C H A P T E R 9 Sequences, Series, and Probability · C H A P T E R 9 Sequences, Series, and Probability Section 9.1 Sequences and Series 819 Vocabulary Check 1. infinite sequence

C H A P T E R 9

Sequences, Series, and Probability

Section 9.1 Sequences and Series . . . . . . . . . . . . . . . . . . . . 819

Section 9.2 Arithmetic Sequences and Partial Sums . . . . . . . . . . 831

Section 9.3 Geometric Sequences and Series . . . . . . . . . . . . . . 840

Section 9.4 Mathematical Induction . . . . . . . . . . . . . . . . . . 852

Section 9.5 The Binomial Theorem . . . . . . . . . . . . . . . . . . . 868

Section 9.6 Counting Principles . . . . . . . . . . . . . . . . . . . . . 877

Section 9.7 Probability . . . . . . . . . . . . . . . . . . . . . . . . . 882

Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 888

Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 898

Practice Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 902

C H A P T E R 9

Sequences, Series, and Probability

Section 9.1 Sequences and Series

819

Vocabulary Check

1. infinite sequence 2. terms

3. finite 4. recursively

5. factorial 6. summation notation

7. index; upper; lower 8. series

9. partial sumnth

n Given the general nth term in a sequence, you should be able to find, or list, some of the terms.

n You should be able to find an expression for the apparent nth term of a sequence.

n You should be able to use and evaluate factorials.

n You should be able to use summation notation for a sum.

n You should know that the sum of the terms of a sequence is a series.

1.

a5 5 3s5d 1 1 5 16

a4 5 3s4d 1 1 5 13

a3 5 3s3d 1 1 5 10

a2 5 3s2d 1 1 5 7

a1 5 3s1d 1 1 5 4

an 5 3n 1 1 2.

a5 5 5s5d 2 3 5 22

a4 5 5s4d 2 3 5 17

a3 5 5s3d 2 3 5 12

a2 5 5s2d 2 3 5 7

a1 5 5s1d 2 3 5 2

an 5 5n 2 3 3.

a5 5 25 5 32

a4 5 24 5 16

a3 5 23 5 8

a2 5 22 5 4

a1 5 21 5 2

an 5 2n

4.

a5 5 s 12d5

5132

a4 5 s 12d4

5116

a3 5 s 12d3

518

a2 5 s 12d2

514

a1 5 s 12d1

512

an 5 s 12dn

5.

a5 5 s22d5 5 232

a4 5 s22d4 5 16

a3 5 s22d3 5 28

a2 5 s22d2 5 4

a1 5 s22d1 5 22

an 5 s22dn 6.

a5 5 s212d5

5 21

32

a4 5 s212d4

51

16

a3 5 s212d3

5 218

a2 5 s212d2

514

a1 5 s212d1

5 212

an 5 s212dn

820 Chapter 9 Sequences, Series, and Probability

7.

a5 57

5

a4 56

45

3

2

a3 55

3

a2 54

25 2

a1 51 1 2

15 3

an 5n 1 2

n9.

a5 56s5d

3s5d2 2 15

15

37

a4 56s4d

3s4d2 2 15

24

47

a3 56s3d

3s3d2 2 15

9

13

a2 56s2d

3s2d2 2 15

12

11

a1 56s1d

3s1d2 2 15 3

an 56n

3n2 2 1

11.

a5 5 0

a4 52

45

1

2

a3 5 0

a2 52

25 1

a1 5 0

an 51 1 s21dn

n

8.

a5 55

5 1 25

5

7

a4 54

4 1 25

2

3

a3 53

3 1 25

3

5

a2 52

2 1 25

1

2

a1 51

1 1 25

1

3

an 5n

n 1 2

10.

a5 53s5d2 2 5 1 4

2s5d2 1 15

74

51

a4 53s4d2 2 4 1 4

2s4d2 1 15

16

11

a3 53s3d2 2 3 1 4

2s3d2 1 15

28

19

a2 53s2d2 2 2 1 4

2s2d2 1 15

14

9

a1 53s1d2 2 1 1 4

2s1d2 1 15 2

an 53n2 2 n 1 4

2n2 1 112.

a5 5 1 1 s21d5 5 0

a4 5 1 1 s21d4 5 2

a3 5 1 1 s21d3 5 0

a2 5 1 1 s21d2 5 2

a1 5 1 1 s21d1 5 0

an 5 1 1 s21dn

13.

a5 5 2 21

2435

485

243

a4 5 2 21

815

161

81

a3 5 2 21

275

53

27

a2 5 2 21

95

17

9

a1 5 2 21

35

5

3

an 5 2 21

3n14.

a5 525

355

32

243

a4 524

345

16

81

a3 523

335

8

27

a2 522

325

4

9

a1 521

315

2

3

an 52n

3n15.

a5 51

53y2

a4 51

43y25

1

8

a3 51

33y2

a2 51

23y2

a1 51

15 1

an 51

n3y2

Section 9.1 Sequences and Series 821

16.

a5 5103!52

5103!25

a4 5103!42

5103!16

a3 5103!32

5103!9

a2 5103!22

5103!4

a1 510

15 10

an 510

n2y35

103!n2

17.

a5 5 21

25

a4 51

16

a3 5 21

9

a2 51

4

a1 5 21

15 21

an 5s21dn

n2 18.

a5 5 s21d55

5 1 15 2

5

6

a4 5 s21d44

4 1 15

4

5

a3 5 s21d33

3 1 15 2

3

4

a2 5 s21d22

2 1 15

2

3

a1 5 s21d11

1 1 15 2

1

2

an 5 s21dn1 n

n 1 12

19.

a5 523

a4 523

a3 523

a2 523

a1 523

an 523 20.

a5 5 0.3

a4 5 0.3

a3 5 0.3

a2 5 0.3

a1 5 0.3

an 5 0.3 21.

a5 5 s5ds4ds3d 5 60

a4 5 s4ds3ds2d 5 24

a3 5 s3ds2ds1d 5 6

a2 5 s2ds1ds0d 5 0

a1 5 s1ds0ds21d 5 0

an 5 nsn 2 1dsn 2 2d 22.

a5 5 5s52 2 6d 5 95

a4 5 4s42 2 6d 5 40

a3 5 3s32 2 6d 5 9

a2 5 2s22 2 6d 5 24

a1 5 1s12 2 6d 5 25

an 5 nsn2 2 6d

23. a25 5 s21d25s3s25d 2 2d 5 273 24.

a16 5 s21d1621f16s16 2 1dg 5 2240

an 5 s21dn21fnsn 2 1dg

25. a11 54s11d

2s11d2 2 35

44

23926.

a13 54s13d2 2 13 1 3

13s13 2 1ds13 1 2d 537

130

an 54n2 2 n 1 3

nsn 2 1dsn 1 2d27.

0

0

10

10

an 5 3

4n

28.

0

−3

10

2

an 5 2 24

n29.

0

−10

10

18

an 5 16s20.5dn21 30.

0

0

10

12

an 5 8s0.75dn21

31.

0

0

10

2

an 52n

n 1 132.

0

0

10

1

an 5n2

n2 1 233.

The sequence decreases.

Matches graph (c).

a10 58

11a1 5 4,

an 58

n 1 1


34.

Matches graph (b).

a1 5 4, a3 524

45 6

an → 8 as n → `

an 58n

n 1 135.

The sequence decreases.

Matches graph (d).

a10 51

128a1 5 4,

an 5 4s0.5dn21 36.

Matches graph (a).

a1 5 4, a4 544

4!5

256

245 10

2

3

an → 0 as n → `

an 54n

n!

37.

an 5 1 1 sn 2 1d3 5 3n 2 2

1, 4, 7, 10,13, . . . 38. 3, 7, 11, 15, 19, . . .

Apparent pattern:

Each term is one less than four

times n, which implies that

an 5 4n 2 1.

n:

Terms:

1

3

2

7

3

11

4

15

5

19

. . .

. . .

n

an

39.

an 5 n2 2 1

0, 3, 8, 15, 24, . . .

40.

Apparent pattern:

Each term is the product of

and twice n, which implies that

an 5 s21dn11s2nd.

s21dn11

n:

Terms:

1

2

2

24

3

6

4

28

5

10

. . .

. . .

n

an

2, 24, 6, 28, 10, . . . 41.

an 5 s21dn1n 1 1

n 1 22

22

3,

3

4, 2

4

5,

5

6, 2

6

7, . . . 42.

Apparent pattern:

Each term is divided by 2

raised to the n, which implies that

an 5s21dn11

2n.

s21dn11

Terms: 1

2

21

4

1

8

21

16 . . . an

n: 1 2 3 4 . . . n

1

2,

21

4,

1

8,

21

16, . . .

43.

an 5n 1 1

2n 2 1

2

1,

3

3,

4

5,

5

7,

6

9, . . . 44.

Apparent pattern:

Each term is divided by 3

raised to the n, which implies that

an 52n21

3n.

2n21

Terms: 1

3

2

9

4

27

8

81 . . . an

n: 1 2 3 4 . . . n

1

3,

2

9,

4

27,

8

81, . . . 45.

an 51

n2

1, 1

4,

1

9,

1

16,

1

25, . . .

46.

Apparent pattern:

Each term is the reciprocal of n!, which implies that

an 51

n!.

Terms: 1 1

2

1

6

1

24

1

120 . . . an

n: 1 2 3 4 5 . . . n

1, 1

2,

1

6,

1

24,

1

120, . . . 47.

an 5 s21dn11

1, 21, 1, 21, 1, . . .


49.

an 5 1 11

n

1 11

1, 1 1

1

2, 1 1

1

3, 1 1

1

4, 1 1

1

5, . . .48.

Apparent pattern:

Each term is divided by which implies that

an 52n21

sn 2 1d!.

sn 2 1d!,2n21

Terms: 1 2 22

2

23

6

24

24

25

120 . . . an

n: 1 2 3 4 5 6 . . . n

1, 2, 22

2,

23

6,

24

24,

25

120, . . .

50.

n: 1 2 3 4 5 . . . n

Terms: . . .

Apparent pattern: Each term is the sum of 1 and the quantity 1 less than divided by which implies that

an 5 1 12n 2 1

2n.

2n,2n

an1 131

321 1

15

161 1

7

81 1

3

41 1

1

2

1 11

2, 1 1

3

4, 1 1

7

8, 1 1

15

16, 1 1

31

32, . . .

54.

a5 512 a 4 5

12s4d 5 2

a4 512a3 5

12s8d 5 4

a3 512 a2 5

12s16d 5 8

a2 512 a1 5

12s32d 5 16

a1 5 32

a1 5 32, ak11 512ak

59.

a4 534

4!5

81

245

27

8

a3 533

3!5

27

65

9

2

a2 532

2!5

9

2

a1 531

1!5 3

a0 530

0!5 1

an 53n

n!

51.

a5 5 a4 2 4 5 16 2 4 5 12

a4 5 a3 2 4 5 20 2 4 5 16

a3 5 a2 2 4 5 24 2 4 5 20

a2 5 a1 2 4 5 28 2 4 5 24

a1 5 28

a1 5 28 and ak11 5 ak 2 4 52.

a5 5 a4 1 3 5 24 1 3 5 27

a4 5 a3 1 3 5 21 1 3 5 24

a3 5 a2 1 3 5 18 1 3 5 21

a2 5 a1 1 3 5 15 1 3 5 18

a1 5 15

a1 5 15, ak11 5 ak 1 3 53.

a5 5 2sa4 2 1d 5 2s10 2 1d 5 18

a4 5 2sa3 2 1d 5 2s6 2 1d 5 10

a3 5 2sa2 2 1d 5 2s4 2 1d 5 6

a2 5 2sa1 2 1d 5 2s3 2 1d 5 4

a1 5 3

a1 5 3 and ak11 5 2sak 2 1d

55.

In general, an 5 2n 1 4.

a5 5 a4 1 2 5 12 1 2 5 14

a4 5 a3 1 2 5 10 1 2 5 12

a3 5 a2 1 2 5 8 1 2 5 10

a2 5 a1 1 2 5 6 1 2 5 8

a1 5 6

a1 5 6 and ak11 5 ak 1 2 56.

In general, an 5 30 2 5n .

a5 5 a4 2 5 5 10 2 5 5 5

a4 5 a3 2 5 5 15 2 5 5 10

a3 5 a2 2 5 5 20 2 5 5 15

a2 5 a1 2 5 5 25 2 5 5 20

a1 5 25

a1 5 25, ak11 5 ak 2 5

57.

In general,

an 5 8111

32n21

581s3d11

32n

5243

3n.

a5 51

3a4 5

1

3s3d 5 1

a4 51

3a3 5

1

3s9d 5 3

a3 51

3a2 5

1

3s27d 5 9

a2 51

3a1 5

1

3s81d 5 27

a1 5 81

a1 5 81 and ak11 51

3ak 58.

In general, an 5 14s22dn21.

a5 5 s22dsa4d 5 s22ds2112d 5 224

a4 5 s22da3 5 s22ds56d 5 2112

a3 5 s22da2 5 s22ds228d 5 56

a2 5 s22da1 5 s22ds14d 5 228

a1 5 14

a1 5 14, ak11 5 s22dak


61.

a4 51

5!5

1

120

a3 51

4!5

1

24

a2 51

3!5

1

6

a1 51

2!5

1

2

a0 51

1!5 1

an 51

sn 1 1d! 62.

a4 542

s4 1 1d! 516

5 ? 4 ? 3 ? 2 ? 15

2

15

a3 532

s3 1 1d! 59

4 ? 3 ? 2 ? 15

3

8

a2 522

s2 1 1d! 54

3 ? 2 ? 15

2

3

a1 512

s1 1 1d! 51

2 ? 15

1

2

a0 502

s0 1 1d! 50

15 0

an 5n2

sn 1 1d!

63.

a4 51

8!5

1

40,320

a3 51

6!5

1

720

a2 51

4!5

1

24

a1 51

2!5

1

2

a0 51

0!5 1

an 5s21d2n

s2nd! 51

s2nd! 64.

a4 5s21d2?411

s2 ? 4 1 1d! 5s21d9

9!5

21

362,8805 2

1

362,880

a3 5s21d2?311

s2 ? 3 1 1d! 5s21d7

7!5

21

50405 2

1

5040

a2 5s21d2?211

s2 ? 2 1 1d! 5s21d5

5!5

21

1205 2

1

120

a1 5s21d2?111

s2 ? 1 1 1d! 5s21d3

3!5

21

65 2

1

6

a0 5s21d2s0d11

s2 ? 0 1 1d! 5s21d1

1!5

21

15 21

an 5s21d2n11

s2n 1 1d!

65.4!

6!5

1 ? 2 ? 3 ? 4

1 ? 2 ? 3 ? 4 ? 5 ? 65

1

5 ? 65

1

3066.

5!

8!5

1 ? 2 ? 3 ? 4 ? 5

1 ? 2 ? 3 ? 4 ? 5 ? 6 ? 7 ? 85

1

6 ? 7 ? 85

1

336

67.10!

8!5

1 ? 2 ? 3 ? 4 ? 5 ? 6 ? 7 ? 8 ? 9 ? 10

1 ? 2 ? 3 ? 4 ? 5 ? 6 ? 7 ? 85

9 ? 10

15 90 68. 5

24 ? 25

15 600

25!

23!5

1 ? 2 ? 3 ? . . . ? 23 ? 24 ? 25

1 ? 2 ? 3 ? . . . ? 23

69.

5 n 1 1

sn 1 1d!

n!5

1 ? 2 ? 3 ? . . . ? n ? sn 1 1d1 ? 2 ? 3 ? . . . ? n

5n 1 1

170.

5 sn 1 1dsn 1 2d

sn 1 2d!

n!5

1 ? 2 ? 3 ? . . . ? n ? sn 1 1d ? sn 1 2d1 ? 2 ? 3 ? . . . ? n

71.

51

2ns2n 1 1d

s2n 2 1d!s2n 1 1d! 5

1 ? 2 ? 3 ? . . . ? s2n 2 1d1 ? 2 ? 3 ? . . . ? s2n 2 1d ? s2nd ? s2n 1 1d

72.

53n 1 1

15 3n 1 1

s3n 1 1d!

s3nd! 51 ? 2 ? 3 ? . . . ? s3nd ? s3n 1 1d

1 ? 2 ? 3 ? . . . ? s3nd

73. o5

i51

s2i 1 1d 5 s2 1 1d 1 s4 1 1d 1 s6 1 1d 1 s8 1 1d 1 s10 1 1d 5 35

74. o6

i51

s3i 2 1d 5 s3 ? 1 2 1d 1 s3 ? 2 2 1d 1 s3 ? 3 2 1d 1 s3 ? 4 2 1d 1 s3 ? 5 2 1d 1 s3 ? 6 2 1d 5 57

75. o4

k51

10 5 10 1 10 1 10 1 10 5 40 76. o5

k51

5 5 5 1 5 1 5 1 5 1 5 5 25

60.

a4 54!

45

4 ? 3 ? 2 ? 1

45 6

a3 53!

