 405 243

x x x x x x x

x x x x x

x x x x x


1312


21. 4 4 3 2

4 3 2 4 3 2

4 4 4 4 4(3 1) (3 ) (3 ) (3 ) (3 )

0 1 2 3 4

81 4 27 6 9 4 3 1 81 108 54 12 1

x x x x x

x x x x x x x x

22. 5 5 4 3 2 2 3 4 5

5 4 3 2

5 4 3 2

5 5 5 5 5 5(2 3) (2 ) (2 ) 3 (2 ) 3 (2 ) 3 2 3 3

0 1 2 3 4 5

32 5 16 3 10 8 9 10 4 27 5 2 81 243

32 240 720 1080 810 243

x x x x x x

x x x x x

x x x x x

23. 5 5 4 3 2 2 3 4 52 2 2 2 2 2 2 2 2 2 2 2

10 8 2 6 4 4 6 2 8 10

5 5 5 5 5 5

0 1 2 3 4 5

5 10 10 5

x y x x y x y x y x y y

x x y x y x y x y y

24.

6 6 5 4 2 3 3 2 42 2 2 2 2 2 2 2 2 2 2

5 62 2 2

12 10 2 8 4 6 6 4 8 2 10 12

6 6 6 6 6

0 1 2 3 4

6 6

5 6

6 15 20 15 6

x y x x y x y x y x y

x y y

x x y x y x y x y x y y

25.

6 6 5 1 4 2 3 3

2 4 5 6

3 5/ 2 2 3/ 2 1/ 2

3 5/ 2 2 3/ 2 1/ 2

6 6 6 62 2 2 2

0 1 2 3

6 6 6 2 2 2

4 5 6

6 2 15 2 20 2 2 15 4 6 4 2 8

6 2 30 40 2 60 24 2 8

x x x x x

x x

x x x x x x

x x x x x x

26. 4 4 3 1 2 2 3 4

2 3/ 2 1/ 2

2 3/ 2 1/ 2

4 4 4 4 43 3 3 3 3

0 1 2 3 4

4 3 6 3 4 3 3 9

4 3 18 12 3 9

x x x x x

x x x x

x x x x

27. 5 5 4 3 2 2 3 4 5

5 5 4 4 3 3 2 2 2 2 3 3 4 4 5 5

5 5 5 5 5 5

0 1 2 3 4 5

5 10 10 5

ax by ax ax by ax by ax by ax by by

a x a x by a x b y a x b y axb y b y

28. 4 4 3 2 2 3 4

4 4 3 3 2 2 2 2 3 3 4 4

4 4 4 4 4( )

0 1 2 3 4

4 6 4

ax by ax ax by ax by ax by by

a x a x by a x b y axb y b y


1313


29. 10, 4, , 3n j x x a

6 4 6 6

6

10 10! 10 9 8 73 81 81

4 4!6! 4 3 2 1

17,010

x x x

x

6The coefficient of is 17,010.x

30. 10, 7, , 3n j x x a

3 7 3

3

3

10 10!( 3) 2187

7 7!3!

10 9 82187

3 2 1

262,440

x x

x

x


31. 12, 5, 2 , 1n j x x a

7 5 7

7

7

12 12!(2 ) ( 1) 128 ( 1)

5 5!7!

12 11 10 9 8( 128)

5 4 3 2 1

101,376

x x

x

x


32. 12, 9, 2 , 1n j x x a

3 9 3

3

3

12 12!(2 ) (1) 8 (1)

9 9!3!

12 11 108

3 2 1

1760

x x

x

x

3The coefficient of is 1760.x

33. 9, 2, 2 , 3n j x x a

7 2 7

7

7

9 9!(2 ) 3 128 (9)

2 2!7!

9 8128 9

2 1

41, 472

x x

x

x


34. 9, 7, 2 , 3n j x x a

2 7 2

2

2

9 9!(2 ) ( 3) 4 ( 2187)

7 7! 2!

9 84 2187

2 1

314,928

x x

x

x


35. 7, 4, , 3n j x x a

3 4 3 3 37 7! 7 6 53 81 81 2835

4 4!3! 3 2 1x x x x

36. 7, 2, , 3n j x x a

5 2 5 5 57 7! 7 6( 3) 9 9 189

2 2!5! 2 1x x x x

37. 9, 2, 3 , 2n j x x a

7 2 7

7 7

9 9!(3 ) ( 2) 2187 4

2 2!7!

