Math Analysisramsteinmath.weebly.com/uploads/7/0/5/4/7054561/chapter_11_ssm… · 524 CHAPTER 11 SEQUENCES, INDUCTION, AND PROBABILITY 31. a1 = 4 a2 = 1 4 a2−1 = 1 4 a1 = 1 4 ·

SECTION 11-1 523

CHAPTER 11 Section 11−1 1. A sequence is a function whose domain is a set of successive integers. A series is the indicated sum of

terms of a sequence. 3. Each term of the Fibonacci sequence after the second is the sum of the two previous terms, thus: an = an−1 + an−2 is the recursion formula, with a1 = a2 = 1. 5. Simply listing the first three terms of a sequence does not specify the sequence or define its general term. 7. a1 = 1 − 2 = −1 a2 = 2 − 2 = 0 a3 = 3 − 2 = 1 a4 = 4 − 2 = 2

9. a1 = 1 1

1 1

= 0 a2 = 2 1

2 1

= 1

3 a3 =

3 1

3 1

= 2

4 =

1

2 a4 =

4 1

4 1

= 3

5

11. a1 = (−2)1+1 = (−2)2 = 4; a2 = (−2)2+1 = (−2)3 = −8; a3 = (−2)3+1 = (−2)4 = 16; a4 = (−2)4+1 = (−2)5 = −32

13. a8 = 8 − 2 = 6 15. a100 = 100 1

100 1

= 99

101 17. S5 = 1 + 2 + 3 + 4 + 5

19. S3 = 1

1

10 +

2

1

10 +

3

1

10 =

1

10 +

1

100 +

1

1000

21. S4 = (−1)1 + (−1)2 + (−1)3 + (−1)4 = (−1) + 1 + (−1) + 1 = −1 + 1 − 1 + 1

23. a1 = (−1)1+1 12 = 1 a2 = (−1)2+1 22 = −4 a3 = (−1)3+1 32 = 9 a4 = (−1)4+1 42 = −16 a5 = (−1)5+1 52 = 25

25. a1 = 1

3 1

11

10

=

1

3 ·

9

10 =

3

10 = 0.3

a2 = 1

3 2

11

10

=

1

3 ·

99

100 =

33

100 = 0.33

a3 = 1

3 3

11

10

=

1

3 ·

999

1,000 =

333

1,000 = 0.333

a4 = 1

3 4

11

10

=

1

3 ·

9,999

10,000 =

3,333

10,000 = 0.3333

a5 = 1

3 5

11

10

=

1

3 ·

99,999

100,000 =

33,333

100,000 = 0.33333

27. a1 = 1

2

1−1

= 1

2

0

= 1

a2 = 1

2

2−1

= −1

2

a3 = 1

2

3−1

= 1

4

29. a1 = 7 a2 = a2−1 − 4 = a1 − 4 = 7 − 4 = 3 a3 = a3−1 − 4 = a2 − 4 = 3 − 4 = −1 a4 = a4−1 − 4 = a3 − 4 = −1 − 4 = −5 a5 = a5−1 − 4 = a4 − 4 = −5 − 4 = −9

Common Error: a2 ≠ a2 – 1 – 4. The 1 is in the subscript: a2−1; that is, a1

a4 = 1

2

4−1

= −1

8

a5 = 1

2

5−1

= 1

16

524 CHAPTER 11 SEQUENCES, INDUCTION, AND PROBABILITY

31. a1 = 4

a2 = 1

4a2−1 =

1

4a1 =

1

4· 4 = 1 a4 =

1

4a4−1 =

1

4a3 =

1

4 ·

1

4 =

1

16

a3 = 1

4a3−1 =

1

4a2 =

1

4 · 1 =

1

4 a5 =

1

4a5−1 =

1

4a4 =

1

4 ·

1

16 =

1

64

33. a1 = 1 a2 = 2 a3 = a1 + 2a2 = 1 + 2(2) = 5 a4 = a2 + 2a3 = 2 + 2(5) = 12 a5 = a3 + 2a4 = 5 + 2(12) = 29 a6 = a4 + 2a5 = 12 + 2(29) = 70 a7 = a5 + 2a6 = 29 + 2(70) = 169

The graphic demonstrates use of the RECUR program on a TI−83.

35. a1 = −1

a2 = 2 a3 = 2a1 + a2 = 2(−1) + 2 = 0 a4 = 2a2 + a3 = 2(2) + 0 = 4 a5 = 2a3 + a4 = 2(0) + 4 = 4 a6 = 2a4 + a5 = 2(4) + 4 = 12 a7 = 2a5 + a6 = 2(4) + 12 = 20

The graphic demonstrates use of the RECUR program on a TI−83.

37. an: −2, −1, 0, 1, …

n: 1, 2, 3, 4, … Comparing an with n, we see that an = n – 3.

39. an: 5, 7, 9, 11, … an – 1: 4, 6, 8, 10, … an − 3: 2, 4, 6, 8, … n: 1, 2, 3, 4, … Comparing an – 3 with n, we see that an – 3 = 2n,

hence an = 2n + 3. 41. an: −1, 1, −1, 1, …

n: 1, 2, 3, 4, … Comparing an with n, we see that an = (−1)n.

43. an: 2, 3

2,

4

3,

5

4, …

n: 1, 2, 3, 4, …

Comparing an with n, we see that an = 1n

n

.

45. an: −3, 9, −27, 81, … n: 1, 2, 3, 4, … Comparing an with n, we see that an consists of successive nth powers of −3. an = (−3)n

47. an: x, 2

2

x,

3

3

x,

4

4

x, …

n: 1, 2, 3, 4, …

Comparing an with n, we see that an = nx

n.

49. (A) a1 = 12 – 1 + 2 = 2; a2 = 22 – 2 + 2 = 4; a3 = 32 – 3 + 2 = 8; a4 = 42 – 4 + 2 = 14 (B) A (simpler) sequence with the first three terms 2, 4, 8 would be bn = 2n. 51. (A) a1 = 6·12 – 11·1 + 6 = 1; a2 = 6·22 – 11·2 + 6 = 8; a3 = 6·32 – 11·3 + 6 = 27; a4 = 6·42 – 11·4 + 6 = 58 (B) A (simpler) sequence with the first three terms 1, 8, 27 would be bn = n3. 53. (A) a1 = 2·12 – 8·1 + 7 = 1; a2 = 2·22 – 8·2 + 7 = −1; a3 = 2·32 – 8·3 + 7 = 1; a4 = 2·42 – 8·4 + 7 = 7 (B) A (simpler) sequence with the first three terms 1, −1, 1 would be bn = (−1)n+1.

SECTION 11-1 525

55.

0 20

-.3

1

57.

0 20

-1.5

1.5

59. S4 =

1 1( 2)

1

+

2 1( 2)

2

+

3 1( 2)

3

+

4 1( 2)

4

= 4

1 −

8

2 +

16

3 −

32

4

61. S3 = 1

1x1+1 +

1

2x2+1 +

1

3x3+1

= x2 + 3

2

x +

4

3

x

63. S5 = 1 1( 1)

1

x1 +

2 1( 1)

2

x2 +

3 1( 1)

3

x3 +

4 1( 1)

4

x4 +

5 1( 1)

5

x5 = x −

2

2

x +

3

3

x −

4

4

x +

5

5

x

65. S4 = 12 + 22 + 32 + 42 k = 1, 2, 3, 4

Clearly, ak = k2, k = 1, 2, 3, 4; S4 = 4

2

1k

k

67. S5 = 1

1

2 +

2

1

2 +

3

1

2 +

4

1

2 +

5

1

2 k = 1, 2, 3, 4, 5

Clearly, ak = 1

2k, k = 1, 2, 3, 4, 5; S5 =

5

1

1

2kk

69. Sn = 1 + 2

1

2 +

2

1

3 + … +

2

1

n k = 1, 2, 3, …, n

Clearly, ak = 2

1

k, k = 1, 2, 3, …, n Sn =

21

1n

k k

71. Sn = 1 − 4 + 9 + … + (−1)n+1n2 k = 1, 2, 3, …, n

Clearly, ak = (−1)k +1k2, k = 1, 2, 3, …, n

Sn = 1

( 1)n

k

k +1k2

73. (A) a1 = 3

a2 = 22 1

2 1

2

2

a

a

=

21

1

2

2

a

a

=

23 2

2 3

≈ 1.83

a3 = 23 1

3 1

2

2

a

a

=

22

2

2

2

a

a

=

2(1.83) 2

2(1.83)

≈ 1.46

a4 = 24 1

4 1

2

2

a

a

=

23

3

2

2

a

a

=

2(1.46) 2

2(1.46)

≈ 1.415

(B) Calculator 2 = 1.4142135… (C) a1 = 1

a2 = 21

1

2

2

a

a

=

21 2

2 1

= 1.5

a3 = 22

2

2

2

a

a

=

2(1.5) 2

2(1.5)

≈ 1.417

a4 = 23

3

2

2

a

a

=

2(1.417) 2

2(1.417)

≈ 1.414

75. The first ten terms of the Fibonacci sequence an are 1, 1, 2, 3, 5, 8, 13, 21, 34, 55 The first ten terms of bn are 1, 3, 4, 7, 11, 18, 29, 47, 76, 123

The first ten terms of cn = n

n

b

a are

1

1,

3

1,

4

2,

7

3,

11

5,

18

8,

29

13,

47

21,

76

34,

123

55

In decimal notation this becomes 1, 3, 2, 2.33…, 2.2, 2.25, 2.23…, 2.238…, 2.235…, 2.236…

77. e0.2 = 1 + 0.2

1! +

2(0.2)

2! +

3(0.2)

3! +

4(0.2)

4! = 1.2214000

e0.2 = 1.2214028 (calculator—direct evaluation)


79. 1

n

kk

ca = ca1 + ca2 + ca3 + … + can = c(a1 + a2 + a3 + … an) = c

1

n

kk

a

81. (A) The height of the fifth bounce is a5:

a5 = 5 1 4

1 1 1 10 510 10 10

2 2 16 16 8

= 0.625 feet

The height of the tenth bounce is a10:

a10 = 10 1 9

1 1 1 10 510 10 10

2 2 512 512 256

= 0.01953 feet

(B) 1 1 1 2 1 3 1 4 1 5 110

1

1 1 1 1 1 110 10 10 10 10 10

2 2 2 2 2 2

n

n

6 1 7 1 8 1 9 1 10 1

1 1 1 1 110 10 10 10 10

2 2 2 2 2

= 10 + 5 + 2.5 + 1.25 + 0.625 + 0.3125 + 0.1563 + 0.0781 + 0.0391 + 0.0195 = 91.98 feet This is the sum of the heights of all ten bounces.

83. (A) First year: $40,000 = a1 Second year: $40,000+ 0.04($40,000) = $41,600 = a2 Third year: $41,600 + 0.04($41,600) = $43,264 = a3 Fourth year: $43,264 + 0.04($43,264) = $44, 995 = a4 Fifth year: $44, 995 + 0.04($44, 995) = $46,795 = a5 Sixth year: $46,795 + 0.04($46,795) = $48,667 = a6

(B) The salary for each year is calculated by multiplying the previous year’s salary by 0.04 and adding to the previous year’s salary. This can easily be written as a recursive sequence: an = an−1 + 0.04an−1, which can be simplified a bit as an = 1.04an−1. But we can also write as a non-recursive formula: at each stage, we multiply by one more factor of 1.04, so we can write an = 40,000(1.04)n−1.

(C) 6

1n

n

a = 40,000 + 41,600 + 43,264 + 44,995 + 46,795 + 48,667 = $265,321

This is the total amount earned in the first six years. Section 11−2 1. A counterexample is a single case or example for which a conjecture fails. 3. For the statement Pn, Pk is the statement associated with the positive integer k; Pk+1 is the next statement,

associated with k + 1.