35

3 ? 2 ? 1

35 2

a2 52!

25

2 ? 1

25 1

a1 51!

15

1

15 1

a0 50!

05 undefined

an 5n!

n


77. o4

i50

i2 5 02 1 12 1 22 1 32 1 42 5 30

79. o 3

k50

1

k2 1 15

1

11

1

1 1 11

1

4 1 11

1

9 1 15

9

5

81. o5

k52

sk 1 1d2sk 2 3d 5 s3d2s21d 1 s4d2s0d 1 s5d2s1d 1 s6d2s2d 5 88

83. o4

i51

2i 5 21 1 22 1 23 1 24 5 30

78.

5 110

o5

i50

2i2 5 2s02d 1 2s12d 1 2s22d 1 2s32d 1 2s42d 1 2s52d

80. o5

j53

1

j2 2 35

1

32 2 31

1

42 2 31

1

52 2 35

124

429

82. o4

i51

fsi 2 1d2 1 si 1 1d3g 5 fs0d2 1 s2d3g 1 fs1d2 1 s3d3g 1 fs2d2 1 s4d3g 1 fs3d2 1 s5d3g 5 238

84.

5 11

o4

j50

s22d j 5 s22d0 1 s22d1 1 s22d2 1 s22d3 1 s22d4

85. o6

j51

s24 2 3jd 5 81 86. o10

j51

3

j 1 1< 6.06 87. o

4

k50

s21dk

k 1 15

47

6088. o

4

k50

s21dk

k!5

3

8

89.1

3s1d1

1

3s2d1

1

3s3d1 . . . 1

1

3s9d5 o

9

i51

1

3i90.

5

1 1 11

5

1 1 21

5

1 1 31 . . . 1

5

1 1 155 o

15

i51

5

1 1 i

91. 3211

82 1 34 1 3212

82 1 34 1 3213

82 1 34 1 . . . 1 3218

82 1 34 5 o8

i51321 i

82 1 34

92. 31 2 11

622

4 1 31 2 12

622

4 1 . . . 1 31 2 16

622

4 5 o6

k513121k

622

4

93. 3 2 9 1 27 2 81 1 243 2 729 5 o6

i51

s21di113i

94. 1 21

21

1

42

1

81 . . . 2

1

1285

1

202

1

211

1

222

1

231 . . . 2

1

275 o

7

n5012

1

22n

95.1

122

1

221

1

322

1

421 . . . 2

1

2025 o

20

i51

s21di11

i2

97.1

41

3

81

7

161

15

321

31

645 o

5

i51

2i 2 1

2i11

99. o4

i51

511

22i

5 511

22 1 511

222

1 511

223

1 511

224

575

16

96.1

1 ? 31

1

2 ? 41

1

3 ? 51 . . . 1

1

10 ? 125 o

10

k51

1

ksk 1 2d

98.1

21

2

41

6

81

24

161

120

321

720

645 o

6

k51

k!

2k

100.

5242

243

o5

i51

211

32i

5 211

321

1 211

322

1 211

323

1 211

324

1 211

325

101. o3

n51

4121

22n

5 4121

22 1 4121

222

1 4121

223

5 23

2102.

5 251

32

o4

n51

8121

42n

5 8121

421

1 8121

422

1 8121

423

1 8121

424


103. o`

i51

6s 110di

5 0.6 1 0.06 1 0.006 1 0.0006 1 . . . 523

105. By using a calculator, we have

The terms approach zero as

Thus, we conclude that o`

k51

71 1

102k

57

9.

n → `.

o100

k51

71 1

102k

<7

9.

o50

k51

71 1

102k

< 0.7777777778

o10

k51

71 1

102k

< 0.7777777777

107.

(a)

(b) A40 5 $11,040.20

A8 5 $5858.30

A7 5 $5743.43

A6 5 $5630.81

A5 5 $5520.40

A4 5 $5412.16

A3 5 $5306.04

A2 5 $5202.00

A1 5 $5100.00

An 5 500011 10.08

4 2n

, n 5 1, 2, 3, . . .

104.

51

9

5 0.11111 . . .

5 0.1 1 0.01 1 0.001 1 0.0001 1 0.00001 1 . . .

o`

k511 1

102k

51

101

1

1021

1

1031

1

1041

1

1051 . . .

106.

52

9

5 0.222 . . .

5 2s0.111 . . .d

5 2s0.1 1 0.01 1 0.001 1 0.0001 1 . . .d

o`

i51

21 1

102i

5 21 1

101

1

1021

1

1031

1

1041 . . .2

108. (a)

(b)

(c) A240 5 100s101dfs1.01d240 2 1g < $99,914.79

A60 5 100s101dfs1.01d60 2 1g < $8248.64

A6 5 100s101dfs1.01d6 2 1g < $621.35

A5 5 100s101dfs1.01d5 2 1g < $515.20

A4 5 100s101dfs1.01d4 2 1g < $410.10

A3 5 100s101dfs1.01d3 2 1g < $306.04

A2 5 100s101dfs1.01d2 2 1g 5 $203.01

A1 5 100s101dfs1.01d1 2 1g 5 $101.00

109. (a) Linear model:

(c)

The quadratic model is a better fit.

an < 60.57n 2 182

Year n Actual Linear Quadratic

Data Model Model

1998 8 311 303 308

1999 9 357 363 362

2000 10 419 424 420

2001 11 481 484 480

2002 12 548 545 544

2003 13 608 605 611

(b) Quadratic model:

(d) For the year 2008 we have the following predictions:

Linear model: 908 stores

Quadratic model: 995 stores

Since the quadratic model is a better fit, the predicted

number of stores in 2008 is 995.

an < 1.61n2 1 26.8n 2 9.5


110. (a)

(b) The number of cases reported fluctuates.

a9 5 44.9

a8 5 50.0

a7 5 56.6

a6 5 64.3

0

5 14

75 a5 5 73.1

an 5 0.0457n3 2 0.3498n2 2 9.04n 1 121.3, n 5 5, . . . , 13.

a13 5 45.1

a12 5 41.4

a11 5 40.4

a10 5 41.6

111. (a)

billion

billion

billion

billion

billion

billion

billion

(b) The federal debt is increasing.

a6 < $5091.8

a5 < $4914.8

a4 < $4698.2

a3 < $4425.3

a2 < $4079.6

a1 < $3644.3

0

0 14

7000a0 5 $3102.9

an 5 2.7698n3 2 61.372n2 1 600.00n 1 3102.9

billion

billion

billion

billion

billion

billion

billiona13 < $6616.3

a12 < $6251.5

a11 < $5963.5

a10 5 $5735.5

a9 < $5551.0

a8 < $5393.2

a7 < $5245.7

112. million

The results from the model and the figure (which are

approximations) are very similar.

o13

n56

s46.609n2 2 119.84n 2 1125.8d 5 $17,495 113. True, by the

Properties of Sums.

o4

i51

si 2 1 2id 5 o4

i51

i 2 1 2o4

i51

i

114.

True, because 21 1 22 1 23 1 24 5 2322 1 2422 1 2522 1 2622.

o4

j51

2j 5 o6

j53

2j22

115.

a12 5 89 1 55 5 144

a11 5 55 1 34 5 89

a10 5 34 1 21 5 55

a9 5 21 1 13 5 34

a8 5 13 1 8 5 21

a7 5 8 1 5 5 13

a6 5 5 1 3 5 8

a5 5 3 1 2 5 5

a4 5 2 1 1 5 3

a3 5 1 1 1 5 2

a2 5 1

a1 5 1

a1 5 1, a2 5 1, ak12 5 ak11 1 ak, k ≥ 1 116.

bn 5 1 11

bn21

b5 5 1 11

b4

5 1 13

55

8

5

b4 5 1 11

b3

5 1 12

35

5

3

b3 5 1 11

b2

5 1 11

25

3

2

b2 5 1 11

b1

5 1 11

15 2

bn 5an11

an

; b1 5 1, b2 5 2, b3 53

2, b4 5

5

3, . . .

b10 589

55

b9 555

34

b8 534

21

b7 521

13

b6 513

8

b5 58

5

b4 55

3

b3 53

2

b2 52

15 2

b1 51

15 1


117.327.15 1 785.69 1 433.04 1 265.38 1 604.12 1 590.30

6< $500.95

119.

5 0

5 1on

i51

xi22n11

n o

n

i51

xi2

5 1on

i51

xi2 2 nx

on

i51

sxi 2 x d 5 on

i51

xi 2 on

i51

x118.

5 $1.943

x 51

n on

i51

xi 51.899 1 1.959 1 1.919 1 1.939 1 1.999

5

120.

5 on

i51

xi2 1 o

n

i51

xion

i51

xi122

n1

1

n2 5 on

i51

xi2 2

1

n1on

i51

xi22

5 on

i51

xi2 2 2 ?

1

non

i51

xion

i51

xi 1 n ?1

non

i51

xi ?1

non

i51

xi

on

i51

sxi 2 xd25 o

n

i51

sxi2 2 2xix 1 x 2d 5 o

n

i51

xi2 2 2xo

n

i51

xi 1 nx2

121.

a5 5x5

5!5

x5

120

a4 5x4

4!5

x4

24

a3 5x3

3!5

x3

6

a2 5x2

2!5

x2

2

a1 5x1

1!5 x

an 5xn

n!122.

a5 5s21d5x2s5d11

2s5d 1 15 2

x11

11

a4 5s21d4x2s4d11

2s4d 1 15

x9

9

a3 5s21d3x2s3d11

2s3d 1 15 2

x7

7

a2 5s21d2x2s2d11

2s2d 1 15

x5

5

a1 5s21d1x2s1d11

2s1d 1 15 2

x3

3

an 5s21dnx2n11

2n 1 1123.

a5 52x10

10!5 2

x10

3,628,800

a4 5x8

8!5

x8

40,320

a3 52x6

6!5 2

x6

720

a2 5x4

4!5

x4

24

a1 52x2

2!5 2

x2

2

an 5s21dnx2n

s2nd!

124.

a5 5s21d5x2s5d11

s2s5d 1 1d! 5 2x11

11!5 2

x11

39,916,800

a4 5s21d4x2s4d11

s2s4d 1 1d! 5x9

9!5

x9

362,880

a3 5s21d3x2s3d11

s2s3d 1 1d! 5 2x7

7!5 2

x7

5040

a2 5s21d2x2s2d11

s2s2d 1 1d! 5x5

5!5

x5

120

a1 5s21d1x2s1d11

s2s1d 1 1d! 5 2x3

3!5 2

x3

6

an 5s21dnx2n11

s2n 1 1d!125. is one-to-one, so it has an inverse.

f 21sxd 5x 1 3

4

x 1 3

45 y

x 5 4y 2 3

y 5 4x 2 3

f sxd 5 4x 2 3


126.

This is a function of x, so f has in inverse.

f 21sxd 53

x, x Þ 0

y 53

x

xy 5 3

x 53

y

y 53

x

gsxd 53

x

128.

This does not represent y as a function of x, so f does not have an inverse.

1 ±!x 5 y

±!x 5 y 2 1

x 5 sy 2 1d2

y 5 sx 2 1d2

f sxd 5 sx 2 1d2

127. is one-to-one, so it has an inverse.

Domain:

Range:

h21sxd 5x2 2 1

55

1

5sx2 2 1d, x ≥ 0

x2 2 1

55 y, x ≥ 0

x2 5 5y 1 1, x ≥ 0

x 5 !5y 1 1, x ≥ 0, y ≥ 21

5

y 5 !5x 1 1, x ≥ 21

5, y ≥ 0

y ≥ 0

x ≥ 21

5

hsxd 5 !5x 1 1

129. (a)

(b)

(c)

(d) BA 5 322

6

4

234 36

3

5

44 5 3212 1 16

36 2 9

210 1 12

30 2 124 5 3 0

27

6

184

AB 5 36

3

5

44 322

6

4

234 5 3212 1 30

26 1 24

24 2 15

12 2 124 5 318

18

9

04

4B 2 3A 5 4322

6

4

234 2 336

3

5

44 5 328 2 18

24 2 9

16 2 15

212 2 124 5 3226

15

1

2244

A 2 B 5 36

3

5

44 2 322

6

4

234 5 36 2 s22d3 2 6

5 2 4

4 2 s23d4 5 3 8

23

1

74

130. (a)

(b)

(c)

(d) 5 348

36

272

1224 BA 5 30

8

212

114310

24

7

64 5 3 0 1 48

80 2 44

0 2 72

56 1 664

5 356

48

243

1144 AB 5 3 10

24

7

6430

8

212

114 5 30 1 56

0 1 48

2120 1 77

48 1 664

5 3230

44

269

264 4B 2 3A 5 430

8

212

114 2 33 10

24

7

64 5 3 0 2 30

32 1 12

248 2 21

44 2 184

5 3 10

212

19

254 A 2 B 5 3 10

24

7

64 2 30

8

212

114 5 3 10 2 0

24 2 8

7 2 s212d6 2 114


131. (a)

(b)

(c)

(d) BA 5 31

0

0

4

1

3

2

6

14 3

22

4

1

23

5

7

6

7

44 5 3

22 1 16 1 2

0 1 4 1 6

0 1 12 1 1

23 1 20 1 14

0 1 5 1 42

0 1 15 1 7

6 1 28 1 8

0 1 7 1 24

0 1 21 1 44 5 3

16

10

13

31

47

22

42

31

254

AB 5 322

4

1

23

5

7

6

7

44 3

1

0

0

4

1

3

2

6

14 5 3

22 1 0 1 0

4 1 0 1 0

1 1 0 1 0

28 2 3 1 18

16 1 5 1 21

4 1 7 1 12

24 2 18 1 6

8 1 30 1 7

2 1 42 1 44 5 3

22

4

1

7

42

23

216

45

484

4B 2 3A 5 431

0

0

4

1

3

2

6

14 2 33

22

4

1

23

5

7

6

7

44 5 3

4 2 s26d0 2 12

0 2 3

16 2 s29d4 2 15

12 2 21

8 2 18

24 2 21

4 2 124 5 3

10

212

23

25

211

29

210

3

284

A 2 B 5 322

4

1

23

5

7

6

7

44 2 3

1

0

0

4

1

3

2

6

14 5 3

22 2 1

4 2 0

1 2 0

23 2 4

5 2 1

7 2 3

6 2 2

7 2 6

4 2 14 5 3

23

4

1

27

4

4

4

1

34

132. (a)

(b)

(c)

(d) 5 320

2

1

4

15

26

8

24

645 3

0 1 20 1 0

23 1 5 1 0

1 1 0 1 0

0 1 4 1 0

12 1 1 1 2

24 1 0 2 2

0 1 8 1 0

0 1 2 2 6

0 1 0 1 64 BA 5 3

0

3

21

4

1

0

0

22

243

21

5

0

4

1

21

0

2

34

5 312

1

26

0

21

21

28

2

84 5 3

0 1 12 1 0

0 1 3 2 2

0 2 3 2 3

24 1 4 1 0

20 1 1 1 0

0 2 1 1 0

0 2 8 1 0

0 2 2 1 4

0 1 2 1 64 AB 5 3

21

5

0

4

1

21

0

2

343

0

3

21

4

1

0

0

22

24

5 30 2 s23d

12 2 15

24 2 0

16 2 12

4 2 3

0 2 s23d

0 2 0

28 2 6

8 2 94 5 3

3

23

24

4

1

3

0

214

214 4B 2 3A 5 43

0

3

21

4

1

0

0

22

24 2 33

21

5

0

4

1

21

0

2

34

5 321 2 0

5 2 3

0 2 s21d

4 2 4

1 2 1

21 2 0

0 2 0

2 2 s22d3 2 2

4 5 321

2

1

0

0

21

0

4

14 A 2 B 5 3

21

5

0

4

1

21

0

2

34 2 3

0

3

21

4

1

0

0

22

24

133. |A| 5 | 3

21

5

7| 5 3s7d 2 5s21d 5 26 134. |22

12

8

15| 5 22s15d 2 8s12d 5 2126

135.

5 2194 5 3f7s21d 2 3s9dg 1 4f4s3d 2 5s7dg

5 3|79 3

21| 1 4|47 5

3| |A| 5 |3

0

4

4

7

9

5

3

21|136.

—CONTINUED—

5 211s12 2 6d 2 1s10 2 4d 2 6s30 2 24d 5 2108

5 211|212

22

23

21| 2 1|210

22

22

21| 2 6|210

212

22

23| C21 5 s21d211| 11

21

6

10

12

2

2

3

1| 5 |211

1

26

210

212

22

22

23

21| 5 8s12 2 6d 2 3s21 2 18d 1 7s22 2 72d 5 2413

5 8|12

2

3

1| 2 3|21

6

3

1| 1 7|21

6

12

2| C11 5 s21d111| 8

21

6

3

12

2

7

3

1| 5 | 8

21

6

3

12

2

7

3

1| |A| 5 16sC11d 1 9sC21d 2 2sC31d 2 4sC41d

Section 9.2 Arithmetic Sequences and Partial Sums 831

136. —CONTINUED—

5 211,758.