9 88748 314,928

2 1

x x

x x

38. 8, 5, 3 , 2n j x x a

3 5 3

3 3

8 8!(3 ) ( 2) 27 32

5 5!3!

8 7 6864 48,384

3 2 1

x x

x x

39. The 0x term in

12 12122 24 3

0 0

12 121j

j j

j j

x xj jx

occurs when: 24 3 0

24 3

8

j

j

j

The coefficient is 12 12! 12 11 10 9

4958! 4! 4 3 2 18


1314


40. The 0x term in

9 9

9 9 32

0 0

9 911

jj j j

j j

x xj jx

occurs when: 9 3 0

9 3

3

j

j

j

The coefficient is

39 9! 9 8 71 84

3 3!6! 3 2 1

41. The 4x term in

310 10 1010 2

0 0

10 1022

jjj j

j j

x xj jx

occurs when: 3

10 423

62

4

j

j

j

The coefficient is

410 10! 10 9 8 72 16 16 3360

4 6! 4! 4 3 2 1

42. The 2x term in

8 88 4

0 0

8 833

jj j j

j j

x xj jx

occurs when: 4 2

2

2

j

j

j

The coefficient is

28 8! 8 73 9 9 252

2 6! 2! 2 1

43. 5 2 35 3 5 4 3 3 3 2 35 5 5 5(1.001) 1 10 1 1 10 1 10 1 10

0 1 2 3

1 5(0.001) 10(0.000001) 10(0.000000001)

1 0.005 0.000010 0.000000010

1.00501 (correct to 5 decimal places)

44.

6 2 36 6 5 4 36 6 6 6(0.998) 1 0.002 1 1 ( 0.002) 1 0.002 1 0.002

0 1 2 3

1 6 0.002 15 0.000004 20 0.000000008 ...

1 0.012 0.00006 0.00000016

0.98806 (correct to 5 decimal places)

45.

1 !! !

1 1 !( 1 )! 1 !(1)! 1 !

n nn n nn

n n n n n n

! ! ! !1

!( )! !0! ! 1 !

n n n n n

n n n n n n n

46.

! ! !

( )! ( ( ))! ! ! !( )!

n nn n n

n j jn j n n j n j j j n j


1315


47. Show that 20 1

nn n n

n

1 2 2

2 (1 1)

1 1 1 1 1 1 10 1 2

0 1

n n

n n n n n nn n n n

n

n n n

n

48. Show that ( 1) 00 1 2

nn n n n

n

1 2 2

0 (1 1)

1 1 ( 1) 1 ( 1) 1 ( 1)0 1 2

( 1)0 1 2

n

n n n n n n

n

n n n n

n

n n n n

n

49. 5 4 3 2 2 3 4 5 5

55 5 5 5 5 51 1 3 1 3 1 3 1 3 3 1 3(1) 1

0 1 2 3 4 54 4 4 4 4 4 4 4 4 4 4 4

50. 812! 479,001,600 4.790016 10 18

25

20! 2.432902008 10

25! 1.551121004 10

12

2018

2525

12 112! 2 12 1 479,013,972.4

12 12 1

20 120! 2 20 1 2.43292403 10

12 20 1

25 125! 2 25 1 1.551129917 10

12 25 1

e

e

e

51. 1

1

6 5

ln 6 ln 5

ln 6 ( 1) ln 5

ln 6 ln 5 ln 5

ln 6 ln 5 ln 5

x x

x x

x x

x x

x x

(ln 6 ln 5) ln 5

ln 58.827

ln 6 ln 5

x

x

52. a. (2)(3) (3)( 2) 0 v w

b. 1cos (0) 90

c. The vectors are orthogonal.

53.