5. (3 + 5)1 = 31 + 51 True (3 + 5)2 = 32 + 52 False Fails at n = 2

7. 12 = 3·1 − 2 True 22 = 3·2 − 2 True 32 = 3·3 − 2 False Fails at n = 3

9. P1: 2 = 2·12 2 = 2 P2: 2 + 6 = 2·22 8 = 8 P3: 2 + 6 + 10 = 2·32 18 = 18

SECTION 11-2 527

11. P1: a5a1 = a5+1 a6 = a6 P2: a5a2 = a5+2 a5a2 = a5(a1a) = (a5a)a = a6a = a7 = a5+2 P3: a5a3 = a5+3 a5a3 = a5(a2a) = a5(a1a)a = [(a5a)a]a = a8 = a5+3

13. P1: 91 − 1 = 8 is divisible by 4 P2: 92 − 1 = 80 is divisible by 4 P3: 93 − 1 = 728 is divisible by 4

15. Pk: 2 + 6 + 10 + … + (4k − 2) = 2k2

Pk+1: 2 + 6 + 10 + … + (4k − 2) + [4(k + 1) − 2] = 2(k + 1)2

or 2 + 6 + 10 + … + (4k − 2) + (4k + 2) = 2(k + 1)2

17. Pk: a5ak = a5+k Pk+1: a5ak+1 = a5+k+1

19. Pk: 9k − 1 = 4r for some integer r Pk+1: 9k+1 − 1 = 4s for some integer s

21. Prove: 2 + 6 + 10 + … + (4n − 2) = 2n2, n N

Write: Pn: 2 + 6 + 10 + … + (4n − 2) = 2n2 Part 1: Show that P1 is true: P1: 2 = 2·12 = 2 P1 is true Part 2: Show that if Pk is true, then Pk+1 is true:

Write out Pk and Pk+1. Pk: 2 + 6 + 10 + … + (4k − 2) = 2k2

Pk+1: 2 + 6 + 10 + … + (4k − 2) + (4k + 2) = 2(k + 1)2

We start with Pk: 2 + 6 + 10 + … + (4k − 2) = 2k2 Adding (4k + 2) to both sides, we get 2 + 6 + 10 + … + (4k − 2) + (4k + 2) = 2k2 + 4k + 2 = 2(k2 + 2k + 1) = 2(k + 1)2 We have shown that if Pk is true, then Pk+1 is true. Conclusion: Pn is true for all positive integers n.

Common Errors: P1 is not the nonsensical statement 2 + 6 + 10 + … + (4·1 − 2) = 2·12 achieved by "substitution" of 1 for n. The left side of Pn has n terms; the left side of P1 must have 1 term. Also, Pk+1 ≠ Pk + 1 Nor is Pk+1 simply its k + 1st term. 4k + 2 ≠ 2(k + 1)2

23. Prove: a5an = a5+n, n N Write: Pn: a5an = a5+n Part 1: Show that P1 is true: P1: a5a1 = a5+1 a5a = a5+1.

True by the recursive definition of an. Part 2: Show that if Pk is true, then Pk+1 is true: Write out Pk and Pk+1. Pk: a5ak = a5+k Pk+1: a5ak+1 = a5+k+1 We start with Pk: a5ak = a5+k Multiply both sides by a: a5aka = a5+ka a5ak+1 = a5+k+1 by the recursive definition of an. We have shown that if Pk is true, then Pk+1 is true. Conclusion: Pn is true for all positive integers n.

25. Prove: 9n − 1 is divisible by 4.

Write: Pn: 9n − 1 is divisible by 4 Part 1: Show that P1 is true: P1: 91 − 1 = 8 = 4·2 is divisible by 4. Clearly true. Part 2: Show that if Pk is true, then Pk+1 is true: Write out Pk and Pk +1. Pk: 9k − 1 = 4r for some integer r Pk +1: 9k+1 − 1 = 4s for some integer s We start with Pk: 9k − 1 = 4r for some integer r

Now, 9k+1 − 1 = 9k+1 − 9k + 9k − 1= 9k(9 – 1) +9k – 1

= 9k ·8 + 9k − 1= 4·9k ·2 + 4r = 4(9k ·2 + r) Therefore, 9k+1 − 1 = 4s for some integer s (= 2·9k + r) 9k+1 − 1 is divisible by 4. We have shown that if Pk is true, then Pk+1 is true. Conclusion: Pn is true for all positive integers n.

27. The polynomial x4 + 1 is a counterexample, since it has degree 4, and no real zeros.


29. 23 is a counterexample, since there are no prime numbers p such that 23 < p < 29.

31. Pn: 2 + 22 + 23 + … + 2n = 2n+1 − 2 Part 1: Show that P1 is true: P1: 2 = 21+1 − 2 = 22 − 2 = 4 − 2 = 2 P1 is true Part 2: Show that if Pk is true, then Pk+1 is true:

Write out Pk and Pk+1. Pk: 2 + 22 + 23 + … + 2k = 2k+1 − 2

Pk +1: 2 + 22 + 23 + … + 2k + 2k +1 = 2k+2 − 2

We start with Pk: 2 + 22 + 23 + … + 2k = 2k+1 − 2 Adding 2k+1 to both sides, 2 + 22 + 23 + … + 2k + 2k+1 = 2k+1 − 2 + 2k+1 = 2k+1 + 2k+1 − 2 = 2·2k+1 − 2 = 21·2k+1 − 2 = 2k+2 − 2

We have shown that if Pk is true, then Pk+1 is true. Conclusion: Pn is true for all positive integers n.

33. Pn: 12 + 32 + 52 + … + (2n − 1)2 = 1

3(4n3 − n)

Part 1: Show that P1 is true: P1: 12 = 1

3(4·13 − 1) =

1

3(4 − 1) = 1 True.

Part 2: Show that if Pk is true, then Pk+1 is true:

Write out Pk and Pk+1. Pk: 12 + 32 + 52 + … + (2k − 1)2 = 1

3(4k3 − k)

Pk +1: 12 + 32 + 52 + … + (2k − 1)2 + (2k + 1)2 = 1

3[4(k + 1)3 − (k + 1)]

We start with Pk: 12 + 32 + 52 + … + (2k − 1)2 = 1

3(4k3 − k)

Adding (2k + 1)2 to both sides,

12 + 32 + 52 + … + (2k − 1)2 + (2k + 1)2 = 1

3(4k3 − k) + (2k + 1)2 =

1

3[4k3 − k + 3(2k + 1)2]

= 1

3[4k3 − k + 3(4k2 + 4k + 1)] =

1

3[4k3 − k + 12k2 + 12k + 3]

= 1

3[4k3 + 12k2 + 12k − k + 3] =

1

3[4k3 + 12k2 + 12k + 4 − k − 1]

= 1

3[4(k3 + 3k2 + 3k + 1) − (k + 1)] =

1

3[4(k + 1)3 − (k + 1)]


35. Pn: 12 + 22 + 32 + … + n2 = ( 1)(2 1)

6

n n n

Part 1: Show that P1 is true: P1: 12 = 1(1 1)(2 1 1)

6

=

1 2 3

6

= 1 P1 is true.


Write out Pk and Pk+1. Pk: 12 + 22 + 32 + … + k2 = ( 1)(2 1)

6

k k k

Pk+1: 12 + 22 + 32 + … + k2 + (k + 1)2 = ( 1)( 2)(2 3)

6

k k k

We start with Pk: 12 + 22 + 32 + … + k2 = ( 1)(2 1)

6

k k k Adding (k + 1)2 to both sides,

SECTION 11-2 529

12 + 22 + 32 + … + k2 + (k + 1)2 = ( 1)(2 1)

6

k k k + (k + 1)2 =

( 1)(2 1)

6

k k k +

26( 1)

6

k

= 2( 1)(2 1) 6( 1)

6

k k k k =

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

6

k k k k

= 2( 1)[2 7 6]

6

k k k =

( 1)( 2)(2 3)

6

k k k


37. Pn: 3

na

a = an−3 n > 3

Part 1: Show that P4 is true: P4: 4

3

a

a = a4−3 But

4

3

a

a =

3

3

a a

a = a = a1 = a4−3 So, P4 is true.


Write out Pk and Pk+1. Pk: 3

ka

a = ak−3

Pk+1: 1

3

ka

a

= ak−2

We start with Pk: 3

ka

a = ak−3

Multiply both sides by a: 3

ka

aa = ak−3a

1

3

ka

a

= ak−3+1 by the recursive definition of an

1

3

ka

a

= ak−2

We have shown that if Pk is true, then Pk+1 is true. Conclusion: Pn is true for all integers n > 3. 39. Write: Pn: aman = am+n m an arbitrary element of N. Part 1: Show that P1 is true: P1: ama1 = am+1 This is true by the recursive definition of an. Part 2: Show that if Pk is true, then Pk+1 is true:

Write out Pk and Pk+1. Pk: amak = am+k Pk+1: amak+1 = am+k+1 We start with Pk: amak = am+k Multiplying both sides by a: amaka = am+ka amak+1 = am+k+1 by the recursive definition of an. We have shown that if Pk is true, then Pk+1 is true.

Conclusion: aman = am+n is true for arbitrary m, n positive integers.

41. Write: Pn: xn − 1 = (x − 1)Qn(x) for some polynomial Qn(x). Part 1: Show that P1 is true: P1: x1 − 1 = (x − 1)Q1(x) for some Q1(x). This is true since we can choose Q1(x) = 1. Part 2: Show that if Pk is true, then Pk+1 is true: Write out Pk and Pk+1. Pk: xk − 1 = (x − 1)Qk(x) for some Qk(x)

Pk+1: xk+1 − 1 = (x − 1)Qk+1(x) for some Qk+1(x)


We start with Pk: xk − 1 = (x − 1)Qk(x) for some Qk(x)

Now, xk+1 − 1 = xk+1 − xk + xk − 1 = xk(x − 1) + xk − 1 = xk(x − 1) + (x − 1)Qk(x) = (x − 1)(xk + Qk(x)) = (x − 1)Qk+1(x) where Qk+1(x) = xk + Qk(x) Conclusion: xn − 1 = (x − 1)Qn(x) for all positive integers n. Thus xn − 1 is divisible by x − 1 for all positive integers n. 43. Write: Pn: x2n − 1 = (x − 1)Qn(x) for some polynomial Qn(x). Part 1: Show that P1 is true: P1: x2 − 1 = (x − 1)Q1(x) for some polynomial Q1(x). This is true since we can choose Q1(x) = x + 1. Part 2: Show that if Pk is true, then Pk+1 is true: Write out Pk and Pk+1. Pk: x2k − 1 = (x − 1)Qk(x) for some Qk(x)

Pk+1: x2k+2 − 1 = (x − 1)Qk+1(x) for some Qk+1(x) Now, x2k+2 − 1 = x2k+2 − x2k + x2k − 1 = x2k(x2 − 1) + x2k − 1 = x2k(x + 1)(x − 1) + (x − 1)Qk(x) = (x − 1)[x2k(x + 1) + Qk(x)] = (x − 1)Qk+1(x) where Qk+1(x) = x2k(x + 1) + Qk(x) Conclusion: x2n − 1 = (x − 1)Qn(x) for all positive integers n. Thus x2n − 1 is divisible by x − 1 for all positive integers n. 45. Pn: 13 + 23 + 33 + … + n3 = (1 + 2 + 3 + … + n)2 Part 1: Show that P1 is true: P1: 13 = 12 True Part 2: Show that if Pk is true, then Pk+1 is true:

Write out Pk and Pk+1. Pk: 13 + 23 + 33 + … + k3 = (1 + 2 + 3 + … + k)2

Pk+1: 13 + 23 + 33 + … + k3 + (k + 1)3 = (1 + 2 + 3 + … + k + k + 1)2

We take it as proved in text Matched Problem 1 that 1 + 2 + 3 + … + n = ( 1)

2

n n for n N

Thus, 1 + 2 + 3 + … + k = ( 1)

2

k k

1 + 2 + 3 + … + k + (k + 1) = ( 1)( 2)

2

k k are known

We start with Pk: 13 + 23 + 33 + … + k3 = (1 + 2 + 3 + … + k)2 Adding (k + 1)3 to both sides:

13 + 23 + 33 + … + k3 + (k + 1)3 = (1 + 2 + 3 + … + k)2 + (k + 1)3 = ( 1)

2

k k

2

+ (k + 1)3

= (k + 1)22

( 1)2

kk

= (k + 1)22 4( 1)

4 4

k k

= (k + 1)2

2 4 4

4

k k

= 2 2( 1) ( 2)

4

k k =

( 1)( 2)

2

k k

2

= [1 + 2 + 3 + … + k + (k + 1)]2

Conclusion: Pn is true for all positive integers n. 47. We note: 2 = 2 = 1·2 n = 1

2 + 4 = 6 = 2·3 n = 2 2 + 4 + 6 = 12 = 3·4 n = 3 2 + 4 + 6 + 8 = 20 = 4·5 n = 4 Hypothesis: Pn: 2 + 4 + 6 + … + 2n = n(n + 1)

SECTION 11-2 531

Proof: Part 1: Show P1 is true P1: 2 = 1·2 True Part 2: Show that if Pk is true, then Pk+1 is true:

Write out Pk and Pk+1. Pk: 2 + 4 + 6 + … + 2k = k(k + 1)

Pk+1: 2 + 4 + 6 + … + 2k + (2k + 2) = (k + 1)(k + 2)

We start with Pk: 2 + 4 + 6 + … + 2k = k(k + 1)

Adding 2k + 2 to both sides: 2 + 4 + 6 + … + 2k + (2k + 2) = k(k + 1) + 2k + 2 = (k + 1)k + (k + 1)2 = (k + 1)(k + 2)

Conclusion: The hypothesis Pn is true for all positive integers n. 49. n = 1: no line is determined.

n = 2: one line is determined.

n = 3: three lines are determined.

n = 4: six lines are determined.

n = 5: ten lines are determined.