So, |A| 5 16s2413d 1 9s2108d 2 2s215d 2 4s937d

5 211s9 2 84d 1 8s30 2 24d 1 1s70 2 6d 5 937

5 211| 23

212

27

23| 2 s28d|210

212

22

23| 1 1|210

23

22

27| C41 5 s21d411| 11

8

21

10

3

12

2

7

3| 5 |211

28

1

210

23

212

22

27

23| 5 11s3 2 14d 2 8s10 2 4d 1 6s70 2 6d 5 215

5 11|32 7

1| 2 8|10

2

2

1| 1 6|10

3

2

7| C31 5 s21d311|11

8

6

10

3

2

2

7

1| 5 |11

8

6

10

3

2

2

7

1|

Section 9.2 Arithmetic Sequences and Partial Sums

n You should be able to recognize an arithmetic sequence, find its common difference, and find its nth term.

n You should be able to find the nth partial sum of an arithmetic sequence by using the formula

Sn 5n

2sa1 1 and.

Vocabulary Check

1. arithmetic; common 2.

3. sum of a finite arithmetic sequence

an 5 dn 1 c

1.

Arithmetic sequence, d 5 22

10, 8, 6, 4, 2, . . . 3.

Not an arithmetic sequence

1, 2, 4, 8, 16, . . .

5.


94, 2,

74,

32,

54, . . .

7.


13,

23, 1,

43,

56, . . . 9.


ln 1, ln 2, ln 3, ln 4, ln 5, . . .

11.


8, 11, 14, 17, 20

an 5 5 1 3n

2. 4, 7, 10, 13, 16, . . .


4. 80, 40, 20, 10, 5, . . .


6. 3, 2, 1, . . .


32,

52,

8. 5.3, 5.7, 6.1, 6.5, 6.9, . . .

Arithmetic sequence, d 5 0.4

10.


12, 22, 32, 42, 52, . . . 12.

97, 94, 91, 88, 85


an 5 100 2 3n


13.


7, 3, 21, 25, 29

an 5 3 2 4sn 2 2d 15.


21, 1, 21, 1, 21

an 5 s21dn14.

1, 5, 9, 13, 17


an 5 1 1 sn 2 1d4

16.

1, 2, 4, 8, 16


an 5 2n21 18.

2, 8, 24, 64, 160


an 5 s2ndn17.


23, 3

2, 21,

3

4, 2

3

5

an 5s21dn3

n

19.

an 5 a1 1 sn 2 1dd 5 1 1 sn 2 1ds3d 5 3n 2 2

a1 5 1, d 5 3 20.

5 4n 1 11

an 5 a1 1 sn 2 1dd 5 15 1 sn 2 1d4

a1 5 15, d 5 4

21.

5 28n 1 108

an 5 a1 1 sn 2 1dd 5 100 1 sn 2 1ds28d

a1 5 100, d 5 28

23.

an 5 a1 1 sn 2 1dd 5 x 1 sn 2 1ds2xd 5 2xn 2 x

a1 5 x, d 5 2x

25.

5 252n 1

132an 5 a1 1 sn 2 1dd 5 4 1 sn 2 1ds25

2dd 5 2

52

4, 32, 21, 2

72, . . .

27.

5103 n 1

53an 5 a1 1 sn 2 1dd 5 5 1 sn 2 1ds10

3 da4 5 a1 1 3d ⇒ 15 5 5 1 3d ⇒ d 5

103

a1 5 5, a4 5 15

29.

5 23n 1 103

an 5 a1 1 sn 2 1dd 5 100 1 sn 2 1ds23d

a1 5 a3 2 2d ⇒ a1 5 94 2 2s23d 5 100

a6 5 a3 1 3d ⇒ 85 5 94 1 3d ⇒ d 5 23

a3 5 94, a6 5 85

22.

5 223n 1

23

an 5 a1 1 sn 2 1dd 5 sn 2 1ds223d

a1 5 0, d 5 223

24.

5 5yn 2 6y

an 5 a1 1 sn 2 1dd 5 2y 1 sn 2 1ds5yd

a1 5 2y, d 5 5y

26.

an 5 a1 1 sn 2 1dd 5 10 1 sn 2 1ds25d 5 25n 1 15

d 5 25

10, 5, 0, 25, 210, . . .

28.

5 5n 2 9

an 5 a1 1 sn 2 1dd 5 24 1 sn 2 1d5

d 5 5

16 5 24 1 4d

an 5 a1 1 sn 2 1dd

a1 5 24, a5 5 16

30.

5 215n 1 265

an 5 a1 1 sn 2 1dd 5 250 1 sn 2 1ds215d

a1 5 a5 2 4d ⇒ a1 5 190 2 4s215d 5 250

a10 5 a5 1 5d ⇒ 115 5 190 1 5d ⇒ d 5 215

a5 5 190, a10 5 115


31.

a5 5 23 1 6 5 29

a4 5 17 1 6 5 23

a3 5 11 1 6 5 17

a2 5 5 1 6 5 11

a1 5 5

a1 5 5, d 5 6 32.

a5 5114 2

34 5

84 5 2

a4 572 2

34 5

114

a3 5174 2

34 5

144 5

72

a2 5 5 234 5

174

a1 5 5

a1 5 5, d 5 234 33.

a5 5 23.8 1 s20.4d 5 24.2

a4 5 23.4 1 s20.4d 5 23.8

a3 5 23.0 1 s20.4d 5 23.4

a2 5 22.6 1 s20.4d 5 23.0

a1 5 22.6

a1 5 22.6, d 5 20.4

34.

a5 5 17.25 1 0.25 5 17.5

a4 5 17 1 0.25 5 17.25

a3 5 16.75 1 0.25 5 17

a2 5 16.5 1 0.25 5 16.75

a1 5 16.5

a1 5 16.5, d 5 0.25 35.

a5 5 14 1 4 5 18

a4 5 10 1 4 5 14

a3 5 6 1 4 5 10

a2 5 2 1 4 5 6

a1 5 2

44 5 d

44 5 11d

46 5 2 1 s12 2 1dd

a1 5 2, a12 5 46 36.

Answer:

a5 5 16 1 5 5 21

a4 5 11 1 5 5 16

a3 5 6 1 5 5 11

a2 5 1 1 5 5 6

a1 5 1

a1 5 1, d 5 5

46 5 a10 5 a1 1 sn 2 1dd 5 a1 1 9d

16 5 a4 5 a1 1 sn 2 1dd 5 a1 1 3d

a4 5 16, a10 5 46

37.

a5 5 10 1 4 5 14

a4 5 6 1 4 5 10

a3 5 2 1 4 5 6

a2 5 22 1 4 5 2

a1 5 22

26 5 a1 1 28 ⇒ a1 5 22

a8 5 a1 1 7d

42 5 26 1 4d ⇒ d 5 4

a12 5 a8 1 4d

a8 5 26, a12 5 42 38.

Answer:

a5 5 17.275 2 1.725 5 15.55

a4 5 19 2 1.725 5 17.275

a3 5 20.725 2 1.725 5 19

a2 5 22.45 2 1.725 5 20.725

a1 5 22.45

a1 5 22.45, d 5 21.725

21.7 5 a15 5 a1 1 sn 2 1dd 5 a1 1 14d

19 5 a3 5 a1 1 sn 2 1dd 5 a1 1 2d

a3 5 19, a15 5 21.7

39.

an 5 4n 1 11

c 5 a1 2 d 5 15 2 4 5 11

d 5 4

a5 5 27 1 4 5 31

a4 5 23 1 4 5 27

a3 5 19 1 4 5 23

a2 5 15 1 4 5 19

a1 5 15, ak11 5 ak 1 4 41.

an 5 210n 1 210

c 5 a1 2 d 5 200 2 s210d 5 210

d 5 210

a5 5 170 2 10 5 160

a4 5 180 2 10 5 170

a3 5 190 2 10 5 180

a2 5 200 2 10 5 190

a1 5 200, ak11 5 ak 2 1040.

So, an 5 5n 1 1.

5 1

5 6 2 5

c 5 a1 2 d

an 5 5n 1 c

an 5 dn 1 c

d 5 5

a5 5 21 1 5 5 26

a4 5 16 1 5 5 21

a3 5 11 1 5 5 16

a2 5 6 1 5 5 11

a1 5 6, ak11 5 ak 1 5


43.

an 5 218n 1

34

c 5 a1 2 d 558 2 s21

8d 534

d 5 218

a5 514 2

18 5

18

a4 538 2

18 5

14

a3 512 2

18 5

38

a2 558 2

18 5

12

a1 558

a1 558, ak11 5 ak 2

18

42.

So, an 5 26n 1 78.

5 78

5 72 2 s26d

c 5 a1 2 d

an 5 26n 1 c

an 5 dn 1 c

d 5 26

a5 5 54 2 6 5 48

a4 5 60 2 6 5 54

a3 5 66 2 6 5 60

a2 5 72 2 6 5 66

a1 5 72, ak11 5 ak 2 6 44.

So, an 5 0.25n 1 0.125.

5 0.125

5 0.375 2 0.25

c 5 a1 2 d

an 5 0.25n 1 c

an 5 dn 1 c

d 5 0.25

a5 5 1.125 1 0.25 5 1.375

a4 5 0.875 1 0.25 5 1.125

a3 5 0.625 1 0.25 5 0.875

a2 5 0.375 1 0.25 5 0.625

a1 5 0.375, ak11 5 ak 1 0.25

45.

an 5 a1 1 sn 2 1dd ⇒ a10 5 5 1 9s6d 5 59

a1 5 5, a2 5 11 ⇒ d 5 11 2 5 5 6 46.

an 5 10n 2 7, a9 5 10s9d 2 7 5 83

c 5 a1 2 d 5 3 2 10 5 27

an 5 dn 1 c, an 5 10n 1 c

d 5 a2 2 a1 5 13 2 3 5 10

a1 5 3, a2 5 13

47.

an 5 a1 1 sn 2 1dd ⇒ a7 5 4.2 1 6s2.4d 5 18.6

a1 5 4.2, a2 5 6.6 ⇒ d 5 6.6 2 4.2 5 2.4 48.

an 5 213.1n 1 12.4, a8 5 292.4

c 5 a1 2 d 5 20.7 1 13.1 5 12.4

an 5 dn 1 c, an 5 213.1n 1 c

d 5 a2 2 a1 5 213.8 1 0.7 5 213.1

a1 5 20.7, a2 5 213.8

49.

so the sequence is decreasing

and

Matches (b).

a1 5 714.

d 5 234

an 5 234n 1 8 50.

so the sequence is increasing

Matches (d).

and a1 5 22.

d 5 3

an 5 3n 2 5 51.

so the sequence is increasing

and

Matches (c).

a1 5 234.

d 534

an 5 2 134n

52.

so the sequence

is decreasing

Matches (a).

and a1 5 22.

d 5 23

an 5 25 2 3n 53.

0

0

10

14

an 5 15 232n 54.

0

−6

10

16

an 5 25 1 2n

55.

0

2

10

6

an 5 0.2n 1 3 56.

0

4

10

9

an 5 20.3n 1 8 57.

S10 5102 s8 1 116d 5 620

a10 5 8 1 9s12d 5 116

a1 5 8, d 5 12, n 5 10

8, 20, 32, 44, . . .


58.

S25 5252 s2 1 146d 5 1850

a1 5 2 and a25 5 146

an 5 6n 2 4

d 5 6, c 5 2 2 6 5 24

2, 8, 14, 20, . . . , n 5 25 59.

S12 5122 f4.2 1 s21.3dg 5 17.4

a12 5 4.2 1 11s20.5d 5 21.3

a1 5 4.2, d 5 20.5, n 5 12

4.2, 3.7, 3.2, 2.7, . . . 60.

S10 5102 s0.5 1 4.1d 5 23

a1 5 0.5 and a10 5 4.1

an 5 0.4n 1 0.1

d 5 0.4, c 5 0.1

0.5, 0.9, 1.3, 1.7, . . . , n 5 10

61.

S10 510

2s40 1 13d 5 265

a10 5 40 1 9s23d 5 13

a1 5 40, d 5 23, n 5 10

40, 37, 34, 31, . . . 63.

S25 525

2s100 1 220d 5 4000

Sn 5n

2fa1 1 ang

a1 5 100, a25 5 220, n 5 2562.

S25 525

2s75 2 45d 5 375

a1 5 75 and a25 5 245

an 5 25n 1 80

d 5 25, c 5 80

75, 70, 65, 60, . . . , n 5 25

64.

S100 5100

2 s15 1 307d 5 16,100

a1 5 15, a100 5 307, n 5 100 65.

o100

n51

s2n 2 1d 51002 s1 1 199d 5 10,000

a1 5 1, a100 5 199

an 5 2n 2 1

66.

5 1200

o60

i50

si 2 10d 5602 s210 1 50d

a0 5 210, a60 5 50, n 5 60 67.

o50

n51

n 5502 s1 1 50d 5 1275

a1 5 1, a50 5 50, n 5 50 68.

o100

n51

2n 51002 s2 1 200d 5 10,100

a1 5 2, a100 5 200, n 5 100

an 5 2n

69.

o100

n510

6n 5912 s60 1 600d 5 30,030

a10 5 60, a100 5 600, n 5 91

71. o30

n511

n 2 o10

n51

n 5202 s11 1 30d 2

102 s1 1 10d 5 355

70.

o100

n551

7n 5502 s357 1 700d 5 26,425

a51 5 357, a100 5 700

an 5 7n

72.

5 3775 2 1275 5 2500

o100

n551

n 2 o50

n51

n 5502 s51 1 100d 2

502 s1 1 50d

73.

o400

n51

s2n 2 1d 54002 s1 1 799d 5 160,000

a1 5 1, a400 5 799, n 5 400

75. o20

n51

s2n 1 5d 5 520

77. o100

n51

n 1 4

25 2725

74.

o250

n51

s1000 2 nd 52502 s999 1 750d 5 218,625

a1 5 999, a250 5 750, n 5 250

an 5 1000 2 n

76.

o50

n50

s1000 2 5nd 5512 s1000 1 750d 5 44,625

a0 5 1000, a50 5 750, n 5 51

78.

o100

n50

8 2 3n

165

101

2 11

22

73

4 2 5 2896.375

a0 51

2, a100 5

273

4, n 5 101


79. o60

i51

s250 283id 5 10,120 80.

o200

j51

s4.5 1 0.025jd 52002 s4.525 1 9.5d 5 1402.5

a1 5 4.525, a200 5 9.5, n 5 200

81. (a)

(b) S6 562f32,500 1 40,000g 5 $217,500

a6 5 a1 1 5d 5 32,500 1 5s1500d 5 $40,000

a1 5 32,500, d 5 1500 82. (a)

(b) S6 562 f36,800 1 45,550g 5 $247,050

a6 5 a1 1 5d 5 36,800 1 5s1750d 5 $45,550

a1 5 36,800, d 5 1750

83.

S30 5302 s20 1 136d 5 2340 seats

a30 5 20 1 29s4d 5 136

3a1 5 20, d 5 4, n 5 30 85.

S18 5182 s14 1 31d 5 405 bricks

a1 5 14, a18 5 31

86.

S28 5282 s14 1 0.5d 5 203 bricks

a1 5 14, a28 5 0.5, n 5 28

84.

S36 5362 s15 1 120d 5 2430 seats

a36 5 15 1 35s3d 5 120

a1 5 15, d 5 3, n 5 36

87.

S10 5102 s4.9 1 93.1d 5 490 meters

a10 5 4.9 1 9s9.8d 5 93.1 meters

d 5 9.8

4.9, 14.7, 24.5, 34.3, . . .

88.

Distance ft5 o7

n51

s32n 2 16d 5 784

an 5 32n 2 16

c 5 a1 2 d 5 16 2 32 5 216

an 5 dn 1 c 5 32n 1 c

d 5 32

a1 5 16, a2 5 48, a3 5 80, a4 5 112 89. (a)

(b)

S8 582s200 1 25d 5 $900

a8 5 225s8d 1 225 5 25

an 5 225n 1 225

c 5 200 2 s225d 5 225

a1 5 200, a2 5 175 ⇒ d 5 225

90. (a)

(b) Total prize money

5 $7800

5122 s1200 1 100d

5 o12

n51

s2100n 1 1300d

an 5 2100n 1 1300

c 5 a1 2 d 5 1200 1 100 5 1300

an 5 2100n 1 c

an 5 dn 1 c

d 5 2100

a1 5 1200, a2 5 1100, a3 5 1000 91.

The cost of gasoline, labor, equipment, insurance, and

maintenance are a few economic factors that could

prevent the company from meeting its goals, but the

biggest unknown variable is the amount of annual

snowfall.

S6 562s8000 1 15,500d 5 $70,500

a1 5 8000, a6 5 15,500

an 5 1500n 1 6500


92.

