Chapter 11 Sequences, Induction, and Probability...2013/07/31 · Chapter 11 Sequences, Induction,...

Chapter 11 Sequences, Induction, and Probability

1324 Copyright © 2014 Pearson Education, Inc.

Section 11.1

Check Point Exercises

1. a.

1

2

3

4

2 5

2(1) 5 7

2(2) 5 9

2(3) 5 11

2(4) 5 13

na n

a

a

a

a

The first four terms are 7, 9, 11, and 13.

b.

1

1 1

2

2 2

3

3 3

4

4 4

( 1)

2 1

( 1) 1 1

3 32 1

( 1) 1

52 1

( 1) 1 1

9 92 1

( 1) 1

172 1

n

n na

a

a

a

a

The first four terms are 1 1 13 5 9, , , 1

17and .

2. 1 13and 2 5n na a a for 2n

2 1

3 2

4 3

2 5

2(3) 5 11

2 5

2(11) 5 27

2 5

2(27) 5 59

a a

a a

a a


3.

1

2

3

20

( 1)!

20 2010

(1 1)! 2!

20 20 20 10

(2 1)! 3! 6 3

20 20 20 5

(3 1)! 4! 24 6

nan

a

a

a

420 20 20 1

(4 1)! 5! 120 6a

The first four terms are 10 5 13 6 610, , ,and .

4. a. 14! 14 13 12! 14 13

912! 12! 2! 12! 2 1

b. ! ( 1)!

( 1)! ( 1)!

n n nn

n n

5. a. 6

2

1

2 2 2

2 2 2

2

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

2(4) 2(5) 2(6)

2 8 18 32 50 72

182

i

i

b.

5

3

3 4 5

2 3

2 3 2 3 2 3

8 3 16 3 32 3

5 13 29

47

k

k

c. 5

1

4

4 4 4 4 4

20

i

6. a. The sum has nine terms, each of the form 2i , starting at 1i and ending at 9i .

92 2 2 2 2

1

1 2 3 9i

i

b. The sum has n terms, each of the form 11

2i ,

starting at 1i and ending at i n .

1 11

1 1 1 1 11

2 4 8 2 2

n

n ii

Section 11.1 Sequences and Summation Notation

Copyright © 2014 Pearson Education, Inc. 1325

Concept and Vocabulary Check 11.1

1. sequence; integers; terms

2. general

3. 4

4. 1

15

5. 2

6. factorial; 5; 1; 1

7. 3n

8. 1 2 3; ; ; na a a a ; index; upper limit; lower limit

Exercise Set 11.1

1.

1

2

3

4

3 2

3 1 2 5

3 2 2 8

3 3 2 11

3 4 2 14

na n

a

a

a

a


2.

1

2

3

4

4 1

4(1) 1 3

4(2) 1 7

4(3) 1 11

4(4) 1 15

na n

a

a

a

a


3. 3nna

11

22

33

44

3 3

3 9

3 27

3 81

a

a

a

a


4. 1

3

n

na

1

1

2

2

3

3

4

4

1 1

3 3

1 1

3 9

1 1

3 27

1 1

3 81

a

a

a

a

The first four terms are 1 1 1 1

, , , and 3 9 27 81

.

5. 3n

na

11

22

33

44

3 3

3 9

3 27

3 81

a

a

a

a

The first four terms are –3, 9, –27, and 81.

6. 1

3

n

na

1

1

2

2

3

3

4

4

1 1

3 3

1 1

3 9

1 1

3 27

1 1

3 81

a

a

a

a


, , , and 3 9 27 81

.

7. 1 3n

na n

11

22

33

44

1 1 3 4

1 2 3 5

1 3 3 6

1 4 3 7

a

a

a

a

The first four terms are –4, 5, –6, and 7.


1326 Copyright © 2014 Pearson Education, Inc

8. 11 4

nna n

1 11

2 12

3 13

4 14

1 1 4 5

1 2 4 6

1 3 4 7

1 4 4 8

a

a

a

a

The first four terms are 5, –6, 7, and –8.

9. 2

4nn

an

1

2

3

4

2 1 2

1 4 52 2 4 2

2 4 6 32 3 6

3 4 72 4 8

14 4 8

a

a

a

a

The first four terms are 62 25 3 7, , , and1.

10. 3

5nn

an

1

2

3

4

3(1) 3 1

1 5 6 2

3(2) 6

2 5 7

3(3) 9

3 5 8

3(4) 12 4

4 5 9 3

a

a

a

a


, , , and 2 7 8 3

.

11. 1

1

2 1

n

n na

1 1

1 1

1 1

12 1a

n = 1

2 1

2 2

1 1

32 1a

3 1

3 3

1 1

72 1a

4 1

4 4

1 1

152 1a

The first four terms are 1 1 13 7 151, , , and .

12. 1( 1)

2 1

n

n na

1 1

1 1

2 1

2 2

3 1

3 3

4 1

4 4

( 1) 1

32 1

( 1) 1

52 1

( 1) 1

92 1

( 1) 1

172 1

a

a

a

a

The first four terms are 1 1 1 13 5 9 17, , , and .

13. 1 17 and +5 for 2n na a a n

2 1

3 2

4 3

5 7 5 12

5 12 5 17

5 17 5 22

a a

a a

a a


14. 1 1

2 1

3 2

4 3

12 and 4 for 2

4 12 4 16

4 16 4 20

4 20 4 24

n na a a n

a a

a a

a a


15. 1 13 and 4 for 2n na a a n

2 1

3 2

4 3

4 4 3 12

4 4 12 48

4 4 48 192

a a

a a

a a


16. 1 1

2 1

3 2

4 3

2 and 5 for 2

5 5(2) 10

5 5(10) 50

5 5(50) 250

n na a a n

a a

a a

a a


17. 1 14 and 2 3n na a a

2

3

4

2 4 3 11

2 11 3 25

2 25 3 53

a

a

a




18. 1 15 and 3 1n na a a

2 3 5 1 14a

3 3 14 1 41a

4 3 41 1 122a


19. 2

!nn

an

2

1

2

2

2

3

2

4

3 22 3

11

1!

22

2!

3 9 3

3! 6 2

4 16 2

4! 24 3

The first four terms are 1, 2, , and .

a

a

a

a

20.

2

1 !n

na

n

1 2

1 1 !2

1a

2 2

2 1 ! 3! 6 3

4 4 22a

3 2

3 1 ! 4! 24 8

9 9 33a

4 2

4 1 ! 5! 120 15

16 16 24a

The first four terms are 3 8 152 3 22, , , and .

21.

1

2

3

4

2( 1)!