Hypothesis: Pn: n points (no three collinear) determine 1 + 2 + 3 + … + (n − 1) = ( 1)

2

n n lines, n ≥ 2.

Proof: Part 1: Show that P2 is true. P2: 2 points determine one line: 1 = 2(2 1)

2

=

2 1

2

= 1 is true

Part 2: Show that if Pk is true, then Pk+1 is true: Write out Pk and Pk+1.

Pk: k points determine 1 + 2 + 3 + … + (k − 1) = ( 1)

2

k k lines

Pk +1: k + 1 points determine 1 + 2 + 3 + … + (k − 1) + k = ( 1)

2

k k lines

We start with Pk: k points determine 1 + 2 + 3 + … + (k − 1) = ( 1)

2

k k lines

Now, the k + 1st point will determine a total of k new lines, one with each of the previously existing k points. These k new lines will be added to the previously existing lines. Hence, k + 1 points determine

1 + 2 + 3 + … + (k − 1) + k = ( 1)

2

k k + k lines = k

11

2

k lines = k

1 2

2

k

lines = ( 1)

2

k k lines

Conclusion: The hypothesis Pn is true for n ≥ 2. 51. Pn: a > 1 an > 1, n N. Part 1: Show that P1 is true: P1: a > 1 a1 > 1. This is automatically true.


Part 2: Show that if Pk is true, then Pk+1 is true: Write out Pk and Pk+1. Pk: a > 1 ak > 1 Pk+1: a > 1 ak+1 > 1

We start with Pk. Further, assume a > 1 and try to derive ak+1 > 1. If this succeeds, we have proved Pk+1. Assume a > 1.

From Pk we know that ak > 1, also 1 > 0, hence a > 0. We may therefore multiply both sides of the inequality ak > 1 by a without changing the sense of the inequality. Hence,

aka > 1a ak+1 > a

But, a > 1 by assumption. Hence, ak+1 > 1. We have derived this from Pk and a > 1. Thus, if Pk is true, then Pk+1 is true. Conclusion: Pn is true for all n N.

53. Pn: n2 > 2n, n ≥ 3. Part 1: Show that P3 is true:

P3: 32 > 2·3 9 > 6 True

Part 2: Show that if Pk is true, then Pk+1 is true: Write out Pk and Pk+1. Pk: k2 > 2k Pk+1: (k + 1)2 > 2(k + 1)

We start with Pk: k2 > 2k Adding 2k + 1 to both sides: k2 + 2k + 1 > 2k + 2k + 1

(k + 1)2 > 2k + 2 + 2k − 1 Now, 2k − 1 > 0, k N, hence 2k + 2 + 2k − 1 > 2k + 2. Therefore, (k + 1)2 > 2k + 2

(k + 1)2 > 2(k + 1) Thus, if Pk is true, then Pk+1 is true. Conclusion: Pn is true for all integers n ≥ 3. 55. The statement is true. Proof by mathematical induction: Pn: 1 – 2 + 3 − … (2n – 1) = n

Show P1 is true. 1 = 1

Show if Pk is true, Pk+1 is true. Write out Pk and Pk+1. Pk: 1 – 2 + 3 − … + (2k – 1) = k Pk+1: 1 – 2 + 3 − … + (2k + 1) = k + 1

We start with Pk: 1 – 2 + 3 − … + (2k – 1) = k Now add −2k + (2k + 1) to both sides:

1 – 2 + 3 … + (2k – 1) − 2k + (2k + 1) = k – 2k + (2k + 1) = k – 2k + 2k + 1 = k + 1 We have shown that if Pk is true, Pk+1 is true. Conclusion: 1 – 2 + 3 − … + (2n – 1) = n for all alternating sums of an odd number of positive integers.

57. The statement is false. For example, if n = 3

34 + 44 + 54 + 64 = 2258 while 74 = 2401.

59. To prove an = bn, n N, write: Pn: an = bn. Proof: Part 1: Show P1 is true. P1: a1 = b1 a1 = 1 b1 = 2·1 − 1 = 1 Thus, a1 = b1

61. To prove: an = bn, n N, write: Pn: an= bn. Proof: Part 1: Show P1 is true. P1: a1 = b1 a1 = 2 b1 = 22·1−1 = 21 = 2

SECTION 11-3 533

Part 2: Show that if Pk is true, then Pk+1 is true: Write out Pk and Pk+1. Pk: ak = bk Pk+1: ak+1 = bk+1 We start with Pk: ak = bk Now, ak+1 = ak+1−1 + 2 = ak + 2 = bk + 2

= 2k − 1 + 2 = 2k + 1 = 2(k + 1) − 1 = bk+1

Therefore, ak+1 = bk+1 Thus, if Pk is true, then Pk+1 is true. Conclusion: Pn is true for all n N.

Hence, {an} = {bn}.

Thus, a1 = b1 Part 2: Show that if Pk is true, then Pk+1 is true: Write out Pk and Pk+1. Pk: ak = bk Pk+1: ak+1 = bk+1 We start with Pk: ak = bk Now, ak+1 = 22ak+1−1 = 22ak = 22bk = 2222k−1

= 22+2k−1 = 22(k+1)−1 = bk+1 Therefore, ak+1 = bk+1 Thus, if Pk is true, then Pk+1 is true.

Conclusion: Pn is true for all n N. Hence, {an} = {bn}.

Section 11−3

1. An arithmetic sequence is a sequence in which each term differs from the previous term by the same amount.

3. The common difference is the difference between any two successive terms of an arithmetic sequence. The common ratio is the ratio of any term after the first in a geometric sequence to the previous term.

5. Only the sequence an = 0 has a sum. 7. (A) Since (−16) − (−11) = −5 and (−21) − (−16) = −5, the given terms can start an arithmetic sequence

with d = −5. The next terms are then −21 + (−5) = −26, and (−26) + (−5) = −31. (B) Since (−4) − 2 = −6 and 8 − (−4) = 12, there is no common difference. Since (−4) ÷ 2 = −2 and 8 ÷ (−4) = −2, the given terms can start a geometric sequence with r = −2. The next terms are then 8·(−2) = −16, and (−16)·(−2) = 32. (C) Since 4 − 1 ≠ 9 − 4, there is no common difference, so the sequence is not an arithmetic sequence.

Since 4 ÷ 1 ≠ 9 ÷ 4, there is no common ratio, so the sequence is not geometric either.

(D) Since 1

6 −

1

2 = −

1

3 and

1

18 −

1

6 = −

1

9, there is no common difference. Since

1

6 ÷

1

2 =

1

3 and

1

18 ÷

1

6 =

1

3, the given terms can start a geometric sequence with r =

1

3. The next terms are then

1

18 ·

1

3 =

1

54 and

1

54 ·

1

3 =

1

162.

9. a2 = a1 + d = −5 + 4 = −1 a3 = a2 + d = −1 + 4 = 3 a4 = a3 + d = 3 + 4 = 7 an = a1 + (n − 1)d

11. a15 = a1 + 14d = −3 + 14·5 = 67

Sn = 2

n[2a1 + (n − 1)d]

S11 = 11

2[2(−3) + (11 − 1)5]

= 11

2(44) = 242

13. a2 − a1 = 5 − 1 = d = 4

Sn = 2

n[2a1 + (n − 1)d]

S21 = 21

2[2·1 + (21 − 1)4]

= 21

2(82) = 861

15. a2 − a1= 5 − 7 = −2 = d

an = a1 + (n − 1)d a15 = 7 + (15 − 1)(−2) = −21

17. a2 = a1r = (−6)1

2

= 3

a3 = a2r = 31

2

= −3

2

a4 = a3r = 3

2

1

2

= 3

4

19. an = a1rn−1

a10 = 811

3

10−1

= 1

243


21. Sn = 1

1na ra

r

S7 = 3 3(2,187)

1 3

= 3,279

23. an = a1 + (n − 1)d a20 = a1 + 19d 117 = 3 + 19d

114 = 19d So, d = 6. Therefore a101 = a1 + (100)d = 3 + 100(6) = 603

25. Sn = 2

n(a1 + an)

S40 = 40

2(−12 + 22)

= 200

27. a2 − a1 = d

1

2 −

1

3 =

1

6 = d

an = a1 + (n − 1)d

a11 = 1

3 + (11 − 1)

1

6= 2

Sn = 2

n(a1 + an)

S11 = 11

2

12

3

= 77

6

29. an = a1 + (n − 1)d a10 = a1 + 9d a3 = a1 + 2d Eliminating d between these two statements by addition, we have

2a10 = 2a1 + 18d −9a3 = −9a1 − 18d

10 3 12 – 9 – 7a a a

a1 = 10 32 9

7

a a

= 2(55) 9(13)

7

= 1

31. a1 = 8, a2 = 2 a2 = a1r 2 = 8r

r = 1

4

33. a1 = 120 a4 = −15 an = a1rn−1 a4 = a1r3 −15 = 120r3

r3 = −1

8

r = −1

2

35. Sn = 1 1

1

na a r

r

S10 = 102

323

9 9( )

1

=

58,0256,561

13

= 58,025

2,187

37. First find r an = a1rn−1 a8 = a1r7

2,187 = 1·r7 r = 3

Therefore S8 = 8

1 1

1

a a r

r

= 81 1 3

1 3

= 6,560

2

= 3,280

39. a3 = a1r2, a6 = a1r5 72 = a1r2, −243 = a1r5 Dividing, we obtain

5

12

1

a r

a r = r3 = −

243

72

r3 = −27

8

r = −3

2

Therefore

72 = a1

23

2

a1 = 722

2

3

= 32

41. a4 = a1r3

−1 = 1·r3 r3 = −1 r = −1 a100 = a1r99 = 1·(−1)99 = −1

SECTION 11-3 535

43. an = a1 + (n − 1)d d = a2 − a1 = (3·2 + 3) − (3·1 + 3) = 3 a51 = a1 + (51 − 1)d = (3·1 + 3) + 50·3 = 156

Sn = 2

n(a1 + an)

S51 = 51

2( a1 + a51) =

51

2 [6 + 156] = 4,131

45. Sn = 1 1

1

na a r

r

First, note a1 = (−3)1−1 = 1

r = 2

1

a

a =

2 1

1 1

( 3)

( 3)

= −3

S7 = 71 1( 3)

1 ( 3)

= 2,188

4= 547

47. First, find n: an = a1 + (n − 1)d 134 = 22 + (n − 1)2 n = 57

Now, find S57

Sn = 2

n(a1 + an)

S57 = 57

2(22 + 134) = 4,446

49. To prove: 1 + 3 + 5 + … + (2n − 1) = n2 The sequence 1, 3, 5, … is an arithmetic sequence,

with d = 2. We are to find Sn.

But, Sn = 2

n (a1 + an) =

2

n [1 + (2n − 1)] =

2

n · 2n

So, Sn = n2

51. r = 2

1

a

a =

1

2 ÷ 2 =

1

4

Since |r| = 1

4 < 1, the series has a sum.

S∞ = 1

1

a

r =

14

2

1 =

8

3

53. r = 2

1

a

a =

1

3

Since |r| = 1

3

=

1


S∞ = 1

1

a

r =

13

3

1 =

9

4

55. r = 2

1

a

a =

0.1

1 = 0.1

Since |r| = 0.1 < 1, the series has a sum.

S∞ = 1

1

a

r =

1

1 0.1 =

10

9

57. r = 2

1

a

a =

1

2 ÷ (−1) = −

1

2

Since |r| = 1

2 =

1


S∞ = 1

1

a

r =

12

1

1

= −2

3

59. r = 2

1

a

a =

1

1 = −1

Since |r| = |−1| = 1 ≥ 1, the series has no sum.