0

2 3 1

4 2 12

x y z

x y z

x y z

Add the first equation and the second equation to eliminate y:

0

2 3 1

3 2 1

x y z

x y z

x z


1316


Multiply each side of the first equation by 2 and add to the third equation to eliminate y:

2 2 2 0

4 2 12

6 3 12

2 4

x y z

x y z

x z

x z

Multiple the second derived equation by 2 and add the two results to eliminate z: 3 2 1

4 2 8

7 7

1

x z

x z

x

x

Substituting and solving for the other variables:

2(1) 4

2

2

z

z

z

1 ( 2) 0

1 2 0

3

3

y

y

y

y

The solution is 1, 3, 2x y z or (1,3, 2) .

54.

0

0

6

2 10

x

y

x y

x y

Graph 0; 0x y . Shaded region is the first

quadrant. Graph the line 6x y . Use a solid

line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 0 0 6 is true, shade the side of the line containing (0, 0). Graph the line 2 10x y .

Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 2(0) 0 10 is true, shade the side of the

line containing (0, 0). The overlapping region is the solution. The graph is bounded. Find the vertices:

The x-axis and y-axis intersect at (0, 0). The intersection of 6x y and the y-axis is (0, 6).

The intersection of 2 10x y and the x-axis is

(5, 0). To find the intersection of 2 10x y

and 6x y , solve the system:

6

2 10

x y

x y

Solve the first equation for x: 6x y .

Substitute and solve: 2(6 ) 10

12 2 10

12 10

2

2

y y

y y

y

y

y

6 (2) 4x

The point of intersection is (4, 2). The four corner points are (0, 0), (0, 6), (5, 0), and (4, 2).

Chapter 12 Review Exercises

1. 1 2 3 4 51 2 3 4 5

1 3 4 2 3 5 3 3 6 4 3 7 5 3 8( 1) , ( 1) , ( 1) , ( 1) , ( 1)

1 2 3 2 2 4 3 2 5 4 2 6 5 2 7a a a a a

2. 1 2 3 4 5

1 2 3 4 52 2 2 2 2

2 2 2 4 2 8 2 16 2 322, 1, , 1,

1 4 9 16 251 2 3 4 5c c c c c

3. 1 2 3 4 52 2 4 2 4 8 2 8 16

3, 3 2, 2 , ,3 3 3 3 3 9 3 9 27

a a a a a


1317


4. 1 2 3 4 52, 2 2 0, 2 0 2, 2 2 0, 2 0 2a a a a a

5. 4

1

(4 2) 4 1 2 4 2 2 4 3 2 4 4 2 6 10 14 18 48k

k

6. 13

1

1

1 1 1 1 11 1

2 3 4 13k

k k

7. 5na n Arithmetic

( 1 5) ( 5) 6 5 1

6 5 ( 11)2 2n

d n n n n

n nS n n

8. 32nc n Examine the terms of the

sequence: 2, 16, 54, 128, 250, ... There is no common difference; there is no common ratio; neither.

9. 32 nns Geometric

3( 1) 3 3

3 3 3 33 3

2 22 2 8

2 2

n nn n

n nr

1 8 1 8 88 8 8 1

1 8 7 7

n nn

nS

10. 0, 4, 8, 12, ... Arithmetic 4 0 4d

2(0) ( 1)4 4( 1) 2 ( 1)2 2nn n

S n n n n

11. 3 3 3 3

3, , , , , ...2 4 8 16

Geometric

33 1 12

3 2 3 2r

1 11 1

12 23 3 6 1

1 1 212 2

n n

n

nS

12. Neither. There is no common difference or common ratio.

13.

30 30 302 2

1 1 1

( 2) 2

30 30 1 2 30 12(30) 9515

6

k k k

k k

14. 40 40 40

1 1 1

40 40

1 1

( 2 8) 2 8

2 8

40(1 40)2 40(8)

2

1640 320 1320

k k k

k k

k k

k

15.

7 7

7

1

1 11 1

1 1 13 31 23 3 313 3

1 11

2 2187

1 2186 10930.49977

2 2187 2187

k

k

16.

1010

1

1 22 2

1 ( 2)

1 1024 22 1023

3 3

682

k

k

17. Arithmetic

1 13, 4, ( 1)na d a a n d

9 3 (9 1)4 3 8(4) 3 32 35a

18. Geometric

11 1

11, , 11;

10n

na r n a a r

11 1 10

111 1

110 10

1

10,000,000,000

a

19. Arithmetic

1 12, 2, 9, ( 1)na d n a a n d

9 2 (9 1) 2 2 8 2

9 2 12.7279

a


1318


20. 7 1 20 16 31 19 96a a d a a d ;

Solve the system of equations:

1

1

6 31

19 96

a d

a d

Subtract the second equation from the first equation and solve for d.