Total sales 5 o10

n51

s5000n 1 10,000d 5102 s15,000 1 60,000d 5 $375,000

an 5 5000n 1 10,000

c 5 a1 2 d 5 15,000 2 5000 5 10,000

an 5 dn 1 c 5 5000n 1 c

n 5 1, . . . , 10

d 5 5,000

a1 5 15,000

93. (a)Monthly Payment Unpaid Balance

$1800

$1600

$1400

$1200

$1000

$800a6 5 200 1 0.01s1000d 5 $210

a5 5 200 1 0.01s1200d 5 $212

a4 5 200 1 0.01s1400d 5 $214

a3 5 200 1 0.01s1600d 5 $216

a2 5 200 1 0.01s1800d 5 $218

a1 5 200 1 0.01s2000d 5 $220

(b)

Interest paid: $110

S10 5102 s220 1 202d 5 $2110

an 5 22n 1 222 ⇒ a10 5 202

94. (a) Borrowed Amount

Amount of Balance Paid Per Month

Unpaid Balance

Interest Balance Before Payment

Total Payment

(b) Total Interest Paid 5 o20

n51

f5000 2 250sn 2 1dg ? 0.01 5202 fs5000ds0.01d 1 s250ds0.01dg 5 $525

5 $250 1 I

? 1% 5 an21 ? 0.015 I 5

5 an 5 5000 2 250n

5 $250

5 a0 5 $5,000

Month (n) 1 2 3 4 5 6

Interest (I) $50 $47.50 $45.00 $42.50 $40.00 $37.50

Total Payment $300 $297.50 $295.00 $292.50 $290.00 $287.50

Unpaid Balance $4750 $4500 $4250 $4000 $3750 $3500sand

s$250 1 Id

Month (n) 7 8 9 10 11 12

Interest (I) $35.00 $32.50 $30.00 $27.50 $25.00 $22.50

Total Payment $285.00 $282.50 $280.00 $277.50 $275.00 $272.50

Unpaid Balance $3250 $3000 $2750 $2500 $2250 $2000sand

s$250 1 Id


96. (a) is 1997.

(b)

(c) Total revenue

million

(d) a18 5 1726.93s18d 2 11,718.43 5 $19,366.31 million

5 $38,856

572s370.08 1 10,731.66d

5 o13

n57

s1726.93n 2 11,718.43d

an 5 Revenue 5 1726.93n 2 11,718.43

2,000

4,000

6,000

8,000

n

an

10,000

12,000

Rev

enue

(in m

illi

ons

of

doll

ars)

Year (7 ↔ 1997)

7 8 9 10 11 12 13

n 5 7

98. True, by the formula for the sum of a finite arithmetic

sequence,

Sn 5n

2sa1 1 and.

99. A sequence is arithmetic if the differences between

consecutive terms are the same.

an11 2 an 5 d for n ≥ 1

100. First term plus times the common differencesn 2 1d

101. (a)

(c) The graph of contains only points at

the positive integers. The graph of is a

solid line which contains these points.

y 5 3x 1 2

an 5 2 1 3n

an

n

33

30

27

24

21

18

15

12

9

6

3

1 2 3 4 5 6 7 8 9 10 11−1

an 5 2 1 3n (b)

(d) The slope is equal to the common difference

In general, these should be equal.d 5 3.

m 5 3

y

x

33

30

27

24

21

18

15

12

963

1 2 3 4 5 6 7 8 9 10 11

y 5 3x 1 2

102. (a)

(b)

(c) Sn 5n

2f1 1 s2n 2 1dg 5

n

2s2nd 5 n2

S7 5 1 1 3 1 5 1 7 1 9 1 11 1 13 5 49 5 72

Sn 5 n2

1 1 3 1 5 1 7 1 9 1 11 5 36

1 1 3 1 5 1 7 1 9 5 25

1 1 3 1 5 1 7 5 16

1 1 3 1 5 5 9

1 1 3 5 4 103.

a1 5 4

2a1 5 8

2a1 1 57 5 65

10s2a1 1 57d 5 650

S20 520

2Ha1 1 fa1 1 s20 2 1ds3dgJ 5 650

97. True; given and then and

an 5 a1 1 sn 2 1dd.

d 5 a2 2 a1a2a1

95. (a) Using (5, 23,078) and (6, 24,176) we have

and

(b)

The models are similar.

(c)

(d) For 2004 use

For 2005 use

(e) Answers will vary.

n 5 15: $34,058

n 5 14: $32,960

20,000

3 13

32,000

an < 1114.95n 1 17,795.07

an < 1098n 1 17,588

c 5 23,078 2 5s1098) 5 17,588.

d 5 1098


104. be the sum

of the first n terms of the original

sequence.

5 Sn 1 5n

5n

2sa1 1 an) 1 5n

5n

2sa1 1 and 1

n

2s10d

5n

2sa1 1 an 1 10d

Sn9 5n

2sa1 1 5 1 an 1 5d

Let Sn 5n

2 sa1 1 an d 105.

Slope:

y-intercept:

y

x−4 −3 −2 −1

−4

−3

−2

1

2

3

4

1 2 3 4

10, 23

42

m 51

2

y 51

2x 2

3

4

2x 2 4y 5 3 106.

Slope:

y-intercept:

y

x−4 −3 −2 −1

−12

−10

−8

−6

2

4

1 2 3 4

s0, 28d

29

y 5 29x 2 8

9x 1 y 5 28

107.

Vertical line

No slope

No y-intercept

x 5 7

y

x−2 2 4 6

−8

−6

−4

−2

2

4

6

8

8 10 12 14

x 2 7 5 0 108.

Slope:

y-intercept: s0, 211d

0

y 5 211

y

x

−8 −6 −4 −2

−12

−10

−8

−6

−4

−2

2

4

2 4 6 8

y 1 11 5 0

109.

Answer: x 5 1, y 5 5, z 5 21

s2172 d

s210dEq.3 1 Eq.1

Eq.3 1 Eq.25x

y

z

5

5

5

1

5

21

s2 199dEq.3

5x

y

1

1

172 z

10z

z

5

5

5

2152

25

21

s12dEq.2 1 Eq.1

s27dEq.2 1 Eq.35

x

y

1

1

2

172 z

10z

99z

5

5

5

2152

25

99

s212dEq.2

2 Eq.35

x 212 y

y

7y

1

1

2

72z

10z

29z

5

5

5

25

25

64

5x 2

2

12y

2y

72y

1

2

2

72z

20z

292 z

5

5

5

25

10

32

s23dEq.1 1 Eq.2

s26dEq.1 1 Eq.35

x 2

2

12 y72 y

2y

1

2

2

72z

292 z

20z

5

5

5

25

32

10

12 Eq.1

5x

3x

6x

2

1

2

12 y

2y

5y

1

2

1

72z

4z

z

5

5

5

25

17

220

Equation 1

Equation 2

Equation 35

2x

3x

6x

2

1

2

y

2y

5y

1

2

1

7z

4z

z

5

5

5

210

17

220

110.

x 5 2, y 5 26, z 5 3

R1 1 4R2 1 10R3 →

31

0

0

0

1

0

0

0

1

:

:

:

2

26

34

R2 2 3R3 → 31

0

0

24

1

0

210

0

1

:

:

:

24

26

34

125R3 →

31

0

0

24

1

0

210

3

1

:

:

:

24

3

34

117R2 → 3

1

0

0

24

1

0

210

3

25

24

3

754

2R1 →

31

0

0

24

17

0

210

51

25

:

:

:

24

51

754

2R2 2 R3 →

321

0

0

4

17

0

10

51

25

:

:

:

4

51

754

8R1 1 R3 →

321

0

0

4

17

34

10

51

77

:

:

:

4

51

274

5R1 1 R2 → 321

0

8

4

17

2

10

51

23

:

:

:

4

51

254

321

5

8

4

23

2

10

1

23

:

:

:

4

31

254

840 Chapter 9 Sequences, Series and Probability

Section 9.3 Geometric Sequences and Series

n You should be able to identify a geometric sequence, find its common ratio, and find the nth term.

n You should know that the nth term of a geometric sequence with common ratio r is given by

n You should know that the nth partial sum of a geometric sequence with common ratio is given by

n You should know that if then

o`

n51

a1rn21 5 o

`

n50

a1rn 5

a1

1 2 r.

|r| < 1,

Sn 5 a111 2 rn

1 2 r 2.

r Þ 1

an 5 a1rn21.

Vocabulary Check

1. geometric; common 2.

3. 4. geometric series

5. S 5a1

1 2 r

Sn 5 a111 2 r n

1 2 r 2an 5 a1r

n21

1.

Geometric sequence, r 5 3

5, 15, 45, 135, . . . 3.

Not a geometric sequence

Note: It is an arithmetic sequence

with d 5 9.

3, 12, 21, 30, . . .2. 3, 12, 48, 192, . . .


4. 36, 27, 18, 9, . . .


6. 5, 1, 0.2, 0.04, . . .

Geometric sequence, r 515 5 0.2

5.


1, 212,

14, 2

18, . . .

7.


18,

14,

12, 1, . . . 9.


1, 12,

13,

14, . . .8.


9, 26, 4, 283, . . .

10.


15,

27,

39,

411, . . . 12.

a5 5 6s2d4 5 96

a4 5 6s2d3 5 48

a3 5 6s2d2 5 24

a2 5 6s2d1 5 12

a1 5 6

a1 5 6, r 5 211.

a5 5 54s3d 5 162

a4 5 18s3d 5 54

a3 5 6s3d 5 18

a2 5 2s3d 5 6

a1 5 2

a1 5 2, r 5 3

111. Answers will vary.

Section 9.3 Geometric Sequences and Series 841

13.

a5 518s1

2d 5116

a4 514s1

2d 518

a3 512s1

2d 514

a2 5 1s12d 5

12

a1 5 1

a1 5 1, r 512 14.

a5 5 1s13d4

51

81

a4 5 1s 13d

35

127

a3 5 1s 13d

25

19

a2 5 1s13d

15

13

a1 5 1

a1 5 1, r 513 15.

a5 5 s2 1200ds2 1

10d 51

2000

a4 5120s2 1

10d 5 21

200

a3 5 s212ds2 1

10d 51

20

a2 5 5s2 110d 5 2

12

a1 5 5

a1 5 5, r 5 2110

16.

a5 5 6s214d4

53

128

a4 5 6s214d3

5 2332

a3 5 6s214d2

538

a2 5 6s214d1

5 232

a1 5 6

a1 5 6, r 5 214 18.

a5 5 3s!5 d45 75

a4 5 3s!5 d35 15!5

a3 5 3s!5 d25 15

a2 5 3s!5 d15 3!5

a1 5 3

a1 5 3, r 5 !517.

a5 5 se3dsed 5 e4

a4 5 se2dsed 5 e3

a3 5 sedsed 5 e2

a2 5 1sed 5 e

a1 5 1

a1 5 1, r 5 e

19.

a5 5 1 x3

3221x

42 5x4

128

a4 5 1x2

8 21x

42 5x3

32

a3 5 1x

221x

42 5x2

8

a2 5 21x

42 5x

2

a1 5 2

a1 5 2, r 5x

420.

a5 5 5s2xd4 5 80x4

a4 5 5s2xd3 5 40x3

a3 5 5s2xd2 5 20x2

a2 5 5s2xd1 5 10x

a1 5 5

a1 5 5, r 5 2x 21.

an 5 6411

22n21

512811

22n

r 51

2

a5 51

2s8d 5 4

a4 51

2s16d 5 8

a3 51

2s32d 5 16

a2 51

2s64d 5 32

a1 5 64

a1 5 64, ak11 51

2ak

22.

an 5 81s 13d

n215 243s 1

3dn

a5 513s3d 5 1

a4 513s9d 5 3

a3 513s27d 5 9

a2 513s81d 5 27

a1 5 81

a1 5 81, ak11 513ak

23.

an 5 7s2dn21 572s2dn

r 5 2

a5 5 2s56d 5 112

a4 5 2s28d 5 56

a3 5 2s14d 5 28

a2 5 2s7d 5 14

a1 5 7

a1 5 7, ak11 5 2ak 24.

an 5 5s22dn215 2

52s22dn

a5 5 22s240d 5 80

a4 5 22s20d 5 240

a3 5 22s210d 5 20

a2 5 22s5d 5 210

a1 5 5

a1 5 5, ak11 5 22ak


25.

an 5 6s232dn21

or an 5 24s232dn

r 5 232

a5 5 232s281

4 d 52438

a4 5 232s27

2 d 5 2814

a3 5 232s29d 5

272

a2 5 232s6d 5 29

a1 5 6

a1 5 6, ak11 5 232ak 26.

an 5 48s212dn21

5 296s212dn

a5 5 212 s26d 5 3

a4 5 212 s12d 5 26

a3 5 212 s224d 5 12

a2 5 212s48d 5 224

a1 5 48

a1 5 48, ak11 5 212ak 27.

a10 5 4s12d

95 s1

2d7

51

128

an 5 a1rn21 5 4s1

2dn21

a1 5 4, r 512, n 5 10

28.

a8 5 5 13

227

510,935

128

an 5 a1rn21 5 513

22n21

a1 5 5, r 53

2, n 5 8 29.

a12 5 6121

3211

5 22

310

an 5 a1rn21 5 612

1

32n21

1a1 5 6, r 5 21

3, n 5 12 30.

a10 5 64121

429

5 264

262,144

an 5 a1rn21 5 6412

1

42n21

a1 5 64, r 5 21

4, n 5 10

31.

a9 5 100sexd8 5 100e8x

an 5 a1rn21 5 100sexdn21

a1 5 100, r 5 ex, n 5 9 32.

a8 5 1s!3d75 27!3

an 5 a1 rn21 5 1s!3 dn21

a1 5 1, r 5 !3, n 5 8 33.

a40 5 500s1.02d39 < 1082.372

an 5 a1rn21 5 500s1.02dn21

a1 5 500, r 5 1.02, n 5 40

34.

a60 5 1000s1.005d59 < 1342.139

an 5 a1rn21 5 1000s1.005dn21

a1 5 1000, r 5 1.005, n 5 60 35.

a9 5 7s3d8 5 45,927

an 5 7s3dn21

7, 21, 63, . . . ⇒ r 5 3 36.

a7 5 s3ds12d6 5 8,957,952

an 5 a1rsn21d

r 5a2

a1

536

35 12

a1 5 3, a2 5 36, a3 5 432

37.

a10 5 5s6d9 5 50,388,480

an 5 5s6dn21

5, 30, 180, . . . ⇒ r 5 6 38.

a22 5 s4ds2d21 5 8,388,608

an 5 a1rn21

r 5a2

a1

58

45 2

a1 5 4, a2 5 8, a3 5 16 39.

a3 5 1613

422

5 9

an 5 1613

42n21

3

45 r

27

645 r3

27

45 16r3

a4 5 a1r3

a1 5 16, a4 527

4


40.

a1 5 12

3 5 a111

42a2 5 a1r

1

r 51

4

1

645 r3

3

645 3r3

a5 5 a2r3

a5 5 a2rs522d

a2 5 3, a5 53

6441.

a6 5a7

r5

2y3

21y35 22

21

35 r

21

275 r3

2

35 218r3

a7 5 a4r3

a4 5 218, a7 52

342.

a7 5 164

27212

322

5256

243

a7 5 a5r2

a7 5 a5rs725d

r 52

3

r 2 54

9

64

275

16

3r 2

a5 5 a3r2

a5 5 a3rs523d

a3 516

3, a5 5

64

27

43.

and

Since the sequence is decreasing.

Matches (a).

0 < r < 1,

r 523a1 5 18

an 5 18s 23d

n2144.

so the sequence alternates as it

approaches 0.

Matches (c).

r 5 s223d > 21,

an 5 18s223dn 21

45.

and so the sequence is increasing.

Matches (b).

r 532 > 1,a1 5 18

an 5 18s32d

n21

46.

so the sequence alternates as it

approaches

Matches (d).

`.

r 5 s232d < 21,

an 5 18s232dn 21

47.

0

−16

10

16

an 5 12s20.75dn21 48.

0

0 10

400

an 5 10s1.5dn21 49.

0

−15

10

15

an 5 12s20.4dn21

50.

−200

0 10

200

an 5 20s21.25dn21 52.

0

0 10

66

an 5 10s1.2dn2151.

0

0

10

24

an 5 2s1.3dn21

53.

S9 51s1 2 29d

1 2 25 511

o9

n51

2n21 5 1 1 21 1 22 1 . . . 1 28 ⇒ a1 5 1, r 5 2


54.

S10 5 131 2 s52d10

1 2 s52d 4 5 2

2

331 2 15

2210

4 53,254,867

512< 6357.162

o10

n5115

22n21

5 1 1 15

221

1 15

222

1 . . . 1 15

229

⇒ a1 5 1, r 5

5

2

55.

S9 5 111 2 s22d9

1 2 s22d 2 5 171

o9

n51

s22dn21 ⇒ a1 5 1, r 5 22, n 5 9

56.

S8 5 531 2 s232d8

1 2 s232d 4 5 231 2 12

3

228

4 5 26305

128< 249.258

o8

n51

5123

22n21

5 5 1 5123

221

1 5123

222

1 . . . 1 5123

227

⇒ a1 5 5, r 5 23

2

57.