61. 0. 7 = 0.777… = 0.7 + 0.07 + 0.007 + 0.0007 + … This is an infinite geometric series with a1 = 0.7 and

r = 0.1. Thus,

S∞ = 1

1

a

r =

0.7

1 0.1 =

0.7

0.9 =

7

9

63. 0. 54 … = 0.54 + 0.0054 + 0.000054 + …

This is an infinite geometric series with a1 = 0.54 and r = 0.01. Thus, S∞ = 1

1

a

r =

0.54

1 0.01 =

0.54

0.99 =

6

11


65. 3. 216 = 3.216216216… = 3 + 0.216 + 0.000216 + 0.000000216 + … Therefore, we note: 0.216 + 0.000216 + 0.000000216 + … is an infinite geometric series with a1 = 0.216

and r = 0.001. Thus, 3. 216 = 3 + S∞ = 3 + 1

1

a

r = 3 +

0.216

1 0.001 = 3 +

0.216

0.999 = 3 8

37 or 119

37

69. This is a geometric sequence with ratio 1

n

n

a

a = −3.

Hence, an = a1rn−1 = (−2)(−3)n−1

67. Write: Pn: an = a1 + (n − 1)d Proof: Part 1: Show that P1 is true:

P1: a1 = a1 + (1 − 1)d = a1 P1 is true. Part 2: Show that if Pk is true, then Pk +1 is true: Write out Pk and Pk +1. Pk: ak = a1 + (k − 1)d Pk+1: ak +1 = a1 + kd

We start with Pk: ak = a1 + (k − 1)d Adding d to both sides: ak + d = a1 + (k − 1)d + d ak+1 = a1 + d[(k − 1) + 1] ak+1 = a1 + kd Thus, if Pk is true, then Pk+1 is true. Conclusion: Pn is true for all positive integers n.

71. Write: Pn: an = a1rn−1, n N Proof: Part 1: Show that P1 is true: P1: a1 = a1r1−1 = a1r0 = a1 True. Part 2: Show that if Pk is true, then Pk+1 is true: Write out Pk and Pk+1. Pk: ak = a1rk−1 Pk+1: ak+1 = a1rk We start with Pk: ak = a1rk−1 Multiply both sides by r: akr = a1rk−1r akr = a1rk But, we are given that ak+1 = rak = akr.

Hence, ak+1 = a1rk Thus, if Pk is true, then Pk+1 is true. Conclusion: Pn is true for all n N.

73. If 1

n

n

a

a = r and an = an−1 + d, then

an = ran−1 = an−1 + d ran−1 – an−1 = d an−1(r – 1) = d Thus, an−1 is a constant for all terms of the sequence. Moreover, if all terms are the same constant, d = 0 and r = 1.

75. With each firm, the salaries form an arithmetic sequence. We are asked for the sum of fifteen terms, or S15, given a1 and d.

Sn = 2

n[2a1 + (n − 1)d ]

S15 = 15

2[2a1 + 14d ] = 15(a1 + 7d)

Firm A: a1 = 25,000 d = 1,200 S15 = 15(25,000 + 7·1,200) = $501,000 Firm B: a1 = 28,000 d = 800 S15 = 15(28,000 + 7·800) = $504,000

77. We are asked for the sum of an infinite geometric series. a1 = $800,000 r = 0.8 |r| ≤ 1, so the series has a sum,

S∞= 1

1

a

r=

$800,000

1 0.8=

$800,000

0.2= $4,000,000

79. After one year, P(1 + r) is present. Hence, the geometric sequence has a1 = P(1 + r). The ratio is given as (1 + r), hence an = a1(1 + r)n−1 = P(1 + r)(1 + r)n−1 = P(1 + r)n is the amount present after n years. The time taken for P to double is represented by n years. We set A = 2P, r = 0.06, then solve 2P = P(1 + 0.06)n 2 = (1.06)n log 2 = n log 1.06

n = log 2

log1.06≈ 12 years

SECTION 11-3 537

81. This involves an arithmetic sequence. Let d = increase in earnings each year. a1 = 7,000 a11 = 14,000 an = a1 + (n − 1)d 14,000 = 7,000 + (11 − 1)d d = $700 The amount of money received over the 11 years is S11 = a1 + a2 + … + a11

Sn = 2

n(a1 + an)

S11 = 11

2(7,000 + 14,000) = $115,500

83. We are asked for the sum of an infinite geometric series. a1 = number of revolutions in the first minute = 300

r = 2

3 |r| < 1, so the series has a sum,

S∞ = 1

1

a

r =

23

300

1 = 900 revolutions.

85. This involves a geometric sequence. Let an = 2,000 calories. There are five stages: n = 5. We require a1 on the assumption

that r = 20% = 1

5.

an = a1rn−1

2000 = a11

5

5−1

a1 = 2,000·54 = 1,250,000 calories

87. We have an arithmetic sequence, 16, 48, 80, …, with a1 = 16, d = 32. (A) This requires a11: an = a1 + (n − 1)d a11 = 16 + (11 − 1)32 = 336 feet

(B) This requires s11: sn = 2

n(a1 + an)

s11 = 11

2(16 + 336) = 1,936 feet

(C) This requires st: sn = 2

n [2a1 + (n − 1)d]

st = 2

t[2a1 + (t − 1)32] =

2

t [32 + (t − 1)32]

= 2

t · 32t = 16t2 feet

89. This involves a geometric sequence. Let a1 = 2A0 = number present after 1 half−hour

period. In t hours, 2t half−hours will have elapsed, hence, n = 2t; r = 2, since the number of bacteria doubles in each period.

an = a1rn−1 a2t = 2A022t−1 = A02122t−1 = A022t

91. If bn is the brightness of an nth−magnitude star, we find r for the geometric progression b1, b2, b3, … given b1 = 100b6.

bn = b1rn−1 b6 = b1r6−1 b6 = b1r5 b6 = 100b6r5

1

100 = r 5

r = 5 0.01 = 10−0.4 = 0.398

93. This involves a geometric sequence. a1 = amount of money on first square; a1 = $0.01, r = 2; an = amount of money on nth square. an = a1rn−1 a64 = 0.01 · 264−1 = 9.22 × 1016 dollars

The amount of money on the whole board = a1 + a2 + a3 + … + a64 = S64.

Sn = 1

1na ra

r

S64 = 160.01 2(9.223 10 )

1 2

= 1.845 × 1017 dollars


95. This involves a geometric sequence. Let a1 = 15 = pressure at sea level, r = 1

10 = factor of decrease. We

require a5 = pressure after four 10−mile increases in altitude.

a5 = 151

10

5−1

= 0.0015 pounds per square inch.

97. This involves an infinite geometric series. a1 = perimeter of first triangle = 1, r = 1

2, |r | < 1, so the series

has a sum. S∞ = 1

1

a

r =

12

1

1 = 2

99.

1 12

1

2 3

1

23

4

From the figure, and ordinary induction, it should be clear that the interior angles of an n + 2-sided

polygon, n = 1, 2, 3, …, are those of n triangles, that is, 180°n. Thus, for the sequence of interior angles {an}, an − an −1 = 180°n − 180°(n − 1) = 180° = d.

A detailed proof by mathematical induction is omitted. For a 21−sided polygon, we use an = a1 + (n − 1)d with a1 = 180°, d = 180°, and n + 2 = 21, hence, n = 19. a19 = 180° + (19 − 1)180° = 3,420°

Section 11−4 1. A permutation is an arrangement or ordering of n objects without repetition. 3. If we write a1 = 1 and an = nan−1, then this sequence recursively generates n! so that an = n!. 5. In a permutation of n objects, order matters. In a combination of n objects taken k at a time, order does

not matter.

7. 9! = 362,880 9. 11! = 39,916,800 11. 11!

8! =

11 10 9 8!

8!

= 990

13. 5!

2!3! =

5 4 3!

2 1 3!

= 10 15.7!

4!(7 4)! =

7!

4!3! =

7 6 5 4!

4!3 2 1

= 35

17. The alleged solution for number 11 is incorrect. The function "11 n Pr8" represents the permutations of

11 objects taken 8 at a time, which is calculated by 11!

(11 8)! =

11!

3!

This is different from problem 11, which is 11!/8!. The alleged solution for number 13 is incorrect also.

The calculator interprets the input 5!/2!3! as 5!

2! · 3!. The correct calculation is 5!/(2!3!). The solution for

number 15 is correct.

19. P13,4 = 13!

(13 4)! =

13!

9! =

13 12 11 10 9!

9!

= 17,160

21. P13,9 = 13!

(13 9)! =

13!

4! =

6,227,020,800

24 = 259,459,200

23. C15,8 = 15!

8!(15 8)! =

15!

8!7! = 6,435 (using ! from the MATH PROB menu of the TI-83)

25. C15,12 = 15!

12!(15 12)! =

15!

12!3! =

15 14 13 12!

12!(3 2 1)

= 2,730

6 = 455

SECTION 11-4 539

27. (A) Order is important. The selection is a permutation. (B) Order is not important. The selection is a combination.

29. O1: Selecting the color N1: 5 ways

O2: Selecting the transmission N2: 3 ways O3: Selecting the interior N3: 4 ways O4: Selecting the engine N4: 2 ways Applying the multiplication principle, there are 5·3·4·2 = 120 variations of the car.

31. Order is important here. We use permutations, selecting, in order, three horses out of ten:

P10,3 = 10·9·8 = 720 different finishes 33. For the subcommittee, order is not important. We use combinations, selecting three persons out of seven:

C7,3 = 7!

3!(7 3)! =

7!

3!4! =

7 6 5 4!

3 2 1 4!

= 35 subcommittees

In choosing a president, a vice-president, and a secretary, we can use permutations, or apply the multiplication principle. O1: Selecting the president N1: 7 ways O2: Selecting the vice-president N2: 6 ways (the president is not considered) O3: Selecting the secretary N3: 5 ways (the president and vice-president are not considered) Thus, there are N1·N2·N3 = 7·6·5 (= P7,3) = 210 ways.

35. For each game, we are selecting two teams out of ten to be opponents. Since the order of the opponents

does not matter (this has nothing to do with the order in which the games might be played, which is not

under discussion here), we use combinations. C10,2 = 10!

2!(10 2)! =

10!

2!8! =

10 9 8!

2 1 8!

= 45 games

37. No letter can be repeated Allowing letters to repeat

O1: Selecting first letter N1: 6 ways N1: 6 ways O2: Selecting second letter N2: 5 ways N2: 6 ways O3: Selecting third letter N3: 4 ways N3: 6 ways O4: Selecting fourth letter N4: 3 ways N4: 6 ways

P6,4 = 6·5·4·3 = 360 6·6·6·6 = 1,296 possible code words possible code words

39. No digit can be repeated Allowing digits to repeat

O1: Selecting first digit N1: 10 ways N1: 10 ways O2: Selecting second digit N2: 9 ways N2: 10 ways

M M M O5: Selecting fifth digit N5: 6 ways N5: 10 ways

P10,5 = 10·9·8·7·6 10·10·10·10·10 = 30,240 lock combinations = 100,000 lock combinations

41. We are selecting five cards out of the 13 hearts in the deck. The order is not important, so we use combinations.

C13,5 = 13!

5!(13 5)! =

13!

5!8! =

13 12 11 10 9 8!

5 4 3 2 1 8!

= 1,287


43. Repeats allowed No repeats allowed O1: Selecting first letter N1: 26 ways N1: 26 ways O2: Selecting second letter N2: 26 ways N2: 25 ways O3: Selecting third letter N3: 26 ways N3: 24 ways O4: Selecting first digit N4: 10 ways N4: 10 ways O5: Selecting second digit N5: 10 ways N5: 9 ways O6: Selecting third digit N6: 10 ways N6: 8 ways

26·26·26·10·10·10 26·25·24·10·9·8 =17,576,000 license plates =11,232,000 license plates

45. O1: Choosing 5 spades out of 13 possible (order is not important) N1: C13,5 O2: Choosing 2 hearts out of 13 possible (order is not important) N2: C13,2 Using the multiplication principle, we have:

Number of hands = C13,5·C13,2 = 13!

5!(13 5)!·

13!

2!(13 2)! = 1,287·78 = 100,386

47. O1: Choosing 3 appetizers out of 8 possible (order is not important) N1: C8,3 O2: Choosing 4 main courses out of 10 possible (order is not important) N2: C10,4 O3: Choosing 2 desserts out of 7 possible (order is not important) N3: C7,2 Using the multiplication principle, we have:

Number of banquets = C8,3·C10,4·C7,2 = 8!

3!(8 3)!·

10!

4!(10 4)!·

7!