2(1 1)! 2(2) 4

2(2 1)! 2(6) 12

2(3 1)! 2(24) 48

2(4 1)! 2(120) 240

na n

a

a

a

a


22. 2 1 !na n

1 2 1 1 ! 2(1) 2a

2 2 2 1 ! 2(1) 2a

3 2 3 1 ! 2(2) 4a

4 2 4 1 ! 2(6) 12a

The first four terms are –2, –2, –4, and –12.

23. 17! 17 16 15!

17 16 27215! 15!

24. 18! 18 17 16!

18 17 30616! 16!

25. 16! 16 15 14! 16 15 8 15

1202! 14! 2!14! 2 1 1

26. 20! 20 19 18! 20 19 10 19

1902! 18! 2!18! 2 1 1

27. ( 2)! ( 2)( 1) !

( 2)( 1)! !

n n n nn n

n n

28.

2 1 ! 2 1 2 !2 1

2 ! 2 !

n n nn

n n

29. 6

1

5 5 1 5 2 5 3 5 4 5 5 5 6

5 10 15 20 25 30

105

i

i

30. 6

1

7 7 1 7 2 7 3 7 4 7 5 7 6i

i

7 14 21 28 35 42 147

31. 4

2 2 2 2 2

1

2 2 1 2 2 2 3 2 4

2 8 18 32

60

i

i

32. 5

3 3 3 3 3 3

1

1 2 3 4 5i

i

1 8 27 64 125 225

33. 5

1

( 4) 1(5) 2(6) 3(7) 4(8) 5(9)

5 12 21 32 45

115

k

k k

34.

4

1

3 2

1 3 1 2 2 3)(2 2

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

( 2)(3) ( 1)(4) (0)(5) (1)(6)

4

k

k k



35. 1 2 3 44

1

1 1 1 1 1

2 2 2 2 2

1 1 1 1

2 4 8 165

16

i

i

36. 2 3 44

2

1 1 1 1

3 3 3 3

1 1 1

9 27 817

81

i

i

37. 9

5

11 11 11 11 11 11 55i

38. 7

3

12 12 12 12 12 12 60i

39. 4

0

0 1 2 3 4

( 1)

!

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

0! 1! 2! 3! 4!1 1 1

1 12 6 24

9 3

24 8

i

ii

40.

14

0

1

1 !

i

ii

1 2 3 4 51 1 1 1 1

1! 2! 3! 4! 5!

1 1 1 1 191

2 6 24 120 30

41. 5

1

! 1! 2! 3! 4! 5!

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

1 2 3 4 5 15i

i

i

42. 5

1

2 !

!i

i

i

5

1

2 1 3 2 4 3 5 4 6 5 7 6i

i i

6 12 20 30 42 110

43. 15

2 2 2 2 2

11 2 3 15

ii

44. 12

4 4 4 4 4

11 2 3 12

ii

45. 11

2 3 4 11

12 2 2 2 2 2i

i

46. 12

2 3 12

15 5 5 5 5i

i

47. 30

11 2 3 30

ii

48. 40

11 2 3 40

ii

49. 14

1

1 2 3 14

2 3 4 14 1 1i

i

i

50. 16

1

1 2 3 16

3 4 5 16 2 2i

i

i

51. 2 3

1

4 4 4 44

2 3

n in

in i

52. 2 3

1

1 2 3

9 9 9 9 9

n

n ii

n i

53. 1

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

in i

54. 2 1 1

1

nn i

ia ar ar ar ar

55. 5 7 9 31

Possible answer: 14

1

(2 3)k

k

56. 6 + 8 + 10 + 12 + . . . + 32 16

=3

Possible answer: 2k

k



57. 2 12a ar ar ar

Possible answer: 12

0

k

k

ar

58. 2 14a ar ar ar 14

0

Possible answer: k

k

ar

59. ( ) ( 2 ) ( )a a d a d a nd

Possible answer: 0

( )n

k

a kd

60. 2( ) ( ) ( )na d a d a d

1

Possible answer:n

k

k

a d

61. 5

2 2 2 2 2 2

1

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

17 5 1 5 17

45

ii

a

62. 5

2 2 2 2 2 2

1

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

15 3 ( 1) 3 15

35

ii

b

63. 5

1

(2 ) 2( 4) 4 2( 2) 2 2(0) 0 2(2) ( 2) 2(4) ( 4)

4 ( 2) 0 2 4 0

i ii

a b

64. 5

1

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

8 4 0 4 8 0

i ii

a b

65. 2 2 25

2 2

4

2 4( 1) ( 1) 1 1 2

2 4i

ii

a

b

66. 3 3 35

3 3

4

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

2 4i

ii

a

b

67.

5 52 2 2 2 2 2 2 2 2 2 2 2

1 1

( 4) ( 2) 0 2 4 4 2 0 ( 2) ( 4)

16 4 0 4 16 16 4 0 4 16 80

i ii i

a b



68.

5 52 2 2 2 2 2 2 2 2 2

1 3

( 4) ( 2) 0 2 4 0 ( 2) ( 4)

16 4 0 4 16 0 4 16 40 20 20

i ii i

a b

69. a. 8

1

100 120 145 170 200 220 260 300 1515ii

a

A total of 1515 thousand, or 1,515,000, autism cases were diagnosed in the United States from 2001 through 2008.

b.

8

1

(28 1 63) (28 2 63) (28 3 63) (28 4 63) (28 5 63) (28 6 63) (28 7 63) (28 8 63)ii

a

91 119 147 175 203 231 259 287

1512

The model underestimates the actual sum by 3 thousand.

70. a. 6

1

1 193 96 112 119 134 145

6 6ii

a

1699

6116.5

Average state cigarette tax per pack each year averaged 116.5¢ for the years 2005 through 2010.

b. 6

1

1 111 1 78 11 2 78 11 3 78 11 4 78 11 5 78 11 6 78

6 6ii

a

189 100 111 122 133 144

61

6996116.5

71. 0.06

6000 1 , 1,2,3,4

n

na n

20

200.06

6000 1 8081.134

a

After five years, the balance is $8081.13.

72. 0.08

10,000 1 , 1,2,3,4

n

na n

24

240.08

10,000 1 16,084.374

a

After six years, the balance is $16,084.37.

73. – 80. Answers will vary.



81. Most calculators give error message if the expression is entered directly.

However,200! 200 199 198!

200 199 39,800198! 198!

82. 300

! 15! 1,307,674,368,00020

However, most calculators give a rounded answer in scientific notation.

83. 20!

8,109,673,360,588,800300

However, most calculators give a rounded answer in scientific notation.