13 65

5

d

d

1 31 6(5) 31 30 1a

1 1

1 1 5

1 5 5

5 4

na a n d

n

n

n

General formula: 5 4na n

21. 10 1 18 19 0 17 8a a d a a d ;

Solve the system of equations:

1

1

9 0

17 8

a d

a d

Subtract the second equation from the first equation and solve for d.

8 8

1

d

d

1 9(1) 9a

1 1

9 1 1

9 1

10

na a n d

n

n

n

General formula: 10na n

22. 11

3,3

a r

1


3 3 91 21 2

13 3

n

r

aS

r

23. 11

2,2

a r

Since 1r , the series converges.

1 2 2 431 31

122

na

Sr

24. 1

1

2a ,

3

2r


25. 11

4,2

a r

Since 1r , the series converges.

1 4 48

1 111

2 2

na

Sr

26. I: 3 1

1: 3 1 3 and (1 1) 32

n

II: If 3

3 6 9 3 ( 1)2

kk k , then

3 6 9 3 3( 1)

3 6 9 3 3( 1)

3( 1) 3( 1)

23 3( 1)

( 1) 3 (( 1) 1)2 2

k k

k k

kk k

k kk k


27. I: 1 1 11: 2 3 2 and 3 1 2n

II: If 12 6 18 2 3 3 1k k , then 1 1 1

1

1

2 6 18 2 3 2 3

2 6 18 2 3 2 3

3 1 2 3 3 3 1 3 1

k k

k k

k k k k



1319


28. I: 1:n

2 21(3 1 2) 1 and 1(6 1 3 1 1) 1

2

II: If

2 2 2 211 4 (3 2) 6 3 1

2k k k k ,

then

22 2 2 2

2 2 2 2 2

2 2

3 2 2

3 2

2

3 2 2

2

1 4 7 (3 2) 3( 1) 2

1 4 7 (3 2) (3 1)

16 3 1 (3 1)

21

6 3 18 12 221

6 15 11 221

( 1) 6 9 221

6 6 9 9 2 221

6 1 9 1 2 12

k k

k k

k k k k

k k k k k

k k k

k k k

k k k k k

k k k k k

2

2

2

1( 1) 6 12 6 3 3 1

21

( 1) 6( 2 1) 3( 1) 121

( 1) 6( 1) 3( 1) 12

k k k k

k k k k

k k k


29. 5 5! 5 4 3 2 1 5 4

102 2!3! 2 1 3 2 1 2 1

30. 5 5 4 3 2 2 3 1 4 5

5 4 3 2

5 4 3 2

5 5 5 5 5 5( 2) 2 2 2 2 2

0 1 2 3 4 5

5 2 10 4 10 8 5 16 1 32

10 40 80 80 32

x x x x x x

x x x x x

x x x x x

31. 4 4 3 2 2 3 4

4 3 2

4 3 2

4 4 4 4 4(3 4) (3 ) (3 ) ( 4) (3 ) ( 4) (3 )( 4) ( 4)

0 1 2 3 4

81 4 27 ( 4) 6 9 16 4 3 ( 64) 1 256

81 432 864 768 256

x x x x x

x x x x

x x x x


1320


32. 9, 2, , 2n j x x a

7 2 7 7 7

7

9 9! 9 82 4 4 144

2 2!7! 2 1

The coefficient of is 144.

x x x x

x

33. 7, 5, 2 , 1n j x x a

2 5 2 2 2

2

7 7! 7 6(2 ) 1 4 (1) 4 84

5 5! 2! 2 1

The coefficient of is 84.

x x x x

x

34. This is an arithmetic sequence with

1 80, 3, 25a d n

a. 25 80 (25 1)( 3) 80 72 8 bricksa

b. 2525

(80 8) 25(44) 1100 bricks2

S

1100 bricks are needed to build the steps.