S7 5 6431 2 s212d7

1 2 s212d 4 5

128

3 31 2 121

227

4 5 43

o7

i51

64121

22i21

5 64 1 64121

221

1 64121

222

1 . . . 1 64121

226

⇒ a1 5 64, r 5 21

2

58.

S10 5 231 2 s14d10

1 2 s14d 4 5

8

331 2 11

4210

4 < 2.667

o10

i51

211

42i21

5 2 1 211

421

1 211

422

1 . . . 1 211

429

⇒ a1 5 2, r 51

4

59.

S6 5 3211 2 s14d6

1 2142 5

1365

32

o6

i51

3211

42i21

5 32 1 3211

421

1 3211

422

1 3211

423

1 3211

424

1 3211

425

⇒ a1 5 32, r 51

4, n 5 6

60.

S12 5 1631 2 s12d12

1 2 s12d 4 5 3231 2 11

2212

4 54095

128< 31.992

o12

i51

1611

22i21

5 16 1 1611

221

1 1611

222

1 . . . 1 1611

2211

⇒ a1 5 16, r 51

2

61.

S21 5 331 2 s32d21

1 232

4 5 2631 2 13

2221

4 < 29,921.311

o20

n50

313

22n

5 o21

n51

313

22n21

5 3 1 313

221

1 313

222

1 . . . 1 313

2220

⇒ a1 5 3, r 53

2

62.

S41 5 5 1 331 2 s35d40

1 2 s35d 4 5 5 1

15

2 31 2 13

5240

4 < 12.500

o40

n50

513

52n

5 5 1 o40

n51 513

52n

5 5 1 3513

521

1 513

522

1 513

523

1 . . . 1 513

5240

4 ⇒ a1 5 3, r 53

5


63.

S16 5 211 2 s43d16

1 2432 < 592.647

o15

n50

214

32n

5 o16

n51

214

32n21

5 2 1 214

321

1 214

322

1 . . . 1 214

3215

⇒ a1 5 2, r 54

3, n 5 16

65.

S6 5 30031 2 s1.06d6

1 2 1.06 4 < 2092.596

5 300 1 300s1.06d1 1 300s1.06d2 1 300s1.06d3 1 300s1.06d4 1 300s1.06d5 ⇒ a1 5 300, r 5 1.06

o5

n50

300s1.06dn 5 o6

n51

300s1.06dn21

67.

< 1.6 58

5S41 5 231 2 s21

4d41

1 2 s214d 4 5

8

531 2 121

4241

4

o40

n50

2121

42n

5 2 1 2121

42 1 2121

422

1 . . . 1 2121

4240

⇒ a1 5 2, r 5 21

4, n 5 41

64.

S21 5 10 1 231 2 s15d20

1 2 s15d 4 5 10 1

5

231 2 11

5220

4 < 12.500

o20

n50

1011

52n

5 10 1 o20

n51 1011

52n

5 10 1 31011

521

1 1011

522

1 1011

523

1 . . . 1 1011

5220

4 ⇒ a1 5 2, r 51

5

66.

S7 5 500 1 52031 2 s1.04d6

1 2 s1.04d 4 5 500 2 13,000f1 2 s1.04d6g < 3949.147

a1 5 520, r 5 1.04

o6

n50

500s1.04dn 5 500 1 o6

n51

500s1.04dn 5 500 1 f500s1.04d1 1 500s1.04d2 1 . . . 1 500s1.04d6g

68.

S51 5 15 1 1031 2 s23d50

1 2 s23d 4 5 15 1 3031 2 12

3250

4 < 45.000

o50

n50

1012

32n21

515 1 o50

n51

1012

32n21

515 1 310 1 1012

321

1 1012

322

1 . . . 1 1012

3249

4 ⇒ a1 5 10, r 5

2

3

69.

< 6.400S10 5 831 2 s214d10

1 2 s214d 4 5

32

5 31 2 121

4210

4

o10

i51

8121

42i215 8 1 812

1

4211 812

1

4221 . . . 1 812

1

429 ⇒ a1 5 8, r 5 2

1

4

70.

S26 5 8 2 431 2 s212d25

1 2 s212d 4 5 8 2

8

331 2 121

2225

4 < 5.333

o25

i50

8121

22i

58 1 o25

i51

8121

22i

58 1 324 1 8121

222

1 8121

223

1 . . . 1 8121

2225

4 ⇒ a1 5 24, r 5 2

1

2

71.

S10 5 511 2 s213d10

1 2 s213d 2 < 3.750

o10

i51

5121

32i21

5 5 1 5121

321

1 5121

322

1 . . . 1 5121

329

⇒ a1 5 5, r 5 21

3, n 5 10


72.

S100 5 1531 2 s23d100

1 2 s23d 4 5 4531 2 12

321004 < 45.000

o100

i51

1512

32i21

515 1 1512

321

11512

322

1 . . . 1 1512

3299

⇒ a1 5 15, r 5

2

3

73.

Thus, the sum can be written as o7

n51

5s3dn21.

6 5 n 2 1 ⇒ n 5 7729 5 3n21 ⇒

r 5 3 and 3645 5 5s3dn21

5 1 15 1 45 1 . . . 1 3645 74.


n51

7s2dn21.

n 5 8

n 2 1 5 7

2n21 5 27

2n21 5 128

7s2dn21 5 896

a1 5 7, r 5 2

7 1 14 1 28 1 . . . 1 896

75.

and

By trial and error, we find that


n51

2121

42n21

.

n 5 7.

1

20485 212

1

42n21

r 5 21

4

2 21

21

1

82 . . . 1

1

204876.

By trial and error, we find that

Thus, the sum can be written as

o6

n51

15121

52n21

.

n 5 6.

121

52n

51

15,625

121

52n21

5 21

3125

15121

52n21

5 23

625

a1 5 15, r 5 21

5

15 2 3 13

52 . . . 2

3

625

77.

and


n51

0.1s4dn21.

1024 5 4n21 ⇒ 5 5 n 2 1 ⇒ n 5 6

102.4 5 0.1s4dn21r 5 4

0.1 1 0.4 1 1.6 1 . . . 1 102.4 78.

Thus, the sum can be written as

o5

n51

3213

42n21

.

n 5 5

n 2 1 5 4

13

42n21

5 13

424

13

42n21

581

256

3213

42n21

5 10.125 581

8

a1 5 32, r 53

4

32 1 24 1 18 1 . . . 1 10.125


79.

o`

n5011

22n

5a1

1 2 r5

1

1 2 s12d 5 2

a1 5 1, r 51

2

o`

n5011

22n

5 1 1 11

221

1 11

222

1 . . . 80.

o`

n50

212

32n

5a1

1 2 r5

2

1 223

5 6

a1 5 2, r 52

3

o`

n50

212

32n

5 2 1 212

321

1 212

322

1 . . .

81.

o`

n5012

1

22n

5a1

1 2 r5

1

1 2 s212d

52

3

a1 5 1, r 5 21

2

o`

n5012

1

22n

5 1 1 121

221

1 121

222

1 . . . 82.

o`

n50

2122

32n

5a1

1 2 r5

2

1 2 s223d

56

5

a1 5 2, r 5 22

3

o`

n50

2122

32n

5 2 1 2122

321

1 2122

322

1 . . .

83.

o`

n50

411

42n

5a1

1 2 r5

4

1 2 s14d 5

16

3

a1 5 4, r 51

4

o`

n50

411

42n

5 4 1 411

421

1 411

422

1 . . .

87.

o`

n50

23s0.9dn 523

1 2 0.95 230

a1 5 23, r 5 0.9

o`

n50

23s0.9dn 5 23 2 3s0.9d1 2 3s0.9d2 2 . . . 88.

o`

n50

210s0.2dn 5210

1 2 0.25 212.5

a1 5 210, r 5 0.2

o`

n50

f210s0.2dng 5 210 2 10s0.2d1 2 10s0.2d2 2 . . .

89. 8 1 6 19

21

27

81 . . . 5 o

`

n50

813

42n

58

1 234

5 32

84.

o`

n501 1

102n

5a1

1 2 r5

1

1 2110

510

9

a1 5 1, r 51

10

o`

n501 1

102n

5 1 1 1 1

1021

1 1 1

1022

1 . . .

85.

o`

n50

s0.4dn 51

1 2 0.45

5

3

a1 5 1, r 5 0.4

o`

n50

s0.4dn 5 1 1 s0.4d1 1 s0.4d2 1 . . . 86.

o`

n50

4s0.2dn 54

1 2 0.25 5

a1 5 4, r 5 0.2

o`

n50

4s0.2dn 5 4 1 4s0.2d1 1 4s0.2d2 1 . . .

90.

o`

n50

912

32n

59

1 223

5 27

a1 5 9, r 52

3

9 1 6 1 4 18

31 . . . 91.

The sum is undefined because

|r| 5 |23| 5 3 > 1.

1

92

1

31 1 2 3 1 . . . 5 o

`

n50

1

9s23dn


92.

The sum is undefined because

|r| 5 |26

5| 56

5> 1.

2125

361

25

62 5 1 6 2 . . . 5 o

`

n50

2125

36 126

52n

93. 0.36 5 o`

n50

0.36s0.01dn 50.36

1 2 0.015

0.36

0.995

36

995

4

11

94.

5297

9995

11

37

0.297 5 o`

n50

0 .297s0.001dn 50.297

1 2 0.0015

0.297

0.99995.

535

1105

7

22

53

101

0.018

0.995

3

101

18

9905

3

101

2

110

0.318 5 0.3 1 o`

n50

0.018s0.01dn 53

101

0.018

1 2 0.01

96. 1.38 5 1.3 1 o`

n50

0 .08s0.1dn 5 1.3 10.08

1 2 0.15 1.3 1

0.08

0.95 1

3

101

4

455 1

7

185

25

18

97.

The horizontal asymptote of

This corresponds to the sum of the series.

f sxd is y 5 12.

−4

−15

10

20

f sxd 5 631 2 s0.5dx

1 2 s0.5d 4, o`

n50

611

22n

56

1 212

5 12 98.

The horizontal asymptote of is

This corresponds to the sum of the series.

−9

−25

18

20

y 5 10.fsxd

fsxd 5 231 2 s0.8dx

1 2 s0.8d 4, o`

n50

214

52n

52

1 245

5 10

99. (a)

(b) The population is growing at a rate of 0.6% per year.

(c) For 2010, let :

million

(d)

This corresponds with the year 2008.

n 5

ln1 1320

1190.882ln 1.006

< 17.21

n ln 1.006 5 ln1 1320

1190.882

ln 1.006n 5 ln1 1320

1190.882

1.006n 51320

1190.88

1190.88s1.006dn 5 1320

< 1342.2

an 5 1190.88s1.006d20n 5 20

an < 1190.88s1.006dn 100.

(a)

(b)

(c)

(d)

(e) n 5 365, A 5 100011 10.06

365 2365s10d

< $1822.03

n 5 12, A 5 100011 10.06

12 212s10d

< $1819.40

n 5 4, A 5 100011 10.06

4 24s10d

< $1814.02

n 5 2, A 5 100011 10.06

2 22s10d

< $1806.11

n 5 1, A 5 1000s1 1 0.06d10 < $1790.85

A 5 P11 1r

n2nt

5 100011 10.06

n 2ns10d


101.

(a)

(b)

(c)

(d)

(e) n 5 365: A 5 250011 10.02

365 2s365ds20d

< $3729.52

n 5 12: A 5 250011 10.02

12 2s12ds20d

< $3728.32

n 5 4: A 5 250011 10.02

4 2s4ds20d

< $3725.85

n 5 2: A 5 250011 10.02

2 2s2ds20d

< $3722.16

n 5 1: A 5 250011 10.02

1 2s1ds20d< $3714.87

A 5 P11 1r

n2nt

5 250011 10.02

n 2ns20d

102. V5 5 135,000s0.70d5 5 $22,689.45

103. A 5 o60

n51

10011 10.06

12 2n

5 o60

n51

100s1.005dn 5 100(1.005d ?f1 2 1.00560gf1 2 1.005g

< $7011.89

104.

< $3698.34

5 50s1.006666667d11 2 s1.006666667d60

1 2 1.006666667 2

A 5 o60

n51

5011 10.08

12 2n

105. Let be the total number of deposits.

5 P311 1r

12212t

2 1411 112

r 2

5 P311 1r

122N

2 1411 112

r 2

5 P112

r1 12321 1 11 1

r

122N

4

5 P11 1r

1221212

r 231 2 11 1r

122N

4

5 P11 1r

12231 2 11 1r

122N

1 2 11 1r

122 4 5 P11 1

r

122oN

n5111 1

r

122n21

5 11 1r

1223P 1 P11 1r

122 1 . . . 1 P11 1r

122N21

4

A 5 P11 1r

122 1 P11 1r

1222

1 . . . 1 P11 1r

122N

N 5 12t

106. Let be the total number of deposits.

5Pery12sert 2 1d

sery12 2 1d

5 Pery12s1 2 sery12d12t d

1 2 ery12

5 Pery12s1 2 sery12dN d

s1 2 ery12d

5 oN

n51

Pery12?n

A 5 Pery12 1 Pe2ry12 1 . . . 1 PeNry12

N 5 12t 107.

(a) Compounded monthly:

(b) Compounded continuously:

A 550e0.07y12se0.07s20d 2 1d

e0.07y12 2 1< $26,263.88

< $26,198.27

A 5 50311 10.07

12 212s20d

2 1411 112

0.072

P 5 $50, r 5 7%, t 5 20 years


108.


(b) Compounded continuously: A 575e0.03y12se0.03s25d 2 1d

e0.03y12 2 1< $33,551.91

A 5 75311 10.03

12 212s25d2 1411 1

12

0.032 < $33,534.21

P 5 $75, r 5 3%, t 5 25 years

109.


(b) Compounded continuously: A 5100e0.10y12ses0.10ds40d 2 1d

e0.10y12 2 1< $645,861.43

A 5 100311 10.10

12 212s40d2 1411 1

12

0.102 < $637,678.02

P 5 $100, r 5 10%, t 5 40 years

110.


(b) Compounded continuously: A 520e0.06y12se0.06s50d 2 1d

e0.06y12 2 1< $76,533.16

A 5 20311 10.06

12 212s50d2 1411 1

12

0.062 < $76,122.54

P 5 $20, r 5 6%, t 5 50 years

111.

5 W112

r 231 2 11 1r

122212t

4

5 W 31 2 11 1

r

122212t4

11 1r

122 2 1

5 W1 1

1 1r

12231 2 11 1

r

122212t4

1 21

11 1r

122

5 W11 1r

1222131 2 11 1

r

122212t

1 2 11 1r

12221 4

P 5 W o12t

n51311 1

r

12221

4n

112.

P 5 20001 12

0.09231 2 11 10.09

12 2212s20d4 < $222,289.91

P 5 W112

r 231 2 11 1r

122212t4

W 5 $2000, t 5 20, r 5 9%

113. o`

n51

400s0.75dn 5300

1 2 0.755 $1200 114.

5 $1000

5200

0.20

5200

1 2 0.80

Amount put back into economy 5 o`

n51

250s0.80dn

r 5 80% 5 0.80

a1 5 250s0.80d 5 200


115. o`

n51

600s0.725dn 5435

1 2 0.725< $1581.82 116.

5 $1550

5348.75

0.225

5348.75

1 2 0.775

Amount put back into economy 5 o`

n51

450s0.775dn

r 5 77.5% 5 0.775

a1 5 450s0.775d 5 348.75

117.

Total area of shaded region is approximately

126 square inches.

64 1 32 1 16 1 8 1 4 1 2 5 126 118.

The total sales over the 10-year period is $2653.80 million.

5 2887.141484 2 233.336893 5 2653.80

5 64.8472111 2 es0.172ds13d

1 2 e0.172 2 2 64.8472111 2 es0.172ds3d

1 2 e0.172 2 S 5 S13 2 S3

Sn 5 on

i51

a1ri21 5 a111 2 r n

1 2 r 2a1 5 54.6e0.172 5 64.84721

r 5 e0.172n

an 5 54.6e0.172n, n 5 4, 5, . . . , 13

119.

T 5 o40

n51

30,000s1.05dn21 5 30,000s1 2 1.0540ds1 2 1.05d

< $3,623,993.23

an 5 30,000s1.05dn21

120. (a)

(b) t 5 1 1 2 o`

n51

s0.9dn 5 1 1 23 0.9

1 2 0.94 5 19 seconds

Total distance 5 3 o`

n50

32s0.81dn4 2 16 532

1 2 0.812 16 < 152.42 feet

121. False. A sequence is geometric if the

ratios of consecutive terms are the same.

122. False. NOT

The nth-term of a geometric sequence can be found by

multiplying its first term by its common ratio raised to

the power.sn 2 1dth

ra1n21an 5 a1r

n21,

123. Given a real number r between and 1,

as the exponent n increases, approaches zero.rn

21 124. Sample answer:

o199

n51

4s21dn21 and o8

n51

24

85s22dn21

125.