84. 20!

684020 3 !

85. 54!

24,80454 3 !3!

86. Answers will vary.


88. 11

n

nan

10

10

100

100

1000

1000

10,000

10,000

100,000

100,000

11 2.5937

10

11 2.7048

100

11 2.7169

1000

11 2.7181

10,000

11 2.7183

100,000

a

a

a

a

a

As n gets larger, an gets closer to e ≈ 2.7183.

89. 1n

na

n

As n gets larger, an approaches 1.



90. 100

nan


91. 2

3

2 5 7n

n na

n


92. 4

4 2

3 1

5 2 1n

n na

n n

As n gets larger, an approaches 3

5.

93. does not make sense; Explanations will vary. Sample explanation: There is nothing that implies that there is a negative number of sheep.

94. does not make sense; Explanations will vary. Sample explanation: Any of the terms of this sequence could be negative and/or include non-integers.

95. makes sense

96. does not make sense; Explanations will vary. Sample explanation: Since 2n must be an even exponent, all the terms of the sequence will be positive.

97. false; Changes to make the statement true will vary. A sample change is: ! ( 1)!

( 1)! ( 1)!

n n nn

n n

98. true

99. false; Changes to make the statement true will vary. A sample change is:

2

1 21 2

1

1 2 1 2 1 2 1 2 1 4 2 4 2i i

i



100. false; Changes to make the statement true will vary. A sample change is: 2 2 2

1 1 1i i i i

i i i

a b a b

2

1 1 2 21

2 2

1 2 1 2 1 1 1 2 2 1 2 21 1

i ii

i ii i

a b a b a b

a b a a b b a b a b a b a b

101. 1

1

1

if is even.2

3 5 if is odd

nn

n

n n

aa

aa a

for 2.n

1 9a

Since 9 is odd, 2 3(9) 5 32a .

Since 32 is even, 332

162

a .

Similarly, 4 516 8

8, 42 2

a a .

The first five terms of the sequence are 9, 32, 16, 8, and 4.


103. 2 1 3 8 5a a

3 2 2 3 5a a

4 3 7 ( 2) 5a a

5 4 12 ( 7) 5a a

The difference between consecutive terms is always 5.

104. 2 1 4(2) 3 4(1) 3 4a a

3 2 4(3) 3 4(2) 3 4a a

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

5 4 4(5) 3 4(4) 3 4a a

The difference between consecutive terms is always 4.

105.

8

4 ( 1)( 7)

4 (8 1)( 7) 4 (7)( 7) 4 49 45na n

a



Section 11.2


1. 1 100a

2 1

3 2

4 3

5 4

6 5

30 100 30 70

30 70 30 40

30 40 30 10

30 10 30 20

30 20 30 50

a a

a a

a a

a a

a a

The first five terms are 100, 70, 40, 10, –20, –50.

2. 1 6, 5a d

To find the ninth term, 9a , replace n in the formula

with 9, 1a with 6, and d with –5.

1

9

( 1)

6 (9 1)( 5)

6 8( 5)

6 ( 40)

34

na a n d

a

3. a. 1 ( 1)

16 ( 1)0.35

0.35 15.65

na a n d

n

n

b.

21

0.35 15.65

0.35(21) 15.65

23

na n

a

In 2030, it is projected that 23% of the U.S. population will be Latino.

4. 3, 6, 9, 12, ... To find the sum of the first 15 terms, 15S , replace n

in the formula with 15.

1

15 1 15

( )215

( )2

n nn

S a a

S a a

Use the formula for the general term of a sequence to find 15a . The common difference, d, is 3, and the

first term, 1a , is 3.

1

15

( 1)

3 (15 1)(3)

3 14(3)

3 42

45

na a n d

a

Thus, 1515 15(3 45) (48) 3602 2

S .

5. 30

1(6 11) (6 1 11) (6 2 11)

(6 3 11) (6 30 11)

5 1 7 169

ii

So the first term, 1a , is –5; the common difference, d,

is 1 ( 5) 6 ; the last term, 30a , is 169.

Substitute 130, 5n a , and 30 169a in the

formula 12 ( )nn nS a a .

3030 2 ( 5 169) 15(164) 2460S

Thus, 30

1

(6 11) 2460i

i

6. 1800 64,130na n

1 1800(1) 64,130 65,930a

10 1800(10) 64,130 82,130a

12n nn

S a a

10 1 1010

25 65,930 82,130

5 148,060

$740,300

S a a

It would cost $740,300 for the ten-year period beginning in 2014.


1. arithmetic; common difference

2. 1 ( 1)a n d ; first term; common difference

3. 1 2( )2

na a ; first term; nth term

4. 2; 116

5. 8; 13; 18; 5

Exercise Set 11.2

1. 1 200, 20a d

The first six terms are 200, 220, 240, 260, 280, and 300.

2. a1 = 300, d = 50 The first six terms are 300, 350, 400, 450, 500, and 550.

Section 11.2 Arithmetic Sequences


3. 1 7, 4a d

The first six terms are –7, –3, 1, 5, 9, and 13.

4. a1 = –8, d = 5 The first six terms are –8, –3, 2, 7, 12, and 17.

5. 1 300, 90a d

The first six terms are 300, 210, 120, 30, –60, and –150.

6. a1 = 200, d = –60 The first six terms are 200, 140, 80, 20, –40, –100.

7. 15 1

,2 2

a d

The first six terms are 5 3 1

, 2, , 1, ,and 0.2 2 2

8. 13 1

,4 4

a d

The first six terms are 3 1 1 1

, , , 0, ,4 2 4 4

and 1

2 .

9. 1 16, 9n na a a


10. 14, 7n na a a


11. 1 110, 30n na a a

The first six terms are 30, 20, 10, 0, –10, and –20.

12. 1 120, 50n na a a

The first six terms are 50, 30, 10, –10, –30, and –50.

13. 1 10.4, 1.6n na a a

The first six terms are 1.6, 1.2, 0.8, 0.4, 0, and –0.4.

14. 1 10.3, 1.7n na a a

The first six terms are –1.7, –2.0, –2.3, –2.6, –2.9, and –3.2.