2!(7 2)! = 56·210·21 = 246,960

49. (A) P10,0 = 1 0! = 1 P10,1 = 10 1! = 1 P10,2 = 90 2! = 2 P10,3 = 720 3! = 6 P10,4 = 5,040 4! = 24 P10,5 = 30,240 5! = 120 P10,6 = 151,200 6! = 720 P10,7 = 604,800 7! = 5,040 P10,8 = 1,814,400 8! = 40,320 P10,9 = 3,628,800 9! = 362,880 P10,10 = 3,628,800 10! = 3,628,800

Thus, P10,r ≥ r! for r = 0, 1, … , 10

(B) If r = 0, P10,0 = 0! = 1. If r = 10, P10,10 = 10! = 3,628,800 (C) Pn,r and r! are each the product of r consecutive integers, the largest of which is n for Pn,r

and r for r!. Thus if r ≤ n, Pn,r ≥ r! 51. O1: Choosing a left glove out of 12 possible N1: 12

O2: Choosing a right glove out of all the right gloves that do not match the left glove already chosen N2: 11

Using the multiplication principle, we have: Number of ways to mismatch gloves = 12·11 = 132. 53. (A) We are choosing 2 points out of the 8 to join by a chord. Order is not important.

C8,2 = 8!

2!(8 2)! = 28 chords

SECTION 11-4 541

(B) No three of the points can be collinear, since no line intersects a circle at more than two points. Thus, we can select any three of the 8 to use as vertices of the triangle. Order is not important.

C8,3 = 8!

3!(8 3)! = 56 triangles

(C) We can select any four of the eight points to use as vertices of a quadrilateral. Order is not important.

C8,4 = 8!

4!(8 4)! = 70 quadrilaterals

55. To seat two people, we can seat one person, then the second person. O1: Seat the first person in any chair N1: 5 ways O2: Seat the second person in any remaining chair N2: 4 ways Thus, applying the multiplication principle, we can seat two persons in N1·N2 = 5·4 = 20 ways. We can continue this reasoning for a third person. O3: Seat the third person in any of the three remaining chairs N3: 3 ways Thus we can seat 3 persons in 5·4·3 = 60 ways.

For a fourth person: O4: Seat the fourth person in any of the two remaining chairs N4: 2 ways Thus we can seat 4 persons in 5·4·3·2 = 120 ways. For a fifth person: O5: Seat the fifth person. There will be only one chair remaining. N5: 1 way Thus we can seat 5 persons in 5·4·3·2·1 = 120 ways

57. (A) Order is important, so we use permutations, selecting, in order, 5 persons out of 8:

P8,5 = 8·7·6·5·4 = 6,720 teams. (B) Order is not important, so we use combinations, selecting 5 persons out of 8

C8,5 = 8!

5!(8 5)! =

8!

5!3! =

8 7 6 5!

5! 3 2 1

= 56 teams

(C) O1: Selecting either Mike or Ken out of {Mike, Ken} N1: C2,1 O2: Selecting the 4 remaining players out of the 6 possibilities that do not

include either Mike or Ken. N2: C6,4 Using the multiplication principle, we have

N1·N2 = C2,1·C6,4 = 2!

1!(2 1)!·

6!

4!(6 4)!=

2!

1!1!·

6!

4!2!= 2·15 = 30 teams

59. The number of ways to deal a hand containing exactly 1 king is computed as follows:

O1: Choosing 1 king out of 4 possible (order is not important) N1: C4,1 O2: Choosing 4 cards out of 48 possible (order is not important) N2: C48,4

Using the multiplication principle, we have

N1·N2 = C4,1·C48,4 = 4!

1!(4 1)!·

48!

4!(48 4)!=

4!

1!3!·

48!

4!44!= 4·194,580 = 778,320 ways

The number of ways to deal a hand containing no hearts is

C39,5 = 39!

5!(39 5)! =

39!

5!34! = 575,757 ways

Thus the hand that contains exactly one king is more likely.


Section 11−5 1. The sample space for an experiment is the set of simple events for that experiment. 3. A theoretical probability is found by making reasonable assumptions about an experiment and assigning

probabilities on this basis. An empirical probability is found by actually performing the experiment.

5. A suitable sample space is S = {1, 2, 3, 4, 5, 6}. Assigning the probability of 1

6 to each simple event in

S, we find that E = {1, 6} has probability 1

6 +

1

6 =

1

3.

7. Since there are 52 equally likely cards and 26 of them are red, the event of drawing a red card has

probability 26

52 =

1

2.

9. A suitable sample space is S = {HH, HT, TH, TT}. Assigning the probability of 1

4 to each simple event in

S, we find that E = {HH} has probability 1

4.

11. A suitable sample space has the 36 elements shown in Example 2 of the text. Assigning the probability of 1

36 to each simple event in this space, we find that the probability of

E = {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)} is 6

36 =

1

6.

13. Since there are 52 equally likely cards and 4 of them are kings and 4 of them are queens, the event of

drawing a king or queen has probability 4

52 +

4

52 =

2

13.

15. A suitable sample space is S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}. Assigning the

probability of 1

8 to each simple event in S, we find that the probability of E = {HTT, THT, TTT} is

3

8.

17. The occurrence of E is certain. 19. (A) Must be rejected. It violates condition 1; no probability can be negative. (B) Must be rejected. It violates condition 2, since P(R) + P(G) + P(Y) + P(B) ≠ 1. (C) Is an acceptable probability assignment. 21. This is a compound event made up of the simple events R and Y. P(E) = P(R) + P(Y) = .26 + .30 = .56

23. P(E) ≈ ( )f E

n =

25

250 = .1 25. P(E) ≈

( )f E

n =

189

420 = .45

27. The sample space S is the set of all possible sequences of three digits with no repeats. n(S) = P10,3 = 10·9·8 = 720. The event is the simple event of one particular sequence. n(E) = 1.

Then P(E) = ( )

( )

n E

n S =

1

720 ≈ .0014.

29. The sample space S is the set of all possible 5-card hands, chosen from a deck of 52 cards, n(S) = C52,5. The event E is the set of all possible 5-card hands that are all black, thus, are chosen from the 26 black

cards. n(E) = C26,5. P(E) = ( )

( )

n E

n S = 26,5

52,5

C

C =

26!

5!(26 5)! ÷

52!

5!(52 5)! ≈ .025

SECTION 11-5 543

31. The sample space S is the set of all possible 5-card hands, chosen from a deck of 52 cards. n(S) = C52,5. The event E is the set of all possible 5-card hands that are all face cards, thus, are chosen from the 4 face cards in each of 4 suits, or 16 face cards. n(E) = C16,5.

P(E) = ( )

( )

n E

n S = 16,5

52,5

C

C =

16!

5!(16 5)! ÷

52!

5!(52 5)! ≈ .0017

33. Yes. There are four aces in a deck so a 5-card hand could have anywhere from zero to four aces. But it is not an equally likely sample space. Drawing four aces in a 5-card hand is far less likely than drawing none.

35. The sample space S is the set of all possible four digit numbers less than 5,000 formed from the digits 1,

3, 5, 7, 9. n(S) = N1·N2·N3·N4 where N1 = 2 (digits 1 or 3 only for the first digit since the number is less than 5,000) N2 = N3 = N4 = 5 (digits 1, 3, 5, 7, 9)

n(S) = 2·5·5·5 = 250 The event E is the set of those numbers in S which are divisible by 5. These must end in 5. n(E) = N1·N2·N3·N4 where

N4 = 1 (digit 5 only) n(E) = 2·5·5·1 = 50

P(E) = ( )

( )

n E

n S =

50

250 = .2

37. The sample space S is the set of all arrangements of the five notes in the five envelopes.

n(S) = P5,5 = 5! = 120. The event E is the one correct arrangement. n(E) = 1. P(E) = ( )

( )

n E

n S =

1

120 ≈ .008

In problems 39-49, using Figure 1 of the text, we have that S = the set of all possible ordered pairs of dots showing, n(S) = 36. 39. The sum of a and b can be 3 in 2 ways. E = {(1, 2), (2, 1)}

n(E) = 2

P(E) = ( )

( )

n E

n S =

2 1

36 18

41. The sum of a and b can be greater than nine in 6 ways. E = {(4, 6), (5, 6), (6, 6), (5, 5), (6, 5), (6, 4)} n(E) = 6

P(E) = ( )

( )

n E

n S =

6 1

36 6

43. The product of a and b can be 12 in 4 ways. E = {(2, 6), (3, 4), (4, 3), (6, 2)}

n(E) = 4

P(E) = ( )

( )

n E

n S =

4

36 =

1

9

45. The product of a and b can be less than 5 in 8 ways. E = {(1, 1), (2, 1), (3, 1), (4, 1), (1, 2), (2, 2), (1, 3), (1, 4)} n(E) = 8

P(E) = ( )

( )

n E

n S =

8

36 =

2

9

47. a can equal b in 6 ways.

E = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)} n(E) = 6

P(E) = ( )

( )

n E

n S =

6

36 =

1

6

49. From the last row together with the right-hand column of the figure, there are 11 possibilities for at least one of a and b to be 6. n(E) = 11

P(E) = ( )

( )

n E

n S =

11

36


51. The sample space is the set of events consisting of choosing a person with birthday on day 1 through day 365 of the year. (February 29 is neglected.) S = {1, 2, 3, … , 365}

Since each of these days is essentially equally likely, P(ei) = 1

365.

53. (A) Use P(E) ≈ ( )f E

n with n = 500.

P(2) = 11

500 = .022 P(3) =

35

500 = .07 P(4) =

44

500 = .088

P(5) = 50

500 = .1 P(6) =

71

500 = .142 P(7) =

89

500 = .178

P(8) = 72

500 = .144 P(9) =

52

500 = .104 P(10) =

36

500 = .072

P(11) = 26

500 = .052 P(12) =

14

500 = .028

(B) Use Figure 1 of the text, with S = the set of all possible ordered pairs of dots showing, n(S) = 36.

P(2) = 1

36 P(3) =

2

36 P(4) =

3

36 P(5) =

4

36

P(6) = 5

36 P(7) =

6

36 P(8) =

5

36 P(9) =

4

36

P(10) = 3

36 P(11) =

2

36 P(12) =

1

36

(C) The expected frequency of the occurrence of E is n·P(E). Thus, the expected frequency of the

occurrence of sum 2 is 500·P(2) = 500·1

36 or 13.9.

The expected frequency of the occurrence of sum 3 is 500·P(3) = 500·2

36 = 27.8.


36 = 41.7.


36 = 55.6.


36 = 69.4.


36 = 83.3.


36 = 69.4.


36 = 55.6.


36 = 41.7.


36 = 27.8.


36 = 13.9.

SECTION 11-5 545

(D) Use the command rand Int (i, k, n). Let 1 to 6 represent the possible outcomes in tossing one die; then rand Int (1, 6, 500) will give a list of 500 possible outcomes in tossing one die.

rand Int (1, 6, 500) + rand Int (1, 6, 500) will give a list of 500 possible outcomes in tossing 2 dice. A possible result is outlined here, but the student's result will be different (with a high probability!)

Generating the random Setting up the Selecting the window numbers. histogram. variables.

Histogram Use trace to display any result.

55. (A) It is possible, but very unlikely, to get 29 heads in 30 flips of a coin. If one thinks of the 30 flips as a

sequence of words with 30 letters chosen from {H, T}, then there are 2 ways to choose each of the letters and therefore 230 such words. Only 30 of these words have exactly 29 H's because only 30 have exactly one T. Thus the probability of getting 29 heads in 30 flips of a fair coin is

P(E) = ( )

( )

n E

n S =

30

30

2 ≈ 2.8 10−8

(B) There is a strong suspicion that the coin is unfair, because this event is unlikely, although not as unlikely as in part A. If the coin is unfair, the empirical probabilities are

P(H) ≈ ( )f H

n =

42

50 = .84 P(T) ≈

( )f T

n =

8

50 = .16

57. The sample space for this experiment = {HTH, HHH, THH, TTH} assuming that the third coin is the one

with a head on both sides. Thus n(S) = 4. E is the event {TTH}. Thus n(E) = 1.