5 x2 1 2x 1 1 2 1 5 x2 1 2x

gsx 1 1d 5 sx 1 1d2 2 1

gsxd 5 x2 2 1 126.

f sx 1 1d 5 3sx 1 1d 1 1 5 3x 1 4

f sxd 5 3x 1 1


127.

5 3x2 1 6x 1 1

5 3sx2 1 2xd 1 1

f sgsx 1 1dd 5 f sx2 1 2xd

f sxd 5 3x 1 1, gsxd 5 x2 2 1 128.

From Exercise 126

5 9x2 1 24x 1 15

5 s3x 1 4d2 2 1

g s f sx 1 1dd 5 gs3x 1 4d

gsxd 5 x2 2 1

129. 9x3 2 64x 5 xs9x2 2 64d 5 xs3x 1 8ds3x 2 8d

131. 6x2 2 13x 2 5 5 s3x 1 1ds2x 2 5d

133.3

x 1 3?

xsx 1 3dx 2 3

53x

x 2 3, x Þ 23

135.x

34

3x

6x 1 35

x

3?

3s2x 1 1d3x

52x 1 1

3, x Þ 0, 2

1

2

130. Does not factorx2 1 4x 2 63

132.

5 4x2s2 1 xds2 2 xd

16x2 2 4x4 5 4x2s4 2 x2d

134. x Þ 27, 2x 2 2

x 1 7?

2x1sx 1 7d

6x3

sx 2 2d 51

3,

136. x Þ 3, 5x 2 5

x 2 34

10 2 2x

2s3 2 xd 5x 2 5

x 2 3?

22sx 2 3d22sx 2 5d 5 1,

137.

55x2 2 20 1 7x 2 14 1 2x 1 4

sx 1 2dsx 2 2d 55x2 1 9x 2 30

sx 1 2dsx 2 2d

55sx2 2 4d 1 7sx 2 2d 1 2sx 1 2d

sx 1 2dsx 2 2d

5 17

x 1 21

2

x 2 25

5sx 1 2dsx 2 2d 1 7sx 2 2d 1 2sx 1 2dsx 1 2dsx 2 2d

138.

57x2 1 21x 2 53

sx 2 1dsx 1 4d

58x2 1 24x 2 32 2 x2 1 2x 2 1 2 4x 2 16 2 x 2 4

sx 2 1dsx 1 4d

58sx2 1 3x 2 4d 2 sx2 2 2x 1 1d 2 4x 2 16 2 x 2 4

sx 2 1dsx 1 4d

58sx 2 1dsx 1 4d 2 sx 2 1d2 2 4sx 1 4d 2 sx 1 4d

sx 2 1dsx 1 4d8 2x 2 1

x 1 42

4

x 2 12

x 1 4

sx 2 1dsx 1 4d

Section 9.4 Mathematical Induction

n You should be sure that you understand the principle of mathematical induction. If is a statement involving

the positive integer n, where is true and the truth of implies the truth of for every positive k, then

is true for all positive integers n.

n You should be able to verify (by induction) the formulas for the sums of powers of integers and be able to use

these formulas.

n You should be able to calculate the first and second differences of a sequence.

n You should be able to find the quadratic model for a sequence, when it exists.

PnPk11PkP1

Pn

139. Answers will vary.

Section 9.4 Mathematical Induction 853

Vocabulary Check

1. mathematical induction 2. first

3. arithmetic 4. second

1.

Pk11 55

sk 1 1dfsk 1 1d 1 1g5

5

sk 1 1dsk 1 2d

Pk 55

ksk 1 1d

3.

Pk11 5sk 1 1d2fsk 1 1d 1 1g2

45

sk 1 1d2sk 1 2d2

4

Pk 5k2sk 1 1d2

4

5. 1. When

2. Assume that

Then,

Therefore, we conclude that the formula is valid for all positive integer values of n.

5 Sk 1 2sk 1 1d 5 ksk 1 1d 1 2sk 1 1d 5 sk 1 1dsk 1 2d.

Sk11 5 2 1 4 1 6 1 8 1 . . . 1 2k 1 2sk 1 1d

Sk 5 2 1 4 1 6 1 8 1 . . . 1 2k 5 ksk 1 1d.

S1 5 2 5 1s1 1 1d.n 5 1,

2.

Pk11 51

2sk 1 1 1 2d 51

2sk 1 3d

Pk 51

2sk 1 2d

4.

Pk11 5k 1 1

3f2sk 1 1d 1 1g 5

k 1 1

3s2k 1 3d

Pk 5k

3s2k 1 1d

6. 1. When

2. Assume that

Then,

Therefore, we conclude that this formula is valid.

5 sk 1 1df2sk 1 1d 1 1g .

5 sk 1 1ds2k 1 3d

5 2k 2 1 5k 1 3

5 ks2k 1 1d 1 s4k 1 3d

Sk11 5 Sk 1 ak115s3 1 7 1 11 1 15 1 . . . 1 s4k 2 1dd 1 f4sk 1 1d 2 1g

Sk 5 3 1 7 1 11 1 15 1 . . . 1 s4k 2 1d 5 ks2k 1 1d.

S1 5 3 5 1s2 ? 1 1 1d.n 5 1,


7. 1. When

2. Assume that

Then,

Therefore, we conclude that this formula is valid for all positive integer values of n.

5sk 1 1ds5k 1 4d

25

sk 1 1d2

f5sk 1 1d 2 1g.

55k2 2 k 1 10k 1 4

25

5k2 1 9k 1 4

2

5 Sk 1 s5k 1 5 2 3d 5k

2s5k 2 1d 1 5k 1 2

Sk11 5 2 1 7 1 12 1 17 1 . . . 1 s5k 2 3d 1 f5sk 1 1d 2 3g

Sk 5 2 1 7 1 12 1 17 1 . . . 1 s5k 2 3d 5k

2s5k 2 1d.

S1 5 2 51

2s5s1d 2 1d.n 5 1,

8. 1. When

2. Assume that

Then,


5k 1 1

2f3sk 1 1d 2 1g.

5sk 1 1ds3k 1 2d

2

53k2 1 5k 1 2

2

53k2 2 k 1 6k 1 2

2

5k

2s3k 2 1d 1 s3k 1 1d

Sk11 5 Sk 1 ak11 5 s1 1 4 1 7 1 10 1 . . . 1 s3k 2 2dd 1 s3sk 1 1d 2 2d

Sk 5 1 1 4 1 7 1 10 1 . . . 1 s3k 2 2d 5k

2s3k 2 1d.

S1 5 1 51

2s3 ? 1 2 1d.

n 5 1,

9. 1. When

2. Assume that

Then,


5 Sk 1 2k 5 2k 2 1 1 2k 5 2s2kd 2 1 5 2k11 2 1.

Sk11 5 1 1 2 1 22 1 23 1 . . . 1 2k21 1 2k

Sk 5 1 1 2 1 22 1 23 1 . . . 1 2k2152k 2 1.

S1 5 1 5 21 2 1.n 5 1,


10. 1. When

2. Assume that

Then,


5 3k11 2 1.

5 3 ? 3k 2 1

5 3k 2 1 1 2 ? 3k

5 f2s1 1 3 1 32 1 33 1 . . . 1 3k21dg 1 2 ? 3k1121

Sk11 5 Sk 1 ak11

Sk 5 2s1 1 3 1 32 1 33 1 . . . 1 3k21d 5 3k 2 1.

n 5 1, S1 5 2 5 31 2 1.

11. 1. When

2. Assume that

Then,


5 Sk 1 sk 1 1d 5ksk 1 1d

21

2sk 1 1d2

5sk 1 1dsk 1 2d

2.

Sk11 5 1 1 2 1 3 1 4 1 . . . 1 k 1 sk 1 1d

Sk 5 1 1 2 1 3 1 4 1 . . . 1 k 5ksk 1 1d

2.

n 5 1, S1 5 1 51s1 1 1d

2.

12.

1. When

2. Assume that

Then,


5sk 1 1d2sk2 1 4k 1 4d

45

sk 1 1d2sk 1 2d2

45

sk 1 1d2fsk 1 1d 1 1g2

4

5 Sk 1 sk 1 1d3 5k2sk 1 1d2

41 sk 1 1d3 5

k2sk 1 1d2 1 4sk 1 1d3

4

Sk11 5 13 1 23 1 33 1 43 1 . . . 1 k3 1 sk 1 1d3

Sk 5 13 1 23 1 33 1 43 1 . . . 1 k3 5k2sk 1 1d2

4.

n 5 1, Sn 5 13 5 1 512s1 1 1d2

4.

Sn 5 13 1 23 1 33 1 43 1 . . . 1 n3 5n2sn 1 1d2

4.


13. 1. When

2. Assume that

Then,


Note: The easiest way to complete the last two steps is to “work backwards.” Start with the desired expression

for and multiply out to show that it is equal to the expression you found for Sk 1 sk 1 1d5.Sk11

5sk 1 1d2sk 1 2d2f2sk 1 1d2 1 2sk 1 1d 2 1g

12.

5sk 1 1d2sk2 1 4k 1 4ds2k2 1 6k 1 3d

12

5sk 1 1d2f2k4 1 14k3 1 35k2 1 36k 1 12g

12

5sk 1 1d2f2k4 1 2k3 2 k2 1 12sk3 1 3k2 1 3k 1 1dg

12

5sk 1 1d2fk2s2k2 1 2k 2 1d 1 12sk 1 1d3g

12

5k2sk 1 1d2s2k2 1 2k 2 1d

121

12sk 1 1d5

12

Sk11 5 ok11

i51

i5 5 1ok

i51

i52 1 sk 1 1d5

Sk 5 ok

i51

i5 5k2sk 1 1d2s2k2 1 2k 2 1d

12.

n 5 1, S1 5 1 5s1d2s1 1 1d2s2s1d2 1 2s1d 2 1d

12.

14. 1. When

2. Assume that

Then,


5sk 1 1dsk 1 2ds2sk 1 1d 1 1ds3sk 1 1d2 1 3sk 1 1d 2 1d

30.

5sk 1 1dsk 1 2ds2k 1 3ds3k2 1 9k 1 5d

30

5sk 1 1ds6k4 1 39k3 1 91k2 1 89k 1 30d

30

5sk 1 1dfks2k 1 1ds3k2 1 3k 2 1d 1 30sk 1 1d3g

30

5ksk 1 1ds2k 1 1ds3k2 1 3k 2 1d 1 30sk 1 1d4

30

5ksk 1 1ds2k 1 1ds3k2 1 3k 2 1d

301 sk 1 1d4

Sk11 5 Sk 1 ak11 5 Sk 1 sk 1 1d4

Sk 5 ok

i51

i4 5ksk 1 1ds2k 1 1ds3k2 1 3k 2 1d

30.

S1 5 14 51s1 1 1ds2 ? 1 1 1ds3 ? 12 1 3 ? 1 2 1d

30.

n 5 1,


15. 1. When

2. Assume that

Then,


5sk 1 1dsk 1 2dsk 1 3d

3.

5 Sk 1 sk 1 1dsk 1 2d 5ksk 1 1dsk 1 2d

31

3sk 1 1dsk 1 2d3

Sk11 5 1s2d 1 2s3d 1 3s4d 1 . . . 1 ksk 1 1d 1 sk 1 1dsk 1 2d

Sk 5 1s2d 1 2s3d 1 3s4d 1 . . . 1 ksk 1 1d 5ksk 1 1dsk 1 2d

3.

n 5 1, S1 5 2 51s2ds3d

3.

16. 1. When

2. Assume that

Then,


5k 1 1

2sk 1 1d 1 1.

5s2k 1 1dsk 1 1d

s2k 1 1ds2k 1 3d

52k2 1 3k 1 1

s2k 1 1ds2k 1 3d

5ks2k 1 3d 1 1

s2k 1 1ds2k 1 3d

5k

2k 1 11

1

s2k 1 1ds2k 1 3d

Sk11 5 Sk 1 ak11 5 Sk 11

s2sk 1 1d 2 1ds2sk 1 1d 1 1d

Sk 5 ok

i51

1

s2i 2 1ds2i 1 1d5

k

2k 1 1.

S1 51

35

1

2 ? 1 1 1.

n 5 1,

17. 1. When thus

2. Assume

Then,

since and

Thus,

Therefore, by extended mathematical induction, the inequality is valid for all integers n such that n ≥ 4.

sk 1 1d! > 2k11.

k 1 1 > 2.k! > 2ksk 1 1d! 5 k!sk 1 1d > 2ks2d

k! > 2k, k > 4.

4! > 24.n 5 4, 4! 5 24 and 24 5 16,


18. 1. When

2. Assume that

Then,

Thus,

Therefore, the inequality is valid for all integers n such that n ≥ 7.14

32n

> n

14

32k11

> k 1 1.

14

32k11

5 14

32k14

32 > k14

32 5 k 1k

3> k 1 1 for k > 7.

14

32k

> k, k > 7.

14

327

< 7.4915 > 7.n 5 7,

19. 1. When thus

2. Assume that

Then,

Now it is sufficient to show that

or equivalently

This is true because

Therefore,

Therefore, by extended mathematical induction, the inequality is valid for all integers n such that n ≥ 2.

1

!11

1

!21

1

!31 . . . 1

1

!k1

1

!k 1 1> !k 1 1.

!k !k 1 1 1 1 > !k !k 1 1 5 k 1 1.

!k !k 1 1 1 1 > k 1 1.

smultiplying by !k 1 1d,

!k 11

!k 1 1> !k 1 1, k > 2,

1

!11

1

!21

1

!31 . . . 1

1

!k 1

1

!k 1 1> !k 1

1

!k 1 1.

1

!11

1

!21

1

!31 . . . 1

1

!k> !k, k > 2.

1

!11

1

!2> !2.n 5 2,

1

!11

1

!2< 1.707 and !2 < 1.414,

20. 1. When

2. Assume that

Therefore, for all integers n ≥ 1.1x

y2n11

< 1x

y2n

1x

y2k11

< 1x

y2k ⇒ 1x

y21x

y2k11

< 1x

y21x

y2k

⇒ 1x

y2k12

< 1x

y2k11

.

1x

y2k11

< 1x

y2k

1x

y22

< 1x

y2 and s0 < x < yd.n 5 1,

21. and

Since a is positive, then all of the terms in the binomial expansion are positive.

s1 1 adn 5 1 1 na 1 . . . 1 nan21 1 an> na

a > 0n ≥ 1s1 1 adn≥ na,


22.

1. For the statement is true, because

2. Assuming that you need to show that

For you have

By the assumption you have

Because or for all

you can say that

It follows that

or

Therefore, for all n ≥ 3.2n2sn 1 1d2

2sk 1 1d2> sk 1 2d2.

sk 1 2d2< 2k2 1 2k 1 3 < 2sk 1 1d2

2k2 1 2k 1 3 < 2k2 1 4k 1 2 5 2sk 1 1d2.

k > 3,1 < 2k2k 1 3 < 4k 1 2,

sk 1 1d2 1 2k 1 3 < 2k2 1 2k 1 3.

sk 1 1d2< 2k2,

5 sk 1 1d2 1 2k 1 3.

5 k2 1 2k 1 1 1 2k 1 3

sk 1 2d2 5 k2 1 4k 1 4

n 5 k,2sk 1 1d2> sk 1 2d2.

2k2> sk 1 1d2

2s3d2 5 18 > s3 1 1d2 5 16.

n 5 3,

2n2> sn 1 1d2, n ≥ 3 23. 1. When

2. Assume that

Then,

Thus, sabdn 5 anbn.

5 ak11bk11.

5 akbkab

sabdk11 5 sabdksabd

sabdk 5 akbk.

n 5 1, sabd1 5 a1b1 5 ab.

24. 1. When

2. Assume that

Thus, 1a

b2n

5an

bn.

5ak11

bk11.5

ak

bk?

a

bThen, 1a

b2k11

5 1a

b2k1a

b2

1a

b2k

5ak

bk.

1a

b21

5a1

b1.n 5 1, 25. 1. When

2. Assume that

Then,

Thus, the formula is valid.

5 x121x2

21x321 . . . xk

21xk1121.

5 sx1 x2 x3 . . . xkd21xk1121

sx1 x2 x3 . . . xk xk11d21 5 fsx1 x2 x3

. . . xkdxk11g21

sx1 x2 x3 . . . xkd21 5 x1

21x221x3

21 . . . xk21.

n 5 2, sx1x2d21 51

x1x2

51

x1

?1

x2

5 x121x2

21.

26. 1. When

2. Assume that

Thus, lnsx1x2 x3 . . . xn d 5 ln x1 1 ln x2 1 ln x3 1 . . . ln xn .

5 ln x1 1 ln x2 1 ln x3 1 . . . 1 ln xk 1 ln xk11.

5 lnsx1x2 x3 . . . xkd 1 ln xk11

Then, lnsx1 x2 x3 . . . xk xk11d 5 lnfsx1x2x3 . . . xkdxk11g

lnsx1 x2 x3 . . . xk d 5 ln x1 1 ln x2 1 ln x3 1 . . . 1 ln xk .

ln x1 5 ln x1.n 5 1,

27. 1. When

2. Assume that

Then,

Hence, the formula holds.