15. 1

6

13, 4

13 ( 1)4

13 5(4) 13 20 33n

a d

a n

a

16. 1

16

9, 2

9 ( 1)2

9 (15)2 9 30 39n

a d

a n

a

17. 1

50

7, 5

7 ( 1)2

7 49(5) 252n

a d

a n

a

18. 1

60

8, 6

8 ( 1)6

8 (59)6 362n

a d

a n

a

19. 1

200

40, 5

40 ( 1)5

40 (199)5 955n

a d

a n

a

20. 1

150

60, 5

60 ( 1)5

60 (149)5 685n

a d

a n

a

21. 1

60

35, 3

35 3( 1)

35 3(59) 142n

a d

a n

a

22. 1

70

32, 4

32 ( 1)4

32 (69)4 244n

a d

a n

a

23. 1, 5, 9, 13, …

20

5 1, 4

1 ( 1)4 1 4 4

4 3

4(20) 3 77

n

n

d

a n n

a n

a

24. 2, 7, 12, 17, …

20

7 2 5

2 ( 1)5 2 5 5

5 3

5(20) 3 97

n

n

d

a n n

a n

a

25. 7, 3, –1, –5, …

20

3 7 4

7 ( 1)( 4) 7 4 4

11 4

11 4(20) 69

n

n

d

a n n

a n

a



26. 6, 1, –4, –9, …

20

1 6 5

6 ( 1)( 5) 6 5 5

11 5

11 5(20) 89

n

n

d

a n n

a n

a

27. 1

20

9, 2

9 ( 1)(2)

7 2

7 2(20) 47

n

n

a d

a n

a n

a

28. 1

20

6, 3

6 ( 1)3 6 3 3 3 3

3(20) 3 63n

a d

a n n n

a

29. 1

20

20, 4

20 ( 1)( 4)

20 4 4

16 4

16 4(20) 96

n

n

n

a d

a n

a n

a n

a

30. 1

20

70, 5

70 5( 1) 70 5 5 65 5

65 5(20) 165n

a d

a n n n

a

31. 1 1

20

3, 4

3

4 ( 1)(3)

1 3

1 3(20) 61

n n

n

n

a a a

d

a n

a n

a

32. 1 1

20

5, 6, 5

6 ( 1)5 6 5 5 5 1

5(20) 1 101

n n

n

a a a d

a n n n

a

33. 1 110, 30, 10n na a a d

20

30 10 1 30 10 10

40 10

40 10 20 160

n

n

a n n

a n

a

34. 1 1

20

12, 24, 12

24 12( 1) 24 12 12

36 12

36 12(20) 204

n n

n

n

a a a d

a n n

a n

a

35. 4, 10, 16, 22, . . .

20

10 4 6

4 1 6

4 19 6 118

n

d

a n

a

2020

4 118 12202

S

36. 7, 19, 31, 43, …

25

25

12

7 ( 1)(12)

7 (24)(12) 295

25(7 295) 3775

2

n

d

a n

a

S

37. –10, –6, –2, 2, . . .

50

50

6 ( 10) 6 10 4

10 1 4

10 49 4 186

5010 186 4400

2

n

d

a n

a

S

38. –15, –9, –3, 3, . . .

50

50

9 ( 15) 9 15 6

15 ( 1)6

15 (49)6 279

50( 15 279) 6600

2

n

d

a n

a

S

39. 1 2 3 4 100

100100

(1 100) 50502

S

40. 2 + 4 + 6 + … + 200

100100

2 200 10,1002

S

41. 2 4 6 120

6060

(2 120) 36602

S

42. a80 = 2 + 79(2) = 160

8080

2 160 64802

S

43. even integers between 21 and 45;

12

22 24 26 44

12(22 44) 396

2S



44. odd integers between 30 and 54: 31 33 35 53

1212

31 53 5042

S

45. 17

1(5 3) (5 3) (10 3) (15 3) (85 3) 8 13 18 88

ii

1717

8 88 8162

S

46. 20

1(6 4) (6 4) (12 4) (18 4) (120 4) 2 8 14 116

ii

2020

2 116 11802

S

47. 30

1( 3 5) ( 3 5) ( 6 5) ( 9 5) ( 90 5) 2 1 4 85

ii

3030

(2 85) 12452

S

48. 40

1( 2 6) ( 2 6) ( 4 6) ( 6 6) ( 80 6) 4 2 0 74

ii

4040

(4 74) 14002

S

49. 100

14 4 8 12 400

ii

100100

4 400 20,2002

S

50. 50

14 4 8 12 200

ii

5050

4 200 51002

S

51. First find 14a and 12b :

14 1 ( 1)

1 (14 1)( 3 1) 51

a a n d

12 1 ( 1)

3 (12 1)(8 3) 58

b b n d

So, 14 12 51 58 7a b .


16 1 ( 1)

1 (16 1)( 3 1) 59

a a n d

18 1 ( 1)

3 (18 1)(8 3) 88

b b n d

So, 16 18 59 88 29a b .



53. 1 1

83 1 ( 1)( 3 1)

83 1 4( 1)

84 4 4

88 4

22

na a n d

n

n

n

n

n

There are 22 terms.

54. 1 1

93 3 ( 1)(8 3)

93 3 5( 1)

93 5 2

95 5

19

nb b n d

n

n

n

n

n

There are 19 terms.

55. 12n nn

S a a

For { } :na 14 1 14

147 1 ( 51) 350

2S a a For { } :nb 14 1 14

147 3 68 497

2S b b

So 14 14

1 1

497 ( 350) 847n nn n

b a


15 1 ( 1)

1 (15 1)( 3 1) 55

a a n d

15 1 ( 1)

3 (15 1)(8 3) 73

b b n d

Using 12n nn

S a a for { } :na 15 1 15

157.5 1 ( 55)

2

405

S a a

And then for { } :nb 15 1 1515

7.5 3 73 5702

S b b So 15 15

1 1

570 ( 405) 975n nn n

b a

57. Two points on the graph are (1, 1) and (2, −3). Finding the slope of the line; 2 1

2 2

3 1 44

2 1 1

y ym

x x

Using the point-slope form of an equation of a line; 2 2( )

1 4( 1)

1 4 4

4 5

y y m x x

y x

y x

y x

Thus, ( ) 4 5f x x .



58. Two points on the graph are (1, 3) and (2, 8). Finding the slope of the line;

2 1

2 2

8 3 55

2 1 1

y ym

x x

Using the point-slope form of an equation of a line; 2 25( )

3 5( 1)

3 5 5

5 2

y y x x

y x

y x

y x

Thus, ( ) 5 2g x x .