35. This is an arithmetic sequence with

1 30, 1, 15na d a

15 30 ( 1)( 1)

15 1

16

16

n

n

n

n

1616

(30 15) 8(45) 360 tiles2

S

360 tiles are required to make the trapezoid.


13

20,4

a r .

a. After striking the ground the third time, the

height is 3

3 13520 8.44 feet

4 16

.

b. After striking the ground the thn time, the

height is 3

20 feet4

n

.

c. If the height is less than 6 inches or 0.5 feet, then:

30.5 20

4

30.025

4

3log 0.025 log

4

log 0.02512.82

3log

4

n

n

n

n

The height is less than 6 inches after the 13th strike.

d. Since this is a geometric sequence with 1r , the distance is the sum of the two

infinite geometric series - the distances going down plus the distances going up. Distance going down:

20 2080

3 11

4 4

downS

feet.

Distance going up: 15 15

603 1

14 4

upS

feet.

The total distance traveled is 140 feet.


12 20 240n payment periods. The


12. Thus,

2400.10

1 112

200 $151,873.770.10

12

A


1 50,000, 1.04, 5a r n . Find the fifth

term of the sequence: 5 1 4

5 50,000(1.04) 50,000(1.04)

58,492.93

a

Her salary in the fifth year will be $58,492.93.

Chapter 12 Test

1321


Chapter 12 Test

1. 2 1

8nn

an

2

1

2

2

2

3

2

4

2

5

1 1 00

1 8 9

2 1 32 8 10

3 1 83 8 11

4 1 15 54 8 12 4

5 1 245 8 13

a

a

a

a

a

The first five terms of the sequence are 0, 3

10,

811

, 54

, and 2413

.

2. 1 14; 3 2n na a a

2 1

3 2

4 3

5 4

3 2 3 4 2 14

3 2 3 14 2 44

3 2 3 44 2 134

3 2 3 134 2 404

a a

a a

a a

a a

The first five terms of the sequence are 4, 14, 44, 134, and 404.

3.

2 2 2

31

21

1 1 2 1 3 1

2 3 4

1 1 2 1 3 1

1 2 3

11

1 1 1

2 3 41 1 1

1 4 9

3 4 612

4 9 36

k

k

k

k

4.

1 2 3 4

4

1

2 2 2 21 2 3 4

3 3 3 3

23

2 4 8 161 2 3 4

3 9 27 81130 680

1081 81

k

k

k

5. 2 3 4 11

...5 6 7 14

Notice that the signs of each term alternate, with the first term being negative. This implies that the general term will include a power of 1 . Also note that the numerator is always 1 more than the term number and the denominator is 4 more than the term number. Thus, each term is in

the form 11

4k k

k

. The last numerator is 11

which indicates that there are 10 terms.

10

1

2 3 4 11 1... 1

5 6 7 14 4k

k

kk

6. 6,12,36,144,... 12 6 6 and 36 12 24 The difference between consecutive terms is not constant. Therefore, the sequence is not arithmetic. 126 2 and 36

12 3

The ratio of consecutive terms is not constant. Therefore, the sequence is not geometric.

7. 1

42

nna

11 12 2

1 11 11 2 2

4 4 44

4 4

n nn

n nn

aa

Since the ratio of consecutive terms is constant, the sequence is geometric with common ratio

4r and first term 11

14 2

2a .

The sum of the first n terms of the sequence is given by

111

1 42

1 42

1 43

n

n

n

n

rS a

r

8. 2, 10, 18, 26,...

10 2 8 , 18 10 8 ,

26 18 8

The difference between consecutive terms is constant. Therefore, the sequence is arithmetic with common difference 8d and first term

1 2a .


1322


1 1

2 1 8

2 8 8

6 8

na a n d

n

n

n


2

2 6 82

4 82

2 4

n nn

S a a

nn

nn

n n

9. 72nn

a

1

17 7

2 2

17 7

2 212

n n

nna a

n n

The difference between consecutive terms is constant. Therefore, the sequence is arithmetic

with common difference 12

d and first term

11 13

72 2

a .


12

137

2 2 2

272 2 2

274

n nn

S a a

n n

n n

nn

10. 8

25,10, 4, ,...5

10 225 5

, 4 2

10 5 ,

85 8 1 24 5 4 5

The ratio of consecutive terms is constant. Therefore, the sequence is geometric with

common ratio 25r and first term 1 25a .