P(E) = ( )

( )

n E

n S =

1

4

59. See problem 57. n(S) = 4. E is the event {HHH}. Thus n(E) = 1

P(E) = ( )

( )

n E

n S =

1

4

61. See problem 57. n(S) = 4. E is the event {HTH, HHH, THH}. Thus n(E) = 3

P(E) = ( )

( )

n E

n S =

3

4


63. Since it is equally likely that each die shows a 1, a 2 or a 3, we can draw the following figure to represent S, the set of all possibilities:

First Die

Second Die

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

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

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

n(S) = 9 The sum of the dots can be 2 in exactly one way. n(E) = 1

P(E) = ( )

( )

n E

n S =

1

9

65. See figure, problem 63, n(S) = 9. The sum of the dots can be 4 in 3 ways: E = {(1, 3), (2, 2), (3, 1)}

n(E) = 3

P(E) = ( )

( )

n E

n S =

3

9 =

1

3

67. See figure, problem 63. n(S) = 9. The sum of the dots can be 6 in exactly one way: E = {(3, 3)}. n(E) = 1

P(E) = ( )

( )

n E

n S =

1

9

69. See figure, problem 63. n(S) = 9. The sum of the dots will be odd if the sum is 3 or 5.

E = {(1, 2), (2, 1), (2, 3), (3, 2)}. n(E) = 4

P(E) = ( )

( )

n E

n S =

4

9

71. The sample space S is the set of all possible 5-card hands, chosen from a deck of 52 cards. n(S) = C52,5 = 2,598,960. The event E is the set of all possible 5-card hands that are all jacks through aces,

thus, are chosen from the 16 cards (4 jacks, 4 queens, 4 kings, 4 aces) of this type. n(E) = C16,5 = 4,368.

P(E) = ( )

( )

n E

n S = 16,5

52,5

C

C =

4,368

2,598,960 ≈ .00168

73. The sample space S is the set of all possible 5-card hands chosen from a deck of 52 cards. n(S) = C52,5.

The event E is the set of all possible 5-card hands of the type AAAAx where x is any of the 52 − 4 = 48 cards that are not aces.

n(E) = 48

P(E) = ( )

( )

n E

n S =

52,5

48

C =

48

2,598,960 ≈ .000 0185

75. The sample space S is the set of all possible 5-card hands, chosen from a deck of 52 cards. n(S) = C52,5.

The event E is the set consisting of the cards ace, king, queen, jack, ten, in each of the 4 suits, abbreviated:

{club AKQJ10, diamond AKQJ10, heart AKQJ10, spade AKQJ10}. Thus n(E) = 4.

P(E) = ( )

( )

n E

n S =

52,5

4

C =

4

2,598,960 ≈ .000 0015

SECTION 11-6 547

77. The sample space S is the set of all possible 5-card hands, chosen from a deck of 52 cards. n(S) = C52,5. The event E is the set of all possible 5-card hands chosen as follows: O1: Choose 2 aces out of the set of 4 aces N1: C4,2

O2: Choose 3 queens out of the set of 4 queens N2: C4,3 Thus, applying the multiplication principle, n(E) = N1N2 = C4,2C4,3

P(E) = ( )

( )

n E

n S = 4,2 4,3

52,5

C C

C

=

6 4

2,598,960

= .000 009

79. (A) P (this event) = 15

1,000 = .015 (B)P (this event) =

130 80 12

1,000

= .222

(C) P (this event) = 30 32 28 25 20 21 12 1

1,000

= .169

(D) P (this event) = 1 − P (owning zero television sets) = 1 − 2 10 30

1,000

= .958

Section 11−6 1. A binomial is any algebraic expression written as a sum of two terms. 3. Starting with a 1 in the first row and 1, then 1 in the second, each entry is the sum of the two entries

above it and to the left and right, with the exception of the outside entries in each row, which are 1s. For Problems 5—11, the first ten lines of Pascal's triangle are as follows: 0 1

1 1 1 2 1 2 1 3 1 3 3 1 4 1 4 6 4 1 5 1 5 10 10 5 1 6 1 6 15 20 15 6 1 7 1 7 21 35 35 21 7 1 8 1 8 28 56 70 56 28 8 1

9 1 9 36 84 126 126 84 36 9 1

5. 8

3

is the fourth entry in line 8 (start counting from 0) of the triangle. 56

7. 9

6

is the seventh entry in line 9 (start counting from 0) of the triangle. 84

9. C7,5 is the sixth entry in line 7 (start counting from 0) of the triangle. 21 11. C9,0 is the first entry in line 9 (start counting from 0) of the triangle. 1

13. 13

3

= 13!

3!(13 3)! =

13!

3!10! =

13 12 11 10!

3 2 1 10!

= 13 12 11

3 2 1

= 286

15. 12

4

= 12!

4!(12 4)! =

12!

4!8! =

12 11 10 9 8!

4 3 2 1 8!

= 12 11 10 9

4 3 2 1

= 495

17. C52,3 = 52!

3!(52 3)! =

52!

3!49! =

52 51 50 49!

3 2 1 49!

= 52 51 50

3 2 1

= 22,100

19. C12,6 = 12!

6!(12 6)! =

12!

6!6! =

479,001,600

(720)(720) = 924


21. (m + n)3 = 3

0k 3

k

m3−knk = 3

0

m3 + 3

1

m2n + 3

2

mn2 + 3

3

n3 = m3 + 3m2n + 3mn2 + n3

23. (2x − 3y)3 = [2x + (−3y)]3 = 3

0k 3

k

(2x)3−k(−3y)k

= 3

0

(2x)3 + 3

1

(2x)2(−3y)1 + 3

2

(2x)1(−3y)2 + 3

3

(−3y)3

= 8x3 + 3(4x2)(−3y) + 3(2x)(9y2) + (−27y3) = 8x3 − 36x2y + 54xy2 − 27y3

25. (x − 2)4 = [x + (−2)]4 = 4

0k 4

k

x4−k(−2)k = 4

0

x4 + 4

1

x3(−2)1 + 4

2

x2(−2)2 + 4

3

x(−2)3 + 4

4

(−2)4

= x4 − 8x3 + 24x2 − 32x + 16

27. (m + 3n)4 = 4

0k 4

k

m4−k(3n)k = 4

0

m4 + 4

1

m3(3n)1 + 4

2

m2(3n)2 + 4

3

m(3n)3 + 4

4

(3n)4

= m4 + 12m3n + 54m2n2 + 108mn3 + 81n4

29. (2x − y)5 = [2x + (−y)]5 = 5

0k 5

k

(2x)5−k(−y)k

= 5

0

(2x)5 + 5

1

(2x)4(−y)1 + 5

2

(2x)3(−y)2 + 5

3

(2x)2(−y)3 + 5

4

(2x)(−y)4 + 5

5

(−y)5

= 32x5 − 80x4y + 80x3y2 − 40x2y3 + 10xy4 − y5

31. (m + 2n)6 = 6

0k 6

k

m6−k(2n)k

= 6

0

m6 + 6

1

m5(2n)1 + 6

2

m4(2n)2 + 6

3

m3(2n)3 + 6

4

m2(2n)4 + 6

5

m(2n)5 + 6

6

(2n)6

= m6 + 12m5n + 60m4n2 + 160m3n3 + 240m2n4 + 192mn5 + 64n6

33. In the expansion (x + 1)7 = 7

0k 7

k

x7− k1k the exponent of x is 4 when k = 3.

Thus the term containing x4 is 7

3

x413 = 35x4

35. In the expansion (2x − 1)11 = [2x + (−1)]11 = 11

0k 11

k

(2x)11− k(−1)k the exponent of x is 6 when k = 5.


5

(2x)6(−1)5 = (462)(64x6)(−1) = −29,568x6

37. In the expansion (2x + 3)18 = 18

0k 18

k

(2x)18− k(3)k the exponent of x is 14 when k = 4.


4

(2x)14(3)4 = 3,060(214x14)(34) = 4,060,938,240x14

SECTION 11-6 549

39. In the expansion (x2 − 1)6 = 6

0k 6

k

(x2)6− k(−1)k = 6

0k 6

k

x12−2 k(−1)k the exponent of x, 12 − 2k, is 8,

when 12 − 2k = 8 and 2k = 4, k = 2. Thus the term containing x8 is 6

2

x8(−1)2 = 15x8

41. In the expansion (x2 + 1)9 = 9

0k 9

k

(x2)9− k(1)k = 9

0k 9

k

x18−2 k(1)k the exponent of x, 18 − 2k, is even

and can never equal 11. 43. In the expansion of (a + b)n, the exponent of b in the rth term is r − 1 and the exponent of a is n − (r − 1).

Here, r = 7, n = 15.

Seventh term = 15

6

u9v6 = 15!

9!6!u9v6 =

15 14 13 12 11 10

6 5 4 3 2 1

u9v6 = 5,005u9v6

45. In the expansion of (a + b)n, the exponent of b in the rth term is r − 1 and the exponent of a is n − (r − 1).

Here, r = 11, n = 12.

Eleventh term = 12

10

(2m)2n10 = 12!

10!2!4m2n10 =

12 11

2 1

4m2n10 = 264m2n10


Here, r = 7, n = 12.

Seventh term = 12

6

2

w

6

(−2)6 = 6

6

12!

6!6! 2

w26 =

12 11 10 9 8 7 6!

6 5 4 3 2 1 6!

w6 = 924w6


Here, r = 6, n = 8.

Sixth term = 8

5

(3x)3(−2y)5 = 8!

5!3!(3x)3(−2y)5 =

8 7 6 5!

5! 3 2 1

(27x3)(−32y5)

= −(56)(27)(32)x3y5 = −48,384x3y5

51. ( ) ( )f x h f x

h

=

3 3( )x h x

h

=

3 2 2 3 33 3x x h xh h x

h

=

2 2 33 3x h xh h

h

=

2 2(3 3 )h x xh h

h

= 3x2 + 3xh + h2 As h approaches 0, this quantity approaches 3x2.

53. ( ) ( )f x h f x

h

=

5 5( )x h x

h

=

5 4 3 2 2 3 4 5 55 10 10 5x x h x h x h xh h x

h

= 4 3 2 2 3 4 55 10 10 5x h x h x h xh h

h

=

4 3 2 2 3 4(5 10 10 5 )h x x h x h xh h

h

= 5x4 + 10x3h + 10x2h2 + 5xh3 + h4 As h approaches 0, this approaches 5x4. 55. Graphing the sequence, we obtain

The solid line indicates

1

2(184,756) = 92,378, that is, half the largest term. The graph

shows five terms greater than half the largest term. The table display confirms this:


0

0

200000

20

20

8

= 125,970 20

9

= 167,960 20

10

= 184,756 20

11

= 167,960 20

12

= 125,970

are all greater than 92,378, but the other terms are less than 92,378.

57. (A) Graphing the sequence, we obtain the graph at the right. From the table display we find the

largest term to be 10

4

(0.6)6(0.4)4 = 0.251 as

displayed in the graph.

0

0

.30

10

(B) According to the binomial formula,0

n

k n

k

an− kbk = (a + b)n

Thus, a0 + a1 + a2 + … + a10 = 10

0k n

k

(0.6)10−k(0.4)k = (0.6 + 0.4)10 = 110 = 1

59. (1.01)10 = (1 + 0.01)10 = 10

0k 10

k

110−k(0.01)k, 110−k = 1

= 10

0k 10

k

(0.01)k

= 10

0

+ 10

1

(0.01)1 + 10

2

(0.01)2 + 10

3

(0.01)3 + 10

4

(0.01)4 + 10

5

(0.01)5

+ 10

6

(0.01)6 +10

7

(0.01)7 + 10

8

(0.01)8 + 10

9

(0.01)9 + 10

10

(0.01)10

= 1 + 0.1 + 0.0045 + 0.00012 + 0.0000021 + (other terms with no effect in fourth decimal place) = 1.1046

61. n

r

= !

!( )!

n

r n r =

!

( )! !

n

n r r =

!

( )![ ( )]!

n

n r n n r =

n

n r

63. 1

k

r

+ k

r

= !

( 1)!( 1)!

k

r k r +

!

!( )!

k

r k r=

! ( 1) !

!( 1)!

rk k r k

r k r

= ( 1) !

!( 1)!

r k r k

r k r

= ( 1) !

!( 1)!

k k

r k r

= ( 1)!

!( 1)!

k

r k r

= 1k

r

65. k

k

= !

!( )!

k

k k k = 1 =

( 1)!

( 1)![( 1) ( 1)]!

k

k k k

= 1

1

k

k

CHAPTER 11 REVIEW 551

67. 2n = (1 + 1)n = 0

n

k n

k

1n−k1k, 1n−k = 1k = 1

= 0

n

k n

k

= 0

n

+ 1

n

+ 2

n

+ … + n

n

CHAPTER 11 REVIEW

1. (A) Since 8

16

=

4

8 = −

1

2, this could start a geometric sequence.

(B) Since 7 − 5 = 9 − 7 = 2, this could start an arithmetic sequence. (C) Since −5 − (−8) = −2 − (−5) = 3, this could start an arithmetic sequence.

(D) Since 3

2 ≠

5

3 and 3 − 2 ≠ 5 − 3, this could start neither an arithmetic nor a geometric

sequence.