59. Using 1 ( 1)na a n d and 2 4 :a

2 1

1

(2 1)

4

a a d

a d

And since 6 16 :a

6 1

1

(6 1)

16 5

a a d

a d

The system of equations is

1

1

4

16 5

a d

a d

Solving the first equation for 1a :

1 4a d

Substituting the value into the second equation and solving for d:

16 (4 ) 5

16 4 4

12 4

3

d d

d

d

d

Back-substitute:

1

1

1

4

4 3)

1

a d

a

a

Then 1 ( 1)

1 ( 1)3

1 3 3

3 2

n

n

n

n

a a n d

a n

a n

a n

60. Using 1 ( 1)na a n d and 3 7 :a

3 1

1

(3 1)

7 2

a a d

a d

And since 8 17 :a

8 1

1

(8 1)

17 7

a a d

a d

The system of equations is

1

1

7 2

17 7

a d

a d

Solving the first equation for 1a :

1 7 2a d

Substituting the value into the second equation and solving for d:

117 7

17 (7 2 ) 7

17 7 5

10 5

2

a d

d d

d

d

d

Back-substitute:

1

1

1

7 2

7 2(2)

3

a d

a

a

Then 1 ( 1)

3 ( 1)2

3 2 2

2 1

n

n

n

n

a a n d

a n

a n

a n

61. a. 1 ( 1)

11.0 ( 1)0.5

11.0 0.5 0.5

0.5 10.5

n

n

a a n d

a n

n

n

b. 0.5 10.5

0.5(50) 10.5

35.5

na n

The percentage is projected to be 35.5% in 2019.

62. a. 1 ( 1)

55.2 ( 1)0.86

55.2 0.86 0.86

0.86 54.34

n

n

a a n d

a n

n

n

b. 0.86 54.34

0.86(50) 54.34

97.34

na n

The percentage is projected to be 97.34% in 2019.



63. Company A

24000 1 1600

24000 1600 1600

1600 22400

na n

n

n

10 1600 10 22400

16000 22400 38400

a

Company B

28000 1 1000

28000 1000 1000

1000 27000

na n

n

n

10 1000 10 27000

10000 27000 37000

a

Company A will pay $1400 more in year 10.

64. Company A:

23000 1 1200

23000 1200 1200 1200 21800na n

n n

10 1200 10 21800

12000 21800 33800

a

Company B:

26000 1 800

26000 800 800 800 25200na n

n n

10 800 10 25200 8000 25200

33200

a

Company A will pay $600 more in year 10.

65. a. Total cost: $5836 $6185 $6585 $7020 $25,626

b. 1

4

395(1) 5419 5814

395(4) 5419 6999

a

a

1

4

24

5814 6999 2 12,813 $25,6262

n nn

S a a

S

The model gives actual sum of $25,626 obtained in part (a).

66. a. Total cost: $22,218 $23,712 $25,143 $26,273 $97,346

b. 1

4

1360(1) 20,938 22,298

1360(4) 20,938 26,378

a

a

1

4

24

22,298 26,378 2 48,676 $97,3522

n nn

S a a

S

The model overestimates the actual sum by $6.


68.

10

33,000 1 2500

33,000 9 2500 55,500

1033,000 55,500 442,500

2

n

n

a n

a

S

The total ten year salary is $442,500.

69. Company A:

10

19,000 ( 1)2600

19,000 (9)2600 $42,200na n

a

1010

(19000 42400) $307,0002

S

Company B:

10

10

27,000 1 1200

27,000 (9)1200 $37,800

10(27,000 37,800) $324,000

2

na n

a

S

Company B pays the greater total amount.

70.

26

26

30 ( 1)2

30 (25)2 80

2630 80 1430

2

na n

a

S

The theater has 1430 seats.

71.

38

38

20 ( 1)3

20 (37)3 131

3820 131 2869

2

na n

a

S

The theater has 2869 seats.


78. does not make sense; Explanations will vary. Sample explanation: The difference between terms is not constant. Thus, this is not an arithmetic sequence.

79. makes sense

Section 11.3 Geometric Sequences and Series


80. makes sense

81. makes sense

82. 21,700, 23,172, 24,644, 26,166, . . . , 314,628 d = 23,172 – 21,700 = 1472 314,628 = 1472n + 20,228 1472n = 294,400 n = 200 It is the 200th term.

83. Degree days: 23, 25, 27, …

1a = 23, d = 2

10 23 9(2) 41a

10 1 1010

( )2

S a a

1010

(23 41) 3202

S

There are 320 degree-days.

84. 1 3 5 (2 1)n

2

1 2 12

22

nn

S n

nn

n

85. 2

1

22

1

a

a

3

2

42

2

a

a

4

3

82

4

a

a

5

4

162

8

a

a

The ratio of a term to the term that directly precedes it is always 2.

86. 2

21

1

3 55

3 5

a

a

33

22

3 55

3 5

a

a

44

33

3 55

3 5

a

a

55

44

3 55

3 5

a

a

The ratio of a term to the term that directly precedes it is always 5.

87. 113n

na a 7 1 6

7 11 3 11 3 11 729 8019a

Section 11.3


1. 11

12,2

a r

1

2

2

3

3

4

4

5

5

6

112 6

2

1 1212 3

2 4

1 12 312

2 8 2

1 12 312

2 16 4

1 12 312

2 32 8

a

a

a

a

a

The first six terms are 3 3 3

12, 6, 3, , ,and .2 4 8

2. 1

1

7 1 67

5, 3

5

5( 3) 5( 3) 5(729) 3645

nn

a r

a r

a

The seventh term is 3645.

3. 3, 6, 12, 24, 48, ...

1

1

8 1 78

62, 3

3

3(2)

3(2) 3(2) 3(128) 384

nn

r a

a

a

The eighth term is 384.

4.

1

1

9

9

62, 3

2

(1 )

1

2 1 ( 3) 2(19,684)9842

1 ( 3) 4

r

n

a r

a rS

r

S

The sum of the first nine terms is 9842.



5. 8

1

2 3i

i

11

1

8

8

2 (3) 6, 3

(1 )

1

6 1 3 6( 6560)19,680

1 3 2

n

n

a r

a rS

r

S

Thus, 8

1

2 3 19,680.i

i

6.

1

1

30

30

30,000, 1.06

(1 )

1

30,000 1 (1.06)2,371,746

1 1.06

n

n

a r

a rS

r

S

The total lifetime salary is $2,371,746.

7. a. 1 1

ntrn

rn

PA

12 35

100, 0.095, 12, 35

0.095100 1 1

12333,946

0.095

12

P r n t

A

The value of the IRA will be $333,946.

b. Interest Value of IRA Total deposits

$333,946 $100 12 35

$333,946 $42,000

$291,946

8. 4 8

3 23 9

1

1

2 13 3

23,

3

13 3

91

a r

aS

r

S

The sum of this infinite geometric series is 9.

9. 9 9 9

0.9 0.999910 100 1000

19 1

,10 10

9 9

10 10 11 9

110 10

a r

S

An equivalent fraction for 0.9 is 1.