The sum of the first n terms of the sequence is

given by

1

22 11 55125 25

2 31 15 5

5 2 125 225 1 1

3 5 3 5

nn

n

n

n n

rS a

r

11. 2 32 1nn

an

1

2 2

2

2

2 1 32 3 2 3 2 52 1 2 1 2 12 1 1

2 3 2 1 2 5 2 1

2 1 2 1

4 8 3 4 8 5

4 18

4 1

n n

nn n na a

n n nn

n n n n

n n

n n n n

n

n

The difference of consecutive terms is not constant. Therefore, the sequence is not arithmetic.

1

2 32 3 2 12 3 2 12 1

2 1 2 52 1 3 2 1 2 5

2 1 1

n

n

nn na n nn

a n nn n n

n

The ratio of consecutive terms is not constant. Therefore, the sequence is not geometric.

12. For this geometric series we have 64 1

256 4r

and 1 256a . Since 1 1

14 4

r , the series

converges and we get

1

5144

256 256 10241 51

aS

r

Chapter 12 Test

1323


13. 5 5 4 3 2 2 3 4 5

5 4 3 2

5 4 3 2

5 5 5 5 5 53 2 3 3 2 3 2 3 2 3 2 2

0 1 2 3 4 5

243 5 81 2 10 27 4 10 9 8 5 3 16 32

243 810 1080 720 240 32

m m m m m m

m m m m m

m m m m m

14. First we show that the statement holds for 1n .

11 1 1 2

1

The equality is true for 1n so Condition I holds. Next we assume that 1 1 1 1

1 1 1 ... 1 11 2 3

nn

is

true for some k, and we determine whether the formula then holds for 1k . We assume that

1 1 1 11 1 1 ... 1 1

1 2 3k

k

.

Now we need to show that 1 1 1 1 11 1 1 ... 1 1 1 1 2

1 2 3 1k k

k k

.

We do this as follows:

1 1 1 1 1 1 1 1 1 11 1 1 ... 1 1 1 1 1 ... 1 1

1 2 3 1 1 2 3 1

11 1 (using the induction assumption)

1

11 1 1 1 1

12

k k k k

kk

k k kk

k

Condition II also holds. Thus, formula holds true for all natural numbers.

15. The yearly values of the car form a geometric sequence with first term 1 31,000a and

common ratio 0.85r (which represents a 15% loss in value).

131,000 0.85

nna

The nth term of the sequence represents the value of the car at the beginning of the nth year. Since we want to know the value after 10 years, we are looking for the 11th term of the sequence. That is, the value of the car at the beginning of the 11th year.

1011 111 1 31,000 0.85 6,103.11a a r

After 10 years, the car will be worth $6,103.11.

16. The weights for each set form an arithmetic sequence with first term 1 100a and common

difference 30d . If we imagine the weightlifter only performed one repetition per set, the total weight lifted in 5 sets would be the sum of the first five terms of the sequence.

1

5

1

100 5 1 30 100 4 30 220

na a n d

a

55 52 2

2

100 220 320 800

n nn

S a a

S

Since he performs 10 repetitions in each set, we multiply the sum by 10 to obtain the total weight lifted.

10 800 8000

The weightlifter will have lifted a total of 8000 pounds after 5 sets.


1324


Chapter 12 Cumulative Review

1. 2 9x

2 29 or 9

3 or 3

x x

x x i

2. a. graphing 2 2 100x y and 23y x .

b. solving

32 2 2 2

2 2

2

100 3 3 300

3 3 0

3 300

x y x y

y x x y

y y

2 23 300 3 300 0y y y y

21 1 4 3 300 1 3601

2 3 6y

Substitute and solve for x:

2

2

1 3601 1 36013

6 6

1 3601 1 3601

18 18

y x

x x

or

2

2

1 3601 1 36013

6 6

1 3601 1 3601

18 18

1 3601undefined since 0

18

y x

x x

Therefore, the system has solutions

1 3601 1 3601,

18 6x y

and

1 3601 1 3601,

18 6x y

.