5 xsy1 1 y2 1 . . . 1 yk 1 yk11d.

5 xfsy1 1 y2 1 . . . 1 ykd 1 yk11g

xy1 1 xy2 1 . . . 1 xyk 1 xyk11 5 xsy1 1 y2 1 . . . 1 ykd 1 xyk11

xsy1 1 y2 1 . . . 1 ykd 5 xy1 1 xy2 1 . . . 1 xyk.

n 5 1, xsy1d 5 xy1.


28. 1. When and are complex conjugates by definition.

2. Assume that and are complex conjugates. That is, if then

Then,

This implies that and are complex conjugates.

Therefore, and are complex conjugates for n ≥ 1.sa 2 bidnsa 1 bidn

sa 2 bidk11sa 1 bidk11

5 sac 2 bd d 2 isbc 1 ad d.

and sa 2 bidk11 5 sa 2 bidk sa 2 bid 5 sc 2 didsa 2 bid

5 sac 2 bd d 1 isbc 1 ad d

sa 1 bidk11 5 sa 1 bidksa 1 bid 5 sc 1 didsa 1 bid

sa 2 bidk 5 c 2 di .sa 1 bidk 5 c 1 di,sa 2 bidksa 1 bidk

a 2 bia 1 bin 5 1,

29. 1. When is a factor.

2. Assume that 3 is a factor of

Then,

Since 3 is a factor of our assumption, and 3 is a factor of

we conclude that 3 is a factor of the whole sum.

Thus, 3 is a factor of for every positive integer n.sn3 1 3n2 1 2nd

3sk2 1 3k 1 2d,sk3 1 3k2 1 2kd,

5 sk3 1 3k2 1 2kd 1 3sk2 1 3k 1 2d.

5 sk3 1 3k2 1 2kd 1 s3k2 1 9k 1 6d

sk 1 1d3 1 3sk 1 1d2 1 2sk 1 1d 5 k3 1 3k2 1 3k 1 1 1 3k2 1 6k 1 3 1 2k 1 2

k3 1 3k2 1 2k.

n 5 1, f13 1 3s1d2 1 2s1dg 5 6 and 3

30. Prove 3 is a factor of for all positive integers n.

1. When , and 3 is a factor.


Then,

Since 3 is a factor of each term, 3 is a factor of the sum.

Thus, 3 is a factor of for all positive integers n.n3 2 n 1 3

5 sk3 2 k 1 3d 1 3ksk 1 1d.

5 sk3 2 k 1 3d 1 3k2 1 3k

5 k3 1 3k2 1 2k 1 3

sk 1 1d3 2 sk 1 1d 1 3 5 k3 1 3k2 1 3k 1 1 2 k 2 1 1 3

k3 2 k 1 3.

13 2 1 1 3 5 3n 5 1

n3 2 n 1 3

31. A factor of is 2.



Then,

Since 2 is a factor of our assumption, and 2 is a factor of

we conclude that 2 is a factor of the entire expression.

Thus, 2 is a factor of for every positive integer n.n4 2 n 1 4

2s2k3 1 3k2 1 2kd,k4 2 k 1 4,

5 sk4 2 k 1 4d 1 2s2k3 1 3k2 1 2kd.

5 sk4 2 k 1 4d 1 s4k3 1 6k2 1 4kd

sk 1 1d4 2 sk 1 1d 1 4 5 k4 1 4k3 1 6k2 1 4k 1 1 2 k 2 1 1 4

k4 2 k 1 4.

14 2 1 1 4 5 4n 5 1

n4 2 n 1 4


32. Prove 3 is a factor of for all positive integers n.


2. Assume 3 is a factor of

Then,

Since 3 is a factor of each term, 3 is a factor of the sum.

Thus, 3 is a factor of for all positive integers n.22n11 1 1

5 4s22k11 1 1d 2 3

5 4 ? 22k11 1 1

5 22k11 ? 2211

5 2s2k11d12 1 1

22sk11d11 1 1 5 22k1211 1 1

22k11 1 1.

22?111 1 1 5 23 1 1 5 8 1 1 5 9n 5 1

22n11 1 1

33. A factor of is 5.

1. When ,

and 5 is a factor.


Then,

Since 5 is a factor of our assumption, and 5 is a factor of we conclude that

5 is a factor of the entire expression.

Thus, 5 is a factor of for every positive integer n.24n22 1 1

15 ? 24k22,24k22 1 1,

5 s24k22 1 1d 1 15 ? 24k22.

5 24k22 ? 16 1 1

5 24k22 ? 24 1 1

24sk11d22 1 1 5 24k1422 1 1

24k22 1 1.

24s1d22 1 1 5 5

n 5 1

24n22 1 1

34. 1. When and 5 is a factor.


Since 5 is a factor of each set in parentheses and 5 is a factor of then 5 is a factor of the whole sum.

Thus, 5 is a factor of for every positive integer n.s22n21 1 32n21d

5 ? 32k21,

1 s22k21 1 32k21d 1 s22k21 1 32k21d 1 5 ? 32k21.

5 s22k21 1 32k21d 1 s22k21 1 32k21d

5 4 ? 22k21 1 9 ? 32k21

5 22k2122 1 32k2132

Then, 22sk11d21 1 32sk11d21 5 22k1221 1 32k1221

s22k21 1 32k21d.

s22s1d21 1 32s1d21d 5 2 1 3 5 5n 5 1,


35.

From this sequence, it appears that This can be verified by mathematical induction.

The formula has already been verified for Assume that the formula is valid for Then,


5 sk 1 1df2sk 1 1d 2 1g.

5 sk 1 1ds2k 1 1d

5 2k2 1 3k 1 1

5 ks2k 2 1d 1 s4k 1 1d

Sk11 5 f1 1 5 1 9 1 13 1 . . . 1 s4k 2 3dg 1 f4sk 1 1d 2 3g

n 5 k.n 5 1.

Sn 5 ns2n 2 1d.

S4 5 1 1 5 1 9 1 13 5 28 5 4 ? 7

S3 5 1 1 5 1 9 5 15 5 3 ? 5

S2 5 1 1 5 5 6 5 2 ? 3

S1 5 1 5 1 ? 1

Sn 5 1 1 5 1 9 1 13 1 . . . 1 s4n 2 3d

36.

From the sequence, it appears that

This can be verified by mathematical induction. The formula has already been verified for

Assume that the formula is valid for Then,


5k 1 1

2f23sk 1 1d 1 53g.

5 21

2sk 1 1ds3k 2 50d

5 21

2s3k2 2 47k 2 50d

51

2s23k2 1 47k 1 50d

5k

2s23k 1 53d 1 s23k 1 25d

Sk11 5 f25 1 22 1 19 1 16 1 . . . 1 s23k 1 28dg 1 f23sk 1 1d 1 28g

n 5 k.

n 5 1.

Sn 5n

2s23n 1 53d.

S4 5 25 1 22 1 19 1 16 5 82 54

2s41d

S3 5 25 1 22 1 19 5 66 53

2s44d

S2 5 25 1 22 5 47 52

2s47d

S1 5 25 51

2s50d

Sn 5 25 1 22 1 19 1 16 1 . . . 1 s23n 1 28d


37.

Since this series is geometric, we have

5 10 2 101 9

102n

.

Sn 5 on

i511 9

102i21

5

1 2 1 9

102n

1 29

10

5 1031 2 1 9

102n

4

Sn 5 1 19

101

81

1001

729

10001 . . . 1 1 9

102n21

38.

Since the series is geometric, we have

Sn 5 on

i51

3123

22i21

5 331 2 123

22n

1 2 123

22 4 56

531 2 123

22n

4.

Sn 5 3 29

21

27

42

81

81 . . . 1 312

3

22n21

39.

From this sequence, it appears that




5k 1 1

2sk 1 2d.

5sk 1 1d2

2sk 1 1dsk 1 2d

5k2 1 2k 1 1

2sk 1 1dsk 1 2d

5ksk 1 2d 1 1

2sk 1 1dsk 1 2d

5k

2sk 1 1d 11

2sk 1 1dsk 1 2d

Sk11 5 31

41

1

121

1

401 . . . 1

1

2ksk 1 1d4 11

2sk 1 1dsk 1 2d

n 5 k.

n 5 1.

Sn 5n

2sn 1 1d.

S4 51

41

1

121

1

241

1

405

16

405

4

105

4

2s5d

S3 51

41

1

121

1

245

9

245

3

85

3

2s4d

S2 51

41

1

125

4

125

2

65

2

2s3d

S1 51

45

1

2s2d

Sn 51

41

1

121

1

241

1

401 . . . 1

1

2nsn 1 1d


40.

From this sequence, it appears that




5k 1 1

2fsk 1 1d 1 2g.

5sk 1 1dsk 1 2d

2sk 1 2dsk 1 3d

5k2 1 3k 1 2

2sk 1 2dsk 1 3d

5ksk 1 3d 1 2

2sk 1 2dsk 1 3d

5k

2sk 1 2d 11

sk 1 2dsk 1 3d

Sk11 5 31

61

1

121

1

201

1

301 . . . 1

1

sk 1 1dsk 1 2d4 11

sk 1 2dsk 1 3d

n 5 k.

n 5 1.

Sn 5n

2sn 1 2d.

S4 51

61

1

121

1

201

1

305

1

35

4

2 ? 6

S3 51

61

1

121

1

205

3

105

3

2 ? 5

S2 51

61

1

125

1

45

2

2 ? 4

S1 51

65

1

2 ? 3

Sn 51

2 ? 31

1

3 ? 41

1

4 ? 51

1

5 ? 61 . . . 1

1

sn 1 1dsn 1 2d

41. o15

n51

n 515s15 1 1d

25 120

45. o5

n51

n4 55s5 1 1df2s5d 1 1gf3s5d2 1 3s5d 2 1g

305 979

43. o6

n51

n2 56s6 1 1df2s6d 1 1g

65 91

42. o30

n51

n 530s30 1 1d

25 465

44. o10

n51

n3 5102s10 1 1d2

45 3025

46. o8

n51

n5 582s8 1 1d2s2s8d2 1 2s8d 2 1d

125 61,776

47.

5 91 2 21 5 70

56s6 1 1df2s6d 1 1g

62

6s6 1 1d2

o6

n51

sn2 2 nd 5 o6

n51

n2 2 o6

n51

n 48.

5s20d2s21d2 2 2s20ds21d

45 43,890

5s20d2s20 1 1d2

42

20s20 1 1d2

o20

n51

sn3 2 nd 5 o20

n51

n3 2 o20

n51

n

49. 5 636s6 1 1d2 4 2 83s6d2s6 1 1d2

4 4 5 6s21d 2 8s441d 5 23402o6

i51

s6i 2 8i3d 5 6o6

i51

i 2 8o6

i51

i3


50.

53s10ds12d 2 3s10ds11d 1 10s11ds21d

125 195

5 3s10d 21

2?

10s10 1 1d2

11

2?

10s10 1 1ds2 ? 10 1 1d6

o10

j5113 2

1

2j 1

1

2j22 5 o

10

j51

3 21

2o10

j51

j 11

2o10

j51

j2

51.

0 3 6 9 12 15

First differences: 3 3 3 3 3

Second differences: 0 0 0 0

Since the first differences are equal, the sequence has a

linear model.

an:

a6 5 a5 1 3 5 12 1 3 5 15

a5 5 a4 1 3 5 9 1 3 5 12

a4 5 a3 1 3 5 6 1 3 5 9

a3 5 a2 1 3 5 3 1 3 5 6

a2 5 a1 1 3 5 0 1 3 5 3

a1 5 a1 5 0

a1 5 0, an 5 an21 1 3 52.

2 4 6 8 10 12



Since the first differences are equal, the sequence has a

linear model.

an:

a6 5 a5 1 2 5 10 1 2 5 12

a5 5 a4 1 2 5 8 1 2 5 10

a4 5 a3 1 2 5 6 1 2 5 8

a3 5 a2 1 2 5 4 1 2 5 6

a2 5 a1 1 2 5 2 1 2 5 4

a1 5 a1 5 2

a1 5 2, an 5 an21 1 2

53.

3 1 22 26 211 217



Since the second differences are all the same, the sequence has a quadratic model.

an:

a6 5 a5 2 6 5 211 2 6 5 217

a5 5 a4 2 5 5 26 2 5 5 211

a4 5 a3 2 4 5 22 2 4 5 26

a3 5 a2 2 3 5 1 2 3 5 22

a2 5 a1 2 2 5 3 2 2 5 1

a1 5 a1 5 3

a1 5 3, an 5 an21 2 n

54.

a7 5 22a6 5 22s248d 5 96

a6 5 22a5 5 22s24d 5 248

a5 5 22a4 5 22s212d 5 24

a4 5 22a3 5 22s6d 5 212

a3 5 22a2 5 22s23d 5 6

a2 5 23

a1 532

a2 5 23 ⇒ 23 5 22a1

a2 5 23, an 5 22an21

First differences:

Second differences:

Since neither the first differences nor the second differences are equal, the

sequence does not have a linear or quadratic model.

an : 23 6 212 24 248 96

9 218 36 272 144

227 54 2108 216272

292

32


55.

First differences:

Second differences:

Since neither the first differences nor the second differences are equal, the sequence does not have a linear or quadratic model.

an :

a5 5 a42 5 65,5362 5 4,294,967,296

a4 5 a32 5 2562 5 65,536

a3 5 a22 5 162 5 256

a2 5 a12 5 42 5 16

a1 5 a02 5 22 5 4

a0 5 2

a0 5 2, an 5 san21d2

2 4 16 256 65,536 4,294,967,296

2 12 240 65,280 4,294,901,760

10 228 65,040 4,294,836,480

56.

0 1 3 6 10 15



Since the second differences are equal, the sequence has a

quadratic model.

an:

a5 5 a4 1 5 5 10 1 5 5 15

a4 5 a3 1 4 5 6 1 4 5 10

a3 5 a2 1 3 5 3 1 3 5 6

a2 5 a1 1 2 5 1 1 2 5 3

a1 5 a0 1 1 5 0 1 1 5 1

a0 5 0

a0 5 0, an 5 an21 1 n

58.

Let Then:

Thus, an 5 n2 2 2n 1 7.

2a

3a

2

1

b

b

2a

a

5

5

5

5

1

1

2

1

⇒

b 5 22

By elimination:

⇒ 9a

9a

3a

1

1

1

3b

3b

b

1 c 5

5

5

10

3

1

a3 5 as3d2 1 bs3d 1 c 5 10

⇒

⇒

a

a

1

1

b

b

1

c

c

5

5

5

7

6

21

a0

a1

5

5

as0d2

as1d2

1

1

bs0dbs1d

1

1

c

c

5

5

7

6

an 5 an2 1 bn 1 c.

a0 5 7, a1 5 6, a3 5 10

57.

Let

⇒ a 1 b 5 0

a1 5 as1d2 1 bs1d 1 c 5 13 ⇒ a 1 b 1 c 5 3

Thus: a0 5 as0d2 1 bs0d 1 c 5 13 ⇒ c 5 3

an 5 an2 1 bn 1 c.

a0 5 3, a1 5 3, a4 5 15

Thus, an 5 n2 2 n 1 3.

a 5 1 ⇒ b 5 21

3a 5 3

4a 1 b 5 3

By elimination: 2a 2 b 5 0

⇒ 4a 1 4b 5 3

⇒ 16a 1 4b 5 12

a4 5 as4d2 1 bs4d 1 c 5 15 ⇒ 16a 1 4b 1 c 5 15

59.

Let

By elimination:

Thus, an 512n2 1 n 2 3.

a 512 ⇒ b 5 1

2a 5 1

4a 1 b 5 3

22a 2 b 5 22

⇒ 4a 1 4b 5 3

⇒ 16a 1 4b 5 12

a4 5 as4d2 1 bs4d 1 c 5 29 ⇒ 16a 1 4b 1 c 5 9

⇒ 2a 1 2b 5 2

⇒ 4a 1 2b 5 4

a2 5 as2d2 1 bs2d 1 c 5 21 ⇒ 4a 1 2b 1 c 5 1

Then: a0 5 as0d2 1 bs0d 1 c 5 23 ⇒ c 5 23

an 5 an2 1 bn 1 c.

a0 5 23, a2 5 1, a4 5 9


60.

Let Then:

Thus, an 574n2 2 5n 1 3.

⇒ b 5 25

24a

12a

8a

2

1

2b

2b

a

5

5

5

5

3

11

1474

By elimination:

⇒ 36a

36a

12a

1

1

1

6b

6b

2b

1 c 5

5

5

36

33

11

a6 5 a(6d2 1 bs6d 1 c 5 36

⇒

⇒

4a

4a

1

1

2b

2b

1

c

c

5

5

5

3

0

23

a0

a2

5

5

as0d2

as2d2

1

1

bs0dbs2d

1

1

c

c

5

5

3

0

an 5 an2 1 bn 1 c.

a0 5 3, a2 5 0, a6 5 36

61. (a) 120.3 122.5 124.9 127.1 129.4 130.3

First differences: 2.2 2.4 2.2 2.3 0.9

(b) The first differences are not equal, but are fairly close to each other, so a linear model can be used.