10. 1 1000(0.8) 800, 0.8

8004000

1 0.8

a r

S

The total amount spent is $4000.


1. geometric; common ratio

2. 11

na r ; first term; common ratio

3. 1(1 );

1

na r

r

first term; common ratio

4. annuity; P; r; n

5. infinite geometric series; 1; 1

1

a

r; 1r

6. 2; 4; 8; 16; 2

7. arithmetic

8. geometric

9. geometric

10. arithmetic

Exercise Set 11.3

1. 1 5, 3a r

First five terms: 5, 15, 45, 135, 405.

2. 1 4, 3a r

First five terms: 4, 12, 36, 108, 324.

3. 11

20,2

a r

First five terms: 5 52 420, 10, 5, , 5

4 .



4. 11

24,3

a r

First five terms:8 8 8

24, 8, , , 3 9 27

.

5. 1 14 , 10n na a a

First five terms: 10, –40, 160, –640, 2560.

6. 1 13 , 10n na a a

First five terms: 10, –30, 90, –270, 810.

7. 1 15 , 6n na a a

First five terms: –6, 30, –150, 750,–3750.

8. 1 16 , 2n na a a

First five terms: –2, 12, –72, 432,–2592.

9. 1

1

78

6, 2

6 2

6 2 768

nn

a r

a

a

10. 1 5, 3a r 1

78

5 3

5 3 10,935

nna

a

11.

1

1

1112

5, 2

5 ( 2)

5 2 10,240

nn

a r

a

a

12. 1 4, 2a r

1

1112

4 2

4 2 8192

nna

a

13. 1

1

39

40

11000,

2

11000

2

11000

2

0.000000002

n

n

a r

a

a

14. 11

8000,2

a r

1

29

30

18000

2

18000 0.000014901

2

n

na

a

15.

1

1

78

1,000,000, 0.1

1,000,000 0.1

1,000,000 0.1 0.1

nn

a r

a

a

16. 1 40,000, 0.1a r

1

78

40,000 0.1

40,000 0.1 0.004

nna

a

17. 3, 12, 48, 192, . . .

1

67

124

3

3(4)

3(4) 12,288

nn

r

a

a

18. 3, 15, 75, 375, ...

1

67

155

3

3 5

3 5 46,875

nn

r

a

a

19. 2 6 1

19,6,2, , 3 18 3

r

7

11

183

61 2

183 81

n

na

a

20. 3

12, 6, 3, ,2

1

6

7

6 1

12 2

112

2

1 312

2 16

n

n

r

a

a



21. 1.5, –3, 6, –12, . . .

1

67

62

3

1.5( 2)

1.5( 2) 96

nn

r

a

a

22. 1 1

5, 1, , ,5 25

1

6

7

1 1

5 5

15

5

1 15

5 3125

n

n

r

a

a

23. 0.0004, –0.004, 0.04, –0.4, . . .

1

67

0.00410

0.0004

0.0004( 10)

0.0004( 10) 400

nn

r

a

a

24. 0.0007, –0.007, 0.07, –0.7, . .

1

67

0.00710

0.0007

0.0007 10

0.0007 10 700

nn

r

a

a

25. 2, 6, 18, 54, . . .

12

12

63

2

2 1 3 2( 531,440)531,440

1 3 2

r

S

26. 3, 6, 12, 24, …

12

12

62

3

3 1 2 3( 4095)12,285

1 2 1

r

S

27. 3, –6, 12, –24, . . .

11

11

62

3

3 1 ( 2) 3(2049)2049

1 ( 2) 3

r

S

28. 4, –12, 36, –108, …

11

11

123

4

4 1 3 4(177,148)177,148

1 3 4

r

S

29. 3

,3, 6,12,2

14

14

32

323 31 ( 2) ( 16,383) 16,3832 2

1 ( 2) 3 2

r

S

30. 1 1 1 1

, , , ,24 12 6 3

14

14

112 2

1241 11 2 ( 16,383) 546124 24

1 2 3 24

r

S

31. 8

1

3i

i

1

8

8

3, 3

3 1 3 3 65609840

1 3 2

r a

S

32. 6

1

4i

i

1

6

6

4, 4

4 1 4 4 40955460

1 4 3

r a

S

33. 10

1

5 2i

i

1

10

10

2, 10

10 1 2 10 102310,230

1 2 1

r a

S



34. 7

1

4 3i

i

1

7

7

3, 12

12 1 ( 3) 12(2188)6564

1 ( 3) 4

r a

S

35. 16

1

1

2

i

i

1

6

6

1 1,

2 4

1 1 1 6314 2 634 64

1 1 12812 2

r a

S

36. 16

1

1

3

i

i

1

6

6

1 1,

3 9

1 1 728119 3 3649 729

1 2 218713 3

r a

S

37. 1

31 1 3

1 2 213 3

r

S

38. 1 1 1

14 16 64

1

41 1 4

1 3 314 4

r

S

39. 1

43 3

41 314 4

r

S

40. 2 3

5 5 55

6 6 6

1, 516

5 56

1 51

6 6

r a

S

41. 1

21 1 2

31 31

22

r

S

42. 1 1

3 13 9

1

33 3 9

41 4133

r

S

43. 0.3

8 86.15385

1 ( 0.3) 1.3

r

S

44. 112 0.7 12 8.4

1

i

i

r = –0.7

12 12

7.058821 0.7 1.7

S

45. 1

105 5

510 101 9 91

10 10

r

S

46. 1 1 1 1

0.110 100 1000 10000

1

101 1

110 101 9 91

10 10

r

S



47. 1

1004747

47100 1001 99 991

100 100

r

S

48. 83 83 83

0.83100 10,000 1,000,000

1

10083 83

83100 1001 99 991

100 100

r

S

49. 6 9

257 257 2570.257

1000 10 101

1000257 257

2571000 10001 999 9991

1000 1000

r

S

50. 6 9

529 529 5290.529

1000 10 10

1

1000529 529

5291000 10001 999 9991

1000 1000

r

S

51. 5

arithmetic, 1na n

d

52. 3

arithmetic, 1na n

d

53. 2

geometric, 2

nna

r

54. 1

2

n

na

1geometric,

2r

55. 2 5

neitherna n

56. 2 3

neitherna n

57. First find 10a and 10b : 1

10 1

10 1910

( 5) ( 5)( 2)5

2560

na a r

10 1 ( 1)

10 (10 1)( 5 10)

10 (9)( 15) 125

b b n d

So, 10 10 2560 ( 125) 2435a b .