1 3601 1 3601, ,

18 6

1 3601 1 3601,

18 6

c. The graphs intersect at the points

1 3601 1 3601, 1.81,9.84

18 6

1 3601 1 3601,

18 61.81,9.84

3. 2 5xe

5

25

ln ln2

5ln 0.916

2

x

x

e

e

x

The solution set is 5

ln2

.

4. slope 5m ; Since the x-intercept is 2, we

know the point 2,0 is on the graph of the line

and is a solution to the equation 5y x b .

5

0 5 2

0 10

10

y x b

b

b

b

Therefore, the equation of the line with slope 5 and x-intercept 2 is 5 10y x .

5. Given a circle with center (–1, 2) and containing the point (3, 5), we first use the distance formula to determine the radius.

2 2

2 2

3 1 5 2

4 3 16 9

25

5

r

Chapter 12 Cumulative Review

1325


Therefore, the equation of the circle is given by

2 2 2

2 2 2

2 2

2 2

1 2 5

1 2 5

2 1 4 4 25

2 4 20 0

x y

x y

x x y y

x y x y

6. 3

2

xf x

x

, 2 1g x x

a.

2 2 2 1 5

3 5 155 5

5 2 3

g

f

2 2 5 5f g f g f

b.

3 4 124 6

4 2 26 2 6 1 13

f

g

4 4 6 13g f g f g

c.

3 2 1

2 1 2

6 3

2 1

f g x f g x

x

x

x

x

d. To determine the domain of the composition

f g x , we start with the domain of g

and exclude any values in the domain of g that make the composition undefined.

g x is defined for all real numbers and

f g x is defined for all real numbers

except 1

2x . Therefore, the domain of the

composite f g x is 1

|2

x x

.

e. 32 1

2

61

26 2

27 2

2

xg f x

x

x

xx x

xx

x

f. To determine the domain of the composition

g f x , we start with the domain of f

and exclude any values in the domain of f that make the composition undefined.

f x is defined for all real numbers except

2x and g f x is defined for all real

numbers except 2x . Therefore, the

domain of the composite g f x is

| 2x x .

g. 2 1g x x

2 1

2 1

1 2

1

2

y x

x y

x y

xy

1 1

2

xg x

The domain of 1g x is the set of all real

numbers.

h. 3

2

xf x

x

3

23

2

2 3

2 3

3 2

3 2

2

3

xy

xy

xy

x y y

xy x y

xy y x

y x x

xy

x

1 2

3

xf x

x

The domain of 1f x is | 3x x .

7. Center: (0, 0); Focus: (0, 3); Vertex: (0, 4); Major axis is the y-axis; 4; 3a c .

Find b: 2 2 2 16 9 7 7b a c b Write the equation using rectangular coordinates:

2 2

17 16

x y


1326


8. The focus is 1,3 and the vertex is 1, 2 .

Both lie on the vertical line 1x . We have a = 1 since the distance from the vertex to the focus is 1 unit, and since 1,3 is above

1, 2 , the parabola opens up. The equation of

the parabola is:

2

2

2

4

1 4 1 2

1 4 2

x h a y k

x y

x y

9. Center point (0, 4); passing through the pole (0,4) implies that the radius = 4 using rectangular coordinates:

2 2 2

2 2 2

2 2

2 2

0 4 4

8 16 16

8 0

x h y k r

x y

x y y

x y y

converting to polar coordinates: 2

2

8 sin 0

8 sin

8sin

r r

r r

r

10. 22sin sin 3 0, 0 2x x x

2sin 3 sin 1 0x x

32sin 3 0 sin , which is impossible

23

sin 1 0 sin 12

x x

x x x

Solution set 3

2

.

11. 1cos 0.5

We are finding the angle , , whose

cosine equals 0.5 .

1

cos 0.5

2 2cos 0.5

3 3

12. 1

sin , is in Quadrant II4

a. is in Quadrant II cos 0 2

2 1cos 1 sin 1

4

1 15 151

16 16 4

b.