(E) Since 2

1 =

4

2

= −2, this could start a geometric sequence. (11−1, 11−3)

2. an = 2n + 3 (A) a1 = 2·1 + 3 = 5 a2 = 2·2 + 3 = 7 a3 = 2·3 + 3 = 9 a4 = 2·4 + 3 = 11

(B) This is an arithmetic sequence with d = 2. Hence an = a1 + (n − 1)d a10 = 5 + (10 − 1)2 = 23

(C)

Sn = 2

n(a1 + an)

S10 = 10

2(5 + 23) = 140

(11−1, 11−3)

3. an = 321

2

n

(A)

a1 = 321

2

1

= 16

a2 = 321

2

2

= 8

a3 = 321

2

3

= 4

a4 = 321

2

4

= 2

(B) This is a geometric sequence with

r = 1

2. Hence

an = a1rn−1

a10 = 161

2

10−1

= 1

32

(C)

Sn = 1

1na ra

r

S10 = 1 1

2 3212

16 =

164

12

16 = 31 31

32

(11−1, 11−3)

4. a1 = −8, an = an−1 + 3, n ≥ 2 (A) a1 = −8

a2 = a1 + 3 = −8 + 3 = −5 a3 = a2 + 3 = −5 + 3 = −2 a4 = a3 + 3 = −2 + 3 = 1

(B) This is an arithmetic sequence with d = 3. Hence an = a1 + (n − 1)d a10 = −8 + (10 − 1)3 = 19

(C)

Sn = 2

n(a1 + an)

S10 = 10

2(−8 + 19) = 55

(11−1, 11−3)


5. a1 = −1, an = (−2)an−1, n ≥ 2 (A) a1 = −1

a2 = (−2)a1 = (−2)(−1) = 2 a3 = (−2)a2 = (−2)2 = −4 a4 = (−2)a3 = (−2)(−4) = 8

(B) This is a geometric sequence with r = −2. Hence an = a1r n−1 a10 = (−1)(−2)10−1 = 512

(C)

Sn = 1

1na ra

r

S10 = 1 ( 2)(512)

1 ( 2)

= 341

(11−1, 11−3)

6. This is a geometric sequence with a1 = 16 and r = 1

2 < 1, so the sum exists: S∞ = 1

1

a

r=

12

16

1= 32 (11−3)

7. 6! = 6·5·4·3·2·1 = 720 (11−4) 8. 22!

19! =

22 21 20 19!

19!

= 9,240 (11−4)

9. 7!

2!(7 2)! =

7!

2!5! =

7 6 5!

2 1 5!

= 21 (11−4)

10. C6,2 = 6!

2!(6 2)! =

6!

2!4! =

6 5 4!

2 1 4!

= 15; P6,2 = 6·5 = 30 (11−5)

11. (A) The outcomes can be displayed in a tree diagram as follows:

H

T

H

TH

TH

T

H

TH

T

1

2

3

4

5

6

(1, H)

(1, T)

(2, H)

(2, T)(3, H)

(3, T)(4, H)

(4, T)(5, H)

(5, T)

(6, H)

(6, T)

(B) O1: Rolling the die N1: 6 outcomes O2: Flipping the coin N2: 2 outcomes Applying the multiplication principle, there are 6·2 = 12 combined outcomes.

12. O1: Seating the first person N1: 6 ways O2: Seating the second person N2: 5 ways O3: Seating the third person N3: 4 ways O4: Seating the fourth person N4: 3 ways O5: Seating the fifth person N5: 2 ways O6: Seating the sixth person N6: 1 way

Applying the multiplication principle, there are 6·5·4·3·2·1 = 720 arrangements. (11−5)

13. Order is important here. We use permutations to determine the number of arrangements of 6 objects. P6,6 = 6! = 720 (11−5) 14. The sample space S is the set of all possible 5-card hands, chosen from a deck of 52 cards. n(S) = C52,5

The event E is the set of all possible 5-card hands that are all clubs, thus, are chosen from the 13 clubs. n(E) = C13,5

P(E) = ( )

( )

n E

n S = 13,5

52,5

C

C =

13!

5!(13 5)! ÷

52!

5!(52 5)! ≈ .0005 (11−5)


15. The sample space S is the set of all possible arrangements of two persons, chosen from a set of 15 persons. Order is important here, so we use permutations. n(S) = P15,2. The event E is one of these arrangements.

n(E) = 1.

P(E) = ( )

( )

n E

n S =

15,2

1

P =

1

15 14 ≈ .005 (11−5)

16. P(E) ≈ ( )f E

n =

50

1,000 = .05 (11−5)

17. P1: 5 = 12 + 4·1 = 5

P2: 5 + 7 = 22 + 4·2 12 = 12

P3: 5 + 7 + 9 = 32 + 4·3 21 = 21 (11−2)

18. P1: 2 = 21+1 − 2 = 4 − 2 = 2 P2: 2 + 4 = 22+1 − 2

6 = 6 P3: 2 + 4 + 8 = 23+1 − 2

14 = 14 (11−2)

19. P1: 491 − 1 is divisible by 6 48 = 6·8 true

P2: 492 − 1 is divisible by 6 2,400 = 6·400 true P3: 493 − 1 is divisible by 6 117,648 = 19,608·6 true (11−2)

20. Pk: 5 + 7 + 9 + … + (2k + 3) = k2 + 4k

Pk+1: 5 + 7 + 9 + … + (2k + 3) + (2k + 5) = (k + 1)2 + 4(k + 1) (11−2)

21. Pk: 2 + 4 + 8 + … + 2k = 2k+1 − 2

Pk+1: 2 + 4 + 8 + … + 2k + 2k+1 = 2k+2 − 2 (11−2)

22. Pk: 49k − 1 = 6r for some integer r Pk+1: 49k+1 − 1 = 6s for some integer s (11−2)

23. Although 1 is less than 4, 1 + 1

2 is less than 4, 1 +

1

2 +

1

3 is less than 4, and so on, the statement is false.

In fact, 1 + 1

2 +

1

3 + … +

1

31 ≈ 4.027245 hence n = 31 is a counterexample. (11−2)

24. S10 = (2·1 − 8) + (2·2 − 8) + (2·3 − 8) + (2·4 − 8) + (2·5 − 8) + (2·6 − 8) + (2·7 − 8) + (2·8 − 8) + (2·9 − 8) + (2·10 − 8) = (−6) + (−4) + (−2) + 0 + 2 + 4 + 6 + 8 + 10 + 12 = 30 (11−3)

25. S7 = 1

16

2 +

2

16

2 +

3

16

2 +

4

16

2 +

5

16

2 +

6

16

2 +

7

16

2= 8 + 4 + 2 + 1 +

1

2 +

1

4 +

1

8 = 15 7

8 (11−3)

26. This is an infinite geometric sequence with a1 = 27.

r = 18

27

= −

2

3

2

3 < 1,

hence the sum exists

S∞ = 1

1

a

r =

23

27

1 =

81

5 (11−3)

27. Sn = 1

1

( 1)

3

n k

kk

This geometric sequence has a1 = 1

3.

r = 1

9

÷ 1

3 = −

1

3;

1

3 < 1, hence the sum exists.

S∞ = 1

1

a

r=

13

131

= 1

4 (11−3)


28. The probability of an event cannot be negative, but P(e2) is given as negative. The sum of the probabilities of the simple events must be 1, but it is given as 2.5. The probability of an event cannot be greater than 1, but P(e4) is given as 2. (11−5)

29. We can select any three of the six points to use as vertices of the triangle. Order is not important.

C6,3 = 6!

3!(6 3)! = 20 triangles

(11−5)

30. First, find d: an = a1 + (n − 1)d 31 = 13 + (7 − 1)d 31 = 13 + 6d d = 3 Hence, a5 = 13 + (5 − 1)3 = 25 (11−3)

31. Case 1 Case 2 Case 3 O1: select the first letter N1: 8 ways 8 ways 8 ways O2: select the second letter N2: 7 ways 8 ways 7 ways O3: select the third letter N3: 6 ways 8 ways 7 ways

(exclude first and second (exclude second letter.) letter) 8·7·6 = 336 words 8·8·8 = 512 words 8·7·7 = 392 words (11−4)

32. (A) P(2 heads) ≈ 210

1,000 = .21

P(1 head) ≈ 480

1,000 = .48

P(0 heads) ≈ 310

1,000 = .31

(B) A sample space of equally likely events is: S = {HH, HT, TH, TT} Let E1 = 2 heads = {HH} E2 = 1 head = {HT, TH}

E3 = 0 heads = {TT} then P(E1) = 1( )

( )

n E

n S =

1

4 = .25

P(E2) = 2( )

( )

n E

n S =

2

4 = .5 P(E3) = 3( )

( )

n E

n S =

1

4 = .25

(C) Expected frequency = P(E)·number of trials Expected frequency of 2 heads = P(E1)·1,000 = .25(1,000) = 250 Expected frequency of 1 head = P(E2)·1,000 = .5(1,000) = 500 Expected frequency of 0 heads = P(E3)·1,000 = .25(1,000) = 250 (11−5)

33. The sample space S is the set of all possible 5-card hands chosen from a deck of 52 cards. n(S) = C52,5. (A) The event E is the set of all possible 4-card hands that are all diamonds, thus, are chosen from the 13

diamonds. n(E) = C13,5

P(E) = ( )

( )

n E

n S = 13,5

52,5

C

C

(B) The event E is the set of all possible 5-card hands chosen as follows: O1: Choose 3 diamonds out of the 13 in the deck. N1: C13,3 O2: Choose 2 spades out of the 13 in the deck. N2: C13,2 Applying the multiplication principle, n(E) = N1·N2 = C13,3·C13,2

P(E) = ( )

( )

n E

n S = 13,3 13,2

52,5

C C

C

(11−5)


34. The sample space S is the set of all possible 4-person choices out of the 10 people. n(S) = C10,4 The event E is the set of all of those choices that include 2 particular people, thus O1: Choose the 2 married people N1: 1 O2: Choose 2 more people out of the 8 remaining N2: C8,2 Applying the multiplication principle, n(E) = N1·N2 = 1·C8,2 = C8,2

P(E) = ( )

( )

n E

n S = 8,2

10,4

C

C =

8!

2!(8 2)! ÷

10!

4!(10 4)! = 28 ÷ 210 =

2

15 (11−5)

35. The sample space S is the set of all possible results of spinning the device twice:

{(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)}. n(S) = 9.

(A) The event E is the set of all possible results in which both spins are the same. E = {(1, 1), (2, 2), (3, 3)}. n(E) = 3.

P(E) = ( )

( )

n E

n S =

3

9 =

1

3

(B) The event E is the set of those spin sequences that add up to 5: E = {(2, 3), (3, 2)}. n(E) = 2.

P(E) = ( )

( )

n E

n S =

2

9 (11−5)

36. 0. 72 = 0.72 + 0.0072 + 0.000072 + … This is an infinite geometric sequence with a1 = 0.72 and r = 0.01.

0. 72 = S∞ = 1

1

a

r=

0.72

1 0.01=

0.72

0.99=

8

11 (11−3)

37. (A) Order is important here. We are selecting 3 digits out of 6 possible. P6,3 = 6·5·4 = 120 lock combinations

(B) Order is not important here. We are selecting 2 players out of 5. C5,2 = 5!

2!(5 2)! = 10 games (11−4)

38. 20!

18!(20 18)!=

20!

18!2!=

20 19 18!

18!2 1

= 190

(11−6)

39. 16

12

= 16!

12!(16 12)! =

16!

12!4!=

16 15 14 13 12!

12! 4 3 2 1

= 1,820

(11−6)

40. 11

11

= 11!

11!(11 11)! =

11!

11!0! = 1 (11−6)

41. (x − y)5 = [x + (−y)]5 = 5

0k 5

k

(x)5−k(−y)k

= 5

0

x5 + 5

1

x4(−y)1 + 5

2

x3(−y)2 + 5

3

x2(−y)3 + 5

4

x(−y)4 + 5

5

(−y)5

= x5 − 5x4y + 10x3y2 − 10x2y3 + 5xy4 − y5 (11−6)

42. In the expansion (x + 2)9 = 9

0k 9

k

x9− k2k the exponent of x is 6 when k = 3.

Thus the term containing x6 is9

3

x623 = 84x6·8 = 672x6 (11−6)


43. In the expansion of (a + b)n, the exponent of b in the rth term is r − 1 and the exponent of a is n − (r − 1). Here, r = 10, n = 12.

Tenth term = 12

9

(2x)3(−y)9 = 3 912!(8 )( )

9!3!x y = 3 912 11 10 9!