58. First find 11a and 11b : 1

11 1

11 11010

( 5) ( 5)( 2)5

5120

na a r

11 1 ( 1)

10 (11 1)( 5 10)

10 (10)( 15) 140

b b n d

So, 11 11 5120 ( 140) 5260a b .

59. For { }na , 10

25

r

and

10

110

( 5) 1 ( 2)(1 )

1 1 2

( 5)( 1023)1705

3

na rS

r

For { }nb ,

10 1 ( 1)

10 (10 1)( 5 10)

10 (9)( 15) 125

b b n d

1

10 1 10

2

210

(10 ( 125))2

5( 115) 575

n nn

S b b

nS b b



60. For { }na , 10

25

r

and :

11

111

( 5) 1 ( 2)(1 )

1 1 2

( 5)(2049)3415

3

na rS

r

For { }nb ,

11 1 ( 1)

10 (11 1)( 5 10)

10 (10)( 15) 140

b b n d

11 111

(10 ( 140))2 25.5( 130) 715

nn

S b b

So, 11 11

1 1

3415 ( 715)

2700

n nn n

a b

61. For { }na ,

6

16

( 5) 1 ( 2)(1 )

1 1 2

( 5)( 63)105

3

na rS

r

For { }nc ,

1 2 2 41 31 312 2

aS

r

So, 64

105 1403

S S

62. For { }na ,

9

19

( 5) 1 ( 2)(1 )

1 1 2

( 5)(513)855

3

na rS

r

For { }nc ,

1 2 2 41 31 312 2

cS

r

So, 94

855 11403

S S

63. It is given that 4 27a . Using the formula 1

1n

na a r when 4n we have 4 1

3

3

27 8

27

8

27 3

8 2

r

r

r

Thus, 1

1

2 1

23 3

8 8 122 2

nna a r

a

3 1 2

33 3 9

8 8 8 182 2 4

a

64. It is given that 4 54a . Using the formula 1

1n

na a r when 4n we have 4 1

3

3

54 2

27

27 3

r

r

r

Thus,

11

2 12 2 3 2 3 6

nna a r

a

3 1 23 2 3 2 3 2 9 18a

65. 1, 2, 4, 8, . . .

1

1415

2

2

2 $16,384

nn

r

a

a

66. 1

2930

2

2 $536,870,912

nna

a

67.

1

1

67

3,000,000

1.04

3,000,000 1.04

3,000,000 1.04 $3,795,957

nn

a

r

a

a

68. a1 = 30,000

1

56

1.05

30,000 1.05

30,000 1.05 $38,288.45

nn

r

a

a



69. a. 2000 to 200134.21

1.0133.87

r

2001 to 2002

2002 to 2003

2003 to 2004

2004 to 2005

2005 to 2006

2006 to 2007

2007 to 2008

34.551.01

34.2134.90

1.0134.5535.25

1.0134.9035.60

1.0135.2536.00

1.0135.6036.36

1.0136.0036.72

136.36

r

r

r

r

r

r

r

2008 to 2009

2009 to 2010

.01

37.091.01

36.7237.25

1.0137.09

r

r

r is approximately 1.01 for all but one division.

b.

11

133.87 1.01

nn

nn

a a r

a

c. Since year 2020 is the 21th term, find 21.a

1

21 121

33.87 1.01

33.87 1.01 41.33

nna

a

The population of California will be approximately 41.33 million in 2020.

70. a. 2000 to 200121.27

1.0220.85

r

2001 to 2002

2002 to 2003

2003 to 2004

2004 to 2005

2005 to 2006

2006 to 2007

2007 to 2008

21.701.02

21.2722.13

1.0221.7022.57

1.0222.1323.02

1.0222.5723.48

1.0223.0223.95

1.0223.4824.43

123.95

r

r

r

r

r

r

r

2008 to 2009

2009 to 2010

.02

24.921.02

24.4325.15

1.0124.92

r

r

r is approximately 1.02 for all but one division.

b.

11

120.85 1.02

nn

nn

a a r

a

c. Since year 2020 is the 21th term, find 21.a

1

21 121

20.85 1.02

20.85 1.02 30.98

nna

a

The population of Texas will be approximately 30.98 million in 2020.

71. 1, 2, 4, 8, . . .

15

15

2

1(1 2 )32,767

1 2

r

S

The total savings is $32,767.

72. 30

301 2

1,073,741,8231 2

S

The total savings is $1,073,741,823.

73. 1

20

20

24,000, 1.05

24,000 1 (1.05)793,582.90

1 1.05

a r

S

The total salary is $793,583.



74. Company A:

1 30,000, 1.06a r

55

30,000 1 1.06$169,112.79

1 1.06S

Company B:

1 32,000, 1.03a r

55

32,000 1 1.03$169,892.35

1 1.03S

Company B offers the better total salary by $780.

75. 10

10

0.9

20(1 0.9 )130.26

1 0.9

r

S

The total length is 130.26 inches.

76. 226 0.96(16) 0.96 (16)

r = 0.96

10

10

16 1 0.96134.07

1 0.96S

The total length is 134.07 inches.

77. a.

50.075

1 1 2000 1 11

$11,6170.075

1

ntr

Pn

Ar

n

b. $11,617 5 $2000 $1617

78. a.

50.0625

1 1 2500 1 11

$14,1630.0625

1

ntr

Pn

Ar

n

b. $14,163 5 $2500 $1663

79. a. 12 400.055

12

0.05512

1 1 50 1 1$87,052

ntr

Pn

Ar

n

b. $87,052 $50 12 40 $63,052

80. a. 12 400.065

12

0.06512

1 1 75 1 1$171,271

ntrn

rn

PA



b. $171,271 $75 12 40 $135,271

81. a. 4 100.105

4

0.1054

1 1 10,000 1 1$693,031

ntrn

rn

PA

b. $693,031 $10,000 4 10 $293,031

82. a. 4 100.09

4

0.094

1 1 15,000 1 1$956,793

ntrn

rn

PA

b. $956,793 $15,000 4 10 $356,793

83. Find the total value of the lump-sum investment.

201 30,000 1 0.05 79,599

tA P r

Find the total value of the annuity. 20

0.051 1 1500 1 1

149,599

0.05

1

ntr

Pn

Ar

n

$79,599 $49,599 $30,000 You will have $30,000 more from the lump-sum investment.

84. Find the total value of the lump-sum investment.

251 40,000 1 0.065 193,108

tA P r

Find the total value of the annuity. 25

0.0651 1 1600 1 1

194,220

0.065

1

ntr

Pn

Ar

n

$193,108 $94,220 $98,888 You will have $98,888 more from the lump-sum investment.