1sin 1 44

tancos 4 1515

4

1 15 15

1515 15

c. sin(2 ) 2sin cos

1 15 152

4 4 8

d. 2 2

2 2

cos(2 ) cos sin

15 1

4 4

15 1 14 7

16 16 16 8

e. 2 4 2 2

is in Quadrant I2

sin 0

2

151

41 cossin

2 2 2

4 15

4 4 15

2 8

4 15

2 2

Chapter 12 Projects

1327


Chapter 12 Projects

Project I – Internet-based Project

Answers will vary based on the year that is used. Data used in these solutions will be from 2008.

1. I = net immigration = 887,168 Population for 2008 = 303,824,640

2. r = 0.01416 – 0.00826 = 0.0059

3. 1

1

0

(1 0.0059) 887168

(1.0059) 887168

303824640

n n

n n

p p

p p

p

4. 1 0

1

1

(1.0059) 887168

(1.0059)(303824640) 887168

306,504,373

p p

p

p

The population is predicted to be 306,504,373 in 2009.

5. Actual population in 2009: 307,212,123. The formula’s prediction was lower but fairly close.

6. Birth rate: 48.12 per 1000 population (0.04812) Death rate: 12.64 per 1000 population (0.01264) Population for 2008: 31,367,972 I = net immigration = 6587

0.04812 0.01264 0.03548r

1

1

0

(1 0.03548) 6587

(1.03548) 6587

31,367,972

n n

n n

p p

p p

p

1 0

1

1

(1.03548) 6587

(1.03548)(31367972) 6587

32,474,321

p p

p

p

The population is predicted to be 32,474,321 in 2009.

Actual population in 2009: 32,369,558. The formula’s prediction was higher but fairly close.

7. Answers will vary. This appears to support the article. The growth rate for the U.S. is much smaller than the growth rate for Uganda.

8. It could be but one must consider trends in each of the pieces of data to find if the growth rate is increasing or decreasing over time. The same thing must be examined with respect to the net immigration.

Project II

1. 2, 4, 8, 16, 32, 64

2. length n 2n levels

This is a geometric sequence: 2nna

Recursive expression: 1 02 , 1n na a a

3. 8

256 2

2 2

8

n

n

n

Project III

1. 1

0

1 1

1 0

3 2 , 18 3

2, 2, 3, 18

3 3

3 18 2 21 2

3 32

7 , 23

st t dt t

t tt

t t

Q P Q P

P b d c

a a

P PP

P P P

2.

tP

1tP

3. 1

1 1

1 1

17

317

3 2(2) 18 33

1 1

s d

s d

P

Q Q

Q Q


1328


2

2 2

2 2

29

917 29

3 2 18 33 9

25 25

3 3

s d

s d

P

Q Q

Q Q

The market (supply and demand) are getting closer to being the same.

4. The equilibrium price is 4.20.

5. It takes 17 time periods.

6. 17

17

18 3(4.20) 5.40

3 2(4.20) 5.40d

s

Q

Q

The equilibrium quantity is 5.4.

Project IV

1. 1, 2, 4, 7, 11, 16, 22, 29

2. It is not arithmetic because there is no common difference. It is not geometric because there is no common ration.

3. Scatter diagram

2

62

4. 2.5 2.5y x

The graph does not pass through any of the points.

6

7

8

12.5

15

17.5

y

y

y

1

2

3

4

5

0

2.5

5

7.5

10

y

y

y

y

y

5

1

( )

(0 1) (2.5 2) (5 4) (7.5 7) (10 11)

1

ir ii

y y

This is the sum of the errors.

5. 20.5 0.5 1y x x

The graph passes through all of the points.

6

7

8

16

22

29

y

y

y

1

2

3

4

5

1

2

4

7

11

y

y

y

y

y

5

1

( ) 0ir i

i

y y

The sum of the errors is zero.

6. When trying to obtain the cubic and quartic polynomials of best fit, the cubic and quartic terms have coefficient zero and the polynomial of best fit is given as the quadratic in part e. For the exponential function of best fit,

(0.59)(1.83)xy .

6 7 822.2 40.6 74.2y y y

The sum of these errors becomes quite large. This error shows that the function does not fit the data very well as x gets larger.

7. The quadratic function is best.

8. The data does not appear to be either logarithmic or sinusoidal in shape, so it does not make sense to try to fit one of those functions to the data.

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Chapter 12 Sequences; Induction; the Binomial Theorem · 2017-12-11 · 102. When the last deposit is made at the …