( 8 )9! 3 2 1

x y

= −1760x3y9 (11−6)

44. Write: Pn: 5 + 7 + 9 + … + (2n + 3) = n2 + 4n Proof: Part 1: Show that P1 is true: P1: 5 = 12 + 4·1 = 1 + 4 Clearly true.


Pk: 5 + 7 + 9 + … + (2k + 3) = k2 + 4k

Pk+1: 5 + 7 + 9 + … + (2k + 3) + (2k + 5) = (k + 1)2 + 4(k + 1)

We start with Pk: 5 + 7 + 9 + … + (2k + 3) = k2 + 4k Adding 2k + 5 to both sides: 5 + 7 + 9 + … + (2k + 3) + (2k + 5) = k2 + 4k + 2k + 5 = k2 + 6k + 5

= k2 + 2k + 1 + 4k + 4 = (k + 1)2 + 4(k + 1) We have shown that if Pk is true, then Pk+1 is true.

Conclusion: Pn is true for all positive integers n. (11−2)

45. Write: Pn: 2 + 4 + 8 + … + 2n = 2n +1 − 2 Proof: Part 1: Show that P1 is true: P1: 2 = 21+1 − 2 = 4 − 2 Clearly true.


Pk: 2 + 4 + 8 + … + 2k = 2k+1 − 2

Pk+1: 2 + 4 + 8 + … + 2k + 2k+1 = 2k+2 − 2

We start with Pk: 2 + 4 + 8 + … + 2k = 2k+1 − 2

Adding 2k+1 to both sides: 2 + 4 + 8 + … + 2k + 2k+1 = 2k+1 − 2 + 2k+1 = 2k+1 + 2k+1 − 2 = 2·2k+1 − 2 = 2k+2 − 2 We have shown that if Pk is true, then Pk+1 is true.

Conclusion: Pn is true for all positive integers n. (11−2) 46. Write: Pn: 49n − 1 = 6r for some r in N.

Part 1: Show that P1 is true: P1: 491 − 1 = 6r is true (r = 8). Part 2: Show that if Pk is true, then Pk+1 is true: Write out Pk and Pk+1.

Pk: 49k − 1 = 6r, for some integer r Pk+1: 49k +1 − 1 = 6s for some integer s

We start with Pk: 49k − 1 = 6r for some integer r Now, 49k+1 − 1 = 49k+1 − 49k + 49k − 1

= 49k(49 − 1) + 49k − 1 = 49k · 8 · 6 + 6r = 6(49k · 8 + r) = 6s with s = 49k · 8 + r

We have shown that if Pk is true, then Pk+1 is true. Conclusion: Pn is true for all positive integers n. (11−2)


47.

Enter the sequences in the sequence editor.

Graph the sequences.

Display the points (in the appropriate interval) in a table.

From the second graph, 29 is the least positive integer n such that an < bn. The table confirms this. (11−6) 48.

Enter the sequences in the sequence editor. 0

1

300

35

Graph the sequences.

Display the points in a table.

From the graph, the least positive integer n such that an < bn is between 25 and 30. The table confirms that n = 26. (11−1)

49. In the first case, the order matters. We have five successive events, each of which can happen two ways

(girl or boy). Applying the multiplication principle, there are 2·2·2·2·2 = 32 possible families. In the second case, the possible families can be listed as {0 girls, 1 girl, 2 girls, 3 girls, 4 girls, 5 girls}, thus there are 6 possibilities. (11−4)

50. An arithmetic sequence is involved, with a1 = 2

g,

d = 3

2

g −

2

g = g.

Distance fallen during the twenty-fifth second = a25. an = a1 + (n − 1)d

a25 = 2

g + (25 − 1)g =

49

2

g feet

Total distance fallen after twenty-five seconds = a1 + a2 + a3 + … + a25 = S25

Sn = 2

n(a1 + an)

S25 = 25 49

2 2 2

g g

= 625

2

g feet

(11−3)

51. To seat two people, we can seat one person, then the second person.

O1: Seat the first person in any chair. N1: 4 ways O2: Seat the second person in any remaining chair. N2: 3 ways Thus, applying the multiplication principle, there are 4·3 = 12 ways to seat two persons. (11−4)

52. (x + i)6 = 6

0k 6

k

x6−kik = 6

0

x6 + 6

1

x5i1 + 6

2

x4i2 + 6

3

x3i3 + 6

4

x2i4 + 6

5

xi5 + 6

6

i6

= x6 + 6ix5 − 15x4 − 20ix3 + 15x2 + 6ix − 1 (11−6)


53. Since there are several ways in which at least one woman can be selected, it is simplest to note: probability of selecting at least one woman = 1 − probability of selecting 0 women. To calculate the probability of selecting 0 women, that is, three men, we note: The sample space S is the set of all 3-person subsets, chosen from the ten people. n(S) = C10,3 The event E is the set of all 3-man subsets, chosen from the 7 men. n(E) = C7,3

Then P(E) = 7,3

10,3

C

C

Hence P(at least one woman) = 1 − P(0 women) = 1 − P(E) = 1 − 7,3

10,3

C

C = 1 −

7

24 =

17

24 (11−5)

54. (A) P(E) ≈ ( )f E

n =

350

1,000 = .350

(B) A sample space of equally likely events is: S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} Let E = 2 heads = {HHT, HTH, THH}

then P(E) = ( )

( )

n E

n S =

3

8 = .375

(C) Expected frequency = P(E)·number of trials = .375(1,000) = 375 (11−5)

55. Write: Pn: 1

n

k k3 =

1

n

k

k

2

Proof: Part 1: Show that P1 is true. P1: 1

1k k3 = 13 = 12 =

1

1k

k

2

Part 2: Show that if Pj is true, then Pj+1 is true. Write out Pj and Pj +1.

Pj: 1

j

k k3 =

1

j

k

k

2

Pj+1: 1

1

j

k

k3 =

1

1

j

k

k

2

We start with

Pj:1

j

k k3 =

1

j

k

k

2

Adding (j + 1)3 to both sides:

1

j

k k3 + (j + 1)3 =

1

j

k

k

2

+ (j + 1)3

1

1

j

k

k3 = (1 + 2 + 3 + … + j)2 + (j + 1)3 =

( 1)

2

j j

2

+ (j + 1)3 using Matched Problem 1, Section 11−2

= (j + 1)2 2

4

j + (j + 1)2(j + 1) = (j + 1)22

14

jj

= (j + 1)2

2 4 4

4

j j

= (j + 1)2 2

2

( 2)

2

j

= ( 1)( 2)

2

j j

2

= [1 + 2 + 3 + … + (j + 1)]2 using Matched Problem 1, Section 11−2

= 1

1

j

k

k

2

We have shown that if Pj is true, then Pj+1 is true. Conclusion: Pn is true for all positive integers n. (11−2)


56. Write: Pn: x2n − y2n = (x − y)Qn(x, y), where Qn(x, y) denotes some polynomial in x and y. Proof: Part 1: Show that P1 is true. P1: x2·1 − y2·1 = (x − y)(x + y) = (x − y)Q1(x, y) P1 is true. Part 2: Show that if Pk is true, then Pk+1 is true. Write out Pk and Pk+1.

Pk: x2k − y2k = (x − y)Qk(x, y) Pk+1: x2k+2 − y2k+2 = (x − y)Qk+1(x, y)

We start with Pk: x2k − y2k = (x − y)Qk(x, y) Now, x2k+2 − y2k+2 = x2k+2 − x2ky2 + x2ky2 − y2k+2 = x2k(x2 − y2) + y2(x2k − y2k)

= (x − y)x2k(x + y) + y2(x − y)Qk(x, y) by Pk = (x − y)[x2k(x + y) + y2Qk(x, y)] = (x − y)Qk+1(x, y)

We have shown that if Pk is true, then Pk+1 is true. Conclusion: Pn is true for all positive integers n. (11−2)

57. Write: Pn: n

m

a

a = an−m, m an arbitrary positive integer, n > m.

Proof: Part 1: Show Pm+1 is true. Pm+1: 1m

m

a

a

= am+1−m

m

m

a a

a = a1 by the recursive definition of an

a = a Pm+1 is true. Part 2: Show that if Pk is true, then Pk+1 is true. Write out Pk and Pk+1.

Pk: k

m

a

a = ak−m

Pk+1: 1k

m

a

a

= ak−m+1

We start with Pk:

k

m

a

a = ak−m

Multiplying both sides by a:

k

m

a

aa = ak−ma

k

m

a a

a = ak−m+1

1k

m

a

a

= ak−m+1

We have shown that if Pk is true, then Pk+1 is true, for m an arbitrary positive integer. Conclusion: Pn is true for all positive integers m, n. (11−2)

58. To prove an = bn, n a positive integer, write: Pn: an = bn Proof: Part 1: Show P1 is true. a1 = −3 b1 = −5 + 2·1 = −3. Thus, a1 = b1

Part 2: Show that if Pk is true, then Pk+1 is true. Write out Pk and Pk+1. Pk: ak = bk Pk+1: ak+1 = bk+1 We start with Pk: ak = bk Now, ak+1 = ak + 2 = bk + 2 = −5 + 2k + 2 = −5 + 2(k + 1) = bk+1 Therefore, ak+1 = bk+1. Thus, if Pk is true, then Pk+1 is true.

Conclusion: Pn is true for all n N. Hence, {an} = {bn} (11−2)


59. Write: Pn: (1!)1 + (2!)2 + (3!)3 + … + (n!)n = (n + 1)! − 1. Proof: Part 1: Show that P1 is true. P1: (1!)1 = 1 = 2 − 1 = 2! − 1 is true.

Part 2: Show that if Pk is true, then Pk+1 is true. Write out Pk and Pk+1.

Pk: (1!)1 + (2!)2 + (3!)3 + … + (k!)k = (k + 1)! − 1

Pk+1: (1!)1 + (2!)2 + (3!)3 + … + (k!)k + (k + 1)!(k + 1) = (k + 2)! − 1

We start with Pk: (1!)1 + (2!)2 + (3!)3 + … + (k!)k = (k + 1)! − 1 Adding (k + 1)!(k + 1) to both sides: (1!)1 + (2!)2 + (3!)3 + …+ (k!)k + (k + 1)!(k + 1) = (k + 1)! − 1 + (k + 1)!(k + 1)

= (k + 1)!(1 + k + 1) − 1 = (k + 2)(k + 1)! − 1 = (k + 2)! − 1 Thus, if Pk is true, then Pk+1 is true. Conclusion: Pn is true for all positive integers n. (11−2)

60. The unpaid balance starts at $7,200, and decreases each month by $300. Thus the interest starts at

7,200(0.01) and decreases each month by 300(0.01). This is an arithmetic sequence with a1 = 7,200(0.01), a24 = 300(0.01), n = 24.

Then the total interest paid is Sn = 2

n(a1 + an)

S24 = 24

2[7,200(0.01) + 300(0.01)] = $900 (11−3)

61. This involves an infinite geometric series. a1 = (0.75)2,400. r = 0.75. |r | < 1, so the series has a sum.

S∞ = 1

1

a

r =

(0.75)2400

1 0.75 = $7,200 (11−3)

62. Since A(n) is a geometric sequence with common ratio 1.06, we can write A(n) = an = (1.06)n 500 If n = 10, a10 = (1.06)10 500 = $895.42 If n = 20, a20 = (1.06)20 500 = $1,603.57 (11−3)

63. A route plan can be regarded as a series of choices of stores, thus an arrangement of the 5 stores. Since

the order matters, we use permutations: P5,5 = 5! = 120 route plans. (11−4)

64. (A) P(this event) = 40

1,000 = .04

(B) P(this event) = 100 60

1,000

= .16

(C) P(this event) = 1 − P (not 12 − 18 and buys 0 or 1 cassette annually)

= 1 − 60 70 70 110 100 50

1,000

= 1 −

460

1,000 = .54 (11−5)

65. P (shipment returned) = 1 − P (no substandard part found).

The sample space S is the set of all possible 4−part subsets of the 12 part set. n(S) = C12,4 The event, no substandard part found, is the set of all possible 4−part subsets of the 10 acceptable parts. n(E) = C10,4.

Then P(shipment returned) = 1 − ( )

( )

n E

n S = 1 − 10,4

12,4

C

C = 1 −

14

33 ≈ .576. (11−5)

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Math Analysisramsteinmath.weebly.com/uploads/7/0/5/4/7054561/chapter_11_ssm… · 524 CHAPTER 11 SEQUENCES, INDUCTION, AND PROBABILITY 31. a1 = 4 a2 = 1 4 a2−1 = 1 4 a1 = 1 4 ·