85. 0.6

6(0.6)9

1 0.6

r

S

The total economic impact is $9 million.

86. 2 310(0.6) 10(0.6) 10(0.6)

1 10(0.6)a , r = 0.6

10 0.615

1 0.6S

An additional $15 billion will be spent.



87. 14

11 4 14

1 4 3 314

r

S


99.

12 1

3

11

3

x

f x

2 2 2 3

2 2 31 2 3 213 3

S

The sum of the series is 3 and the asymptote of the function is y = 3.

100. 4 1 0.6

1 0.6

x

f x

4 4

101 0.6 0.4

S

The sum of the series is 10 and the asymptote of the function is y = 10.

101. makes sense

102. makes sense

103. makes sense

104. does not make sense; Explanations will vary. Sample explanation: It is not possible to add each term individually of an infinite series.

105. false; Changes to make the statement true will vary. A sample change is: The sequence is not geometric. There is not a common ratio.

106. false; Changes to make the statement true will vary. A sample change is: We do not need to know the

terms between 1 1

and 8 512

, but we do need to

know how many terms there are between 1 1

and 8 512

.

107. false; Changes to make the statement true will vary. A sample change is: The sum of the sequence is

10

11

2

.

108. true

109. Let 1a equal the number of flies released each day.

On any day, the total number of flies is the number released that day, plus 90% of those released the day before, plus 90% of 90% of those released two days before, etc.:

1

1

1

1

1

20,0001 .9

20,000.1

2000

aS

ra

a

a

2000 flies to be released each day.

110.

3600.1

1 112

1,000,0000.1

12

P

1,000,000 2260.49

442.38

P

P

You must deposit $442 monthly.


112. 3(3 1)

1 2 32

3(4)6

26 6

113. 5(5 1)

1 2 3 4 52

5(6)15

215 15



114. 2( 1)(2 1)( 1)

6

k k kk

2

2

2

2

( 1)(2 1) 6( 1)

6 6

( 1)(2 1) 6( 1)

6( 1)[ (2 1) 6( 1)]

6

( 1)[2 6 6]

6

( 1)[2 7 6]

6( 1)( 2)(2 3)

6

k k k k

k k k k

k k k k

k k k k

k k k

k k k

Mid-Chapter 11 Check Point

1. 1( 1)( 1)!

nn

na

n

1 1 21

2 1 32

3 1 43

4 1 54

5 1 65

1 1( 1) ( 1) 1 1 1

(1 1)! 0!

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

(2 1)! 1!

3 3 3 3( 1) ( 1) 1

(3 1)! 2! 2 2

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

(4 1)! 3! 6 3

5 5 5 5( 1) ( 1) 1

(5 1)! 4! 24 24

a

a

a

a

a

2. Using 1 ( 1)na a n d ;

1

2

3

4

5

5

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

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

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

5 (5 1)( 3) 5 4( 3) 5 12 7

a

a

a

a

a

3. Using 11

nna a r ;

1

2 1 12

3 1 23

4 1 34

5 1 45

5

5( 3) 5( 3) 5( 3) 15

5( 3) 5( 3) 5(9) 45

5( 3) 5( 3) 5( 27) 135

5( 3) 5( 3) 5(81) 405

a

a

a

a

a

4. 1 4n na a

1

2 1

3 2

4 3

5 4

3

4 3 4 1

4 1 4 3

4 3 4 1

4 1 4 3

a

a a

a a

a a

a a

5. 2 1 6 2 4d a a

1

20

( 1)

2 ( 1)4

2 4 4

4 2

4(20) 2 78

na a n d

n

n

n

a

6. 2

1

11

1

10 110

9

62

3

3(2)

3(2)

3(2)

1536

nn

n

ar

a

a a r

a

7. 2 13 1

12 2

d a a

1

30

( 1)

3 1( 1)

2 2

3 1 1

2 2 21

22

1(30) 2

215 2

13

na a n d

n

n

n

a

8. 1 2

1

(1 ) 10; 2

1 5

n

na r a

S rr a

10

105(1 2 ) 5( 1023)

51151 2 1

S

Mid-Chapter Check Point


9. First find 10a ;

2 1

50 1

50 1

0 ( 2) 2

( 1) 2 (50 1)(2) 2 49(2) 96

50( ) ( 2 96) 25(94) 2350

2 2n

d a a

a a n d

nS a a

10. 2

1

402

20

ar

a

101

10(1 ) 20( 1 ( 2) ) 20( 1023) 20460

68201 1 ( 2) 3 3

na rS

r

11. First find 100a ;

2 1

100 1

100 1

2 4 6

( 1) 4 (100 1)( 6) 4 99( 6) 590

100( ) (4 590) 50( 586) 29,300

2 2n

d a a

a a n d

nS a a

12. 4

1

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

5(0) 6(1) 7(2) 8(3) 0 6 14 24 44i

i i

13. 50

1

(3 2) (3 1 2) (3 2 2) (3 3 3) ... (3 50 2)

(3 2) (6 2) (9 3) ... (150 2)

1 4 6 ... 148

i

i

The sum of this arithmetic sequence is given by 1( )2n nn

S a a ;

5050

(1 148) 25(149) 37252

S

14. 1 2 3 4 5 66

1

3 3 3 3 3 3 3

2 2 2 2 2 2 2

3 9 27 81 243 729 1995

2 4 8 16 32 64 64

i

i



15. 1 1 1 2 1 3 1

1

0 1 2

2 2 2 2...

5 5 5 5

2 2 2...

5 5 5

2 41 ...

5 25

i

i

This is an infinite geometric sequence with 252

1

2

1 5

ar

a

.

1

7255

1 1 5

1 71

aS

r

16. 1

45 45

100 1000.451 991 1

100 10045 99 45 100 45 5

100 100 100 99 99 11

a

r

17. Answers will vary. An example is 18

12

i

i

i .

18. The arithmetic sequence is 16, 48, 80, 112, …. First find 15a where 2 1 48 16 32d a a .

15 1 ( 1) 16 (15 1)(32) 16 14(32) 16 448 464a a n d

The distance the skydiver falls during the 15th second is 464 feet.

15 115

( ) (16 464) 7.5(480) 36002 2nn

S a a

The total distance the skydiver falls in 15 seconds is 3600 feet.

19. 0.10r

10

(1 )

120000(1 0.10)

311249

tA P r

The value of the house after 10 years is $311,249.

Chapter 11 Sequences, Induction, and Probability...2013/07/31 · Chapter 11 Sequences, Induction,...

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Transcript of Chapter 11 Sequences, Induction, and Probability...2013/07/31 · Chapter 11 Sequences, Induction,...