If we let then

(c) is obtained by using the regression feature of a graphing utility.

(d) For 2008, let

These are very similar.

an 5 2.08s18d 1 103.9 5 141.34

an < 2.2s18d 1 102.7 5 142.3

n 5 18.

an < 2.08n 1 103.9

an < 2.2n 1 102.7

b 5 120.3 2 2.2s8d 5 102.7m 5 2.2,

62. Answers will vary. See page 626. 63. True. may be false.P7 64. False. must be proven to be true.P1

65. True. If the second differences are all zero, then the first

differences are all the same, so the sequence is arithmetic.

66. False. It has second differences.n 2 2

67. s2x2 2 1d2 5 s2x2 2 1ds2x2 2 1d 5 4x 4 2 4x2 1 1 68. s2x 2 yd2 5 4x2 2 4xy 1 y2

69. s5 2 4xd3 5 264x3 1 240x2 2 300x 1 125 70. s2x 2 4yd3 5 8x3 2 48x2y 1 96xy2 2 64y3

71.

(a) Domain: All real numbers x except

(b) Intercept:

(c) Vertical asymptote:

Horizontal asymptote:

(d)

y 5 1

x 5 23

s0, 0d

x 5 23

f sxd 5x

x 1 3

y

x

−12−10 −8 −6 −4

−6

−4

2

4

6

8

10

2 4

(0, 0)

x 1

4142

1222

52f sxd

21222425


72.


(b) Intercept:

(c) Vertical asymptotes:


(d)

y 5 1

x 5 22, x 5 2

s0, 0d

x 5 ±2

y

x−8 −6 −4

2

4

6

8

4 6 8

(0, 0)

gsxd 5x2

x2 2 4

x 0 1.5 3 4

043

952

972

97

95

43gsxd

21.52324

74.


(b) x-intercept:

y-intercept:



(d)

y

x

−2

−4

−6

−8

4

6

8

2 4 6 8(−5, 0)

(0, 5)

y 5 21

x 5 1

s0, 5d

s25, 0d

x 5 1

f sxd 55 1 x

1 2 x73.

(a) Domain: All real numbers t except

(b) Intercept:



(d)

y

t−8 −6 −4 −2

−8

−6

−4

2

4

62 8

(7, 0)

y 5 1

t 5 0

s7, 0d

t 5 0

h std 5t 2 7

t

t 1 2 3

8 2432

5226

92hstd

2122

x 0 2 3 5 7

0 1 5 2225224272

13f sxd

222528

Section 9.5 The Binomial Theorem

Vocabulary Check

1. binomial coefficients 2. Binomial Theorem/Pascal’s Triangle

3. or 4. expanding a binomialnCr1n

r2

n You should be able to use the formula

where to expand Also,

n You should be able to use Pascal’s Triangle in binomial expansion.

nCr 5 1n

r2.sx 1 ydn.nCr 5n!

sn 2 rd!r!,

sx 1 ydn 5 xn 1 nxn21y 1nsn 2 1d

2!xn22y2 1 . . . 1 nCr xn2ryr 1 . . . 1 yn

Section 9.5 The Binomial Theorem 869

5. 20C15 520!

15!5!5

20 ? 19 ? 18 ? 17 ? 16

5 ? 4 ? 3 ? 2 ? 15 15,504

1. 5C3 55!

3!2!5

5 ? 4

2 ? 15 10 2. 8C6 5

8!

6! ? 2!5

8 ? 7

2 ? 15 28 3. 12C0 5

12!

0!12!5 1

4. 20C20 520!

20! ? 0!5 1

6. 12C5 512!

5! ? 7!5

s12 ? 11 ? 10 ? 9 ? 8d ? 7!

5!7!5

12 ? 11 ? 10 ? 9 ? 8

5 ? 4 ? 3 ? 2 ? 15 792

7. 110

42 510!

6!4!5

10 ? 9 ? 8 ? 7 ? 6!

6!s24d 5 210

8. 110

6 2 510!

6! ? 4!5

s10 ? 9 ? 8 ? 7d ? 6!

6! ? 4!5

10 ? 9 ? 8 ? 7

4 ? 3 ? 2 ? 15 210

9. 1100

982 5100!

2!98!5

100 ? 99

2 ? 15 4950 10.

5 4950

1100

2 2 5100!

98! ? 2!5

s100 ? 99d ? 98!

98! ? 2!5

100 ? 99

2 ? 1

11.

the 6th entry in the 8th row.18

52 5 56,

1

1

8

1

1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

1 6 15 20 15 6 1

1

1

7

8

21

28

35

56

35

5670

21

28

7 1

13.

the 5th entry in the 7th row.7C4 5 35,

1

11

1

1

1 1

15101051

1 6 15 20 15 6 1

172135352171

464

3 3 1

12

12.

the 8th entry in the 8th row.18

72 5 8,

1

11

1

1

1 1

15101051

1 6 15 20 15 6 1

1

18285670562881

72135352171

464

3 3 1

12

14.

the 4th entry in the 6th row.6C3 5 20,

1

11

1

1

1 1

15101051

1 6 15 20 15 6 1

464

3 3 1

12

15.

5 x4 1 4x3 1 6x2 1 4x 1 1

sx 1 1d4 5 4C0x4 1 4C1x

3s1d 1 4C2x2s1d2 1 4C3xs1d3 1 4C4s1d4

16.

5 x6 1 6x5 1 15x 4 1 20x3 1 15x2 1 6x 1 1

sx 1 1d6 5 6C0 x6 1 6C1x5s1d 1 6C2x

4s1d2 1 6C3x3s1d3 1 6C4x

2s1d4 1 6C5xs1d5 1 6C6s1d6

17.

5 a4 1 24a3 1 216a2 1 864a 1 1296

5 1a4 1 4a3s6d 1 6a2s6d2 1 4as6d3 1 1s6d4

sa 1 6d4 5 4C0a4 1 4C1a

3s6d 1 4C2a2s6d2 1 4C3as6d3 1 4C4s6d4


18.

5 a5 1 25a4 1 250a3 1 1250a2 1 3125a 1 3125

sa 1 5d5 5 5C0 a5 1 5C1a4s5d 1 5C2a

3s5d2 1 5C3a2s5d3 1 5C4as5d4 1 5C5s5d5

19.

5 y3 2 12y2 1 48y 2 64

5 1y3 2 3y2s4d 1 3ys4d2 2 1s4d3

sy 2 4d3 5 3C0y3 2 3C1y

2s4d 1 3C2ys4d2 2 3C3s4d3

20.

5 y5 2 10y4 1 40y3 2 80y2 1 80y 2 32

s y 2 2d5 5 5C0 y5 2 5C1y4s2d 1 5C2 y3s2d2 2 5C3y

2s2d3 1 5C4ys2d4 2 5C5s2d5

21.

5 x5 1 5x4y 1 10x3y2 1 10x2y3 1 5xy4 1 y5

sx 1 yd5 5 5C0x5 1 5C1x

4y 1 5C2x3y2 1 5C3x

2y3 1 5C4xy4 1 5C5y5

22.

5 c3 1 3c2d 1 3cd2 1 d3

sc 1 dd3 5 3C0c3 1 3C1c

2d 1 3C2cd2 1 3C3d3

23.

5 r6 1 18r5s 1 135r4s2 1 540r3s3 1 1215r2s4 1 1458rs5 1 729s6

5 1r6 1 6r5s3sd 1 15r4s3sd2 1 20r3s3sd3 1 15r2s3sd4 1 6rs3sd5 1 1s3sd6

sr 1 3sd6 5 6C0r6 1 6C1r

5s3sd 1 6C2r4s3sd2 1 6C3r

3s3sd3 1 6C4r2s3sd4 1 6C5rs3sd5 1 6C6s3sd6

24.

5 x4 1 8x3y 1 24x2y2 1 32xy3 1 16y4

5 x4 1 4x3s2yd 1 6x2s4y2d 1 4xs8y3d 1 16y4

sx 1 2yd4 5 4C0 x4 1 4C1x3s2yd 1 4C2x

2s2yd2 1 4C3xs2yd3 1 4C4s2yd4

25.

5 243a5 2 1620a4b 1 4320a3b2 2 5760a2b3 1 3840ab4 2 1024b5

5 s1ds243a5d 2 5s81a4ds4bd 1 10s27a3ds16b2d 2 10s9a2ds64b3d 1 5s3ads256b4d 2 s1ds1024b5d

s3a 2 4bd5 5 5C0s3ad5 2 5C1s3ad4s4bd 1 5C2s3ad3s4bd2 2 5C3s3ad2s4bd3 1 5C4s3ads4bd4 2 5C5s4bd5

26.

5 32x5 2 400x4y 1 2000x3y2 2 5000x2y3 1 6250xy4 2 3125y5

5 s2xd5 1 5s2xd4s25yd 1 10s2xd3s25yd2 1 10s2xd2s25yd3 1 5s2xds25yd4 1 s25yd5

s2x 2 5yd5 5 5C0s2xd5 1 5C1s2xd4s25yd 1 5C2s2xd3s25yd2 1 5C3s2xd2s25yd3 1 5C4s2xds25yd4 1 5C5s25yd5

27.

5 8x3 1 12x2y 1 6xy2 1 y3

5 s1ds8x3d 1 s3ds4x2ds yd 1 s3ds2xdsy2d 1 s1ds y3d

s2x 1 yd3 5 3C0s2xd3 1 3C1s2xd2s yd 1 3C2s2xds y2d 1 3C3s y3d

28.

5 343a3 1 147a2b 1 21ab2 1 b3

5 s7ad3 1 3s7ad2sbd 1 3s7adsbd2 1 sbd3

s7a 1 bd3 5 3C0s7ad3 1 3C1s7ad2sbd 1 3C2s7adsbd2 1 3C3sbd3


29.

5 x8 1 4x6y2 1 6x4y4 1 4x2y6 1 y8

5 s1dsx8d 1 s4dsx6y2d 1 s6dsx4y4d 1 s4dsx2y6d 1 s1ds y8d

sx2 1 y2d4 5 4C0sx2d4 1 4C1sx2d3sy2d 1 4C2sx2d2sy2d2 1 4C3sx2dsy2d3 1 4C4sy2d4

30.

5 x12 1 6x10y2 1 15x8y4 1 20x6y6 1 15x4y8 1 6x2y10 1 y12

1 6C5sx2ds y2d5 1 6C6s y2d6sx2 1 y2d6 5 6C0sx2d6 1 6C1sx2d5s y2d 1 6C2sx2d4s y2d2 1 6C3sx2d3s y2d3 1 6C4sx2d2s y2d4

31.

51

x51

5y

x41

10y2

x31

10y3

x21

5y4

x1 y5

11

x1 y2

5

5 5C011

x25

1 5C111

x24

y 1 5C211

x23

y2 1 5C311

x22

y3 1 5C411

x2y4 1 5C5y5

32.

51

x61

12y

x51

60y2

x41

160y3

x31

240y4

x21

192y5

x1 64y6

5 111

x26

1 6s2d11

x25

y 1 15s4d11

x24

y2 1 20s8d11

x23

y3 1 15s16d11

x22

y4 1 6s32d11

x2y5 1 1s64dy6

1 6C411

x22

s2yd4 1 6C511

x2s2yd5 1 6C6s2yd611

x1 2y26

5 6C011

x26

1 6C111

x25

s2yd 1 6C211

x24

s2yd2 1 6C311

x23

s2yd3

33.

5 2x4 2 24x3 1 113x2 2 246x 1 207

5 2sx4 2 12x3 1 54x2 2 108x 1 81d 1 5sx2 2 6x 1 9d

1 5fx2 2 2sxds3d 1 32g2sx 2 3d4 1 5sx 2 3d2 5 2fx4 2 4sx3ds3d 1 6sx2ds32d 2 4sxds33d 1 34g

34.

5 3x5 1 15x 4 1 26x3 1 18x2 1 3x 2 1

5 3fs1dx5 1 5x4 1 10x3 1 10x2 1 5x 1 1g 2 4fs1dx3 1 3x2 1 3x 1 1g

2 4f3C0 x3 1 3C1x2s1d 1 3C2xs1d2 1 3C3s1d3g

3sx 1 1d5 2 4sx 1 1d3 5 3f5C0 x5 1 5C1x4s1d 1 5C2x

3s1d2 1 5C3x2s1d3 1 5C4xs1d4 1 5C5s1d5g

36. 4th Row of Pascal’s Triangle: 1 4 6 4 1

5 81 2 216z 1 216z2 2 96z3 1 16z4

s3 2 2zd4 5 34 2 4s3d3s2zd 1 6s3d2s2zd2 2 4s3ds2zd3 1 s2zd4

35. 5th Row of Pascal’s Triangle: 1 5 10 10 5 1

5 32t5 2 80t4s 1 80t3s2 2 40t2s3 1 10ts4 2 s5

s2t 2 sd5 5 1s2td5 2 5s2td4ssd 1 10s2td3ssd2 2 10s2td2ssd3 1 5s2tdssd4 2 1(sd5

37. 5th Row of Pascal’s Triangle: 1 5 10 10 5 1

5 x5 1 10x4y 1 40x3y2 1 80x2y3 1 80xy4 1 32y5

sx 1 2yd5 5 1x5 1 5x4s2yd 1 10x3s2yd2 1 10x2s2yd3 1 5xs2yd4 1 1s2yd5

38. 6th Row of Pascal’s Triangle: 1 6 15 20 15 6 1

5 64v6 1 576v5 1 2160v4 1 4320v3 1 4860v2 1 2916v 1 729

s2v 1 3d6 5 s2vd6 1 6s2vd5s3d 1 15s2vd4s3d2 1 20s2vd3s3d3 1 15s2vd2s3d4 1 6s2vds3d5 1 s3d6


39. The 4th term in the expansion of is

10C3x1023y3 5 120x7y 3.

sx 1 yd10


5 1,259,712x2y7.

9C7s4xd927s3yd7 5 36s16x2ds2187y7d

s4x 1 3yd9

41. The 3rd term in the expansion of is

5C2x522s26yd2 5 10x3s36y 2d 5 360x3y2.

sx 2 6yd5


5 32,476,950,000x4y8.

12C8s10xd1228s23yd8 5 495s10,000x4ds6561y 8d

s10x 2 3yd12

47. The term involving in the expansion of is

The coefficient is 1,732,104.

12C7x5s3d7 5

12!

7!5!? 37x5 5 1,732,104x5.

sx 1 3d12x5


The coefficient is 180.

10C2x8s22yd2 5

10!

2!8!? 4x8y2 5 180x8y2.

sx 2 2yd10x8y2


6C6 x 626s2yd6 5 1 ? x 0y 6 5 y 6.

sx 2 yd6


7C3 x723s210zd3 5 35 ? x4s21000z3d 5 235,000x4z3.

sx 2 10zd7


5C4s5ad524s6bd4 5 5 ? s5ads1296b4d 5 32,400ab4.

s5a 1 6bd5


< 1.293 3 1013x9y 6.

15C6s7xd1526s2yd6 5 5005 ? s40,353,607x9ds64y6d

s7x 1 2yd15


The coefficient is 3,247,695.

12C8sx2d4s3d8 512!

s12 2 8d!8!? 38x8 5 3,247,695x8.

sx2 1 3d12x8



10C8s4xd2s2yd8 510!

s10 2 8d!8!? 16x2y8 5 720x2y8.

s4x 2 yd10x2y8


The coefficient is 2326,592.

9C5s3xd4s22yd5 59!

5!4!s81x4ds232y5d 5 2326,592x4y5.

s3x 2 2yd9x4y5 52. The term involving in the expansion of is

The coefficient is 16,128.

8C2s2xd6s23yd2 58!

s8 2 2d!2!s64x6ds9y2d 5 16,128x6y2.

s2x 2 3yd8x6y2

53. The term involving in the expansion of

is


10C6 sx2d4y6 510!

4!6!sx2d4y6 5 210x8y6.sx2 1 yd10

5 sx2d4y6x8y6 54. The term involving in the expansion of is


10C8sz2d2s2td8 510!

s10 2 8d!8!z4t8 5 45z4t8.

sz2 2 td10z4t8

55.

5 x2 1 12x3y2 1 54x 1 108x1y2 1 81

5 x2 1 12x!x 1 54x 1 108!x 1 81

s!x 1 3d45 s!x d4

1 4s!x d3s3d 1 6s!xd2

s3d2 1 4s!x ds3d3 1 s3d4

56.

5 8t3y2 2 12t 1 6t1y2 2 1

s2!t 2 1d35 s2!t d3

1 3s2!t d2s21d 1 3s2!t ds21d2 1 s21d3

57.

5 x2 2 3x4y3y1y3 1 3x2y3y2y3 2 y

sx2y3 2 y1y3d3 5 sx2y3d3 2 3sx2y3d2 sy1y3d 1 3sx2y3d sy1y3d2 2 sy1y3d3


58.

5 u3 1 10u12y5 1 40u9y5 1 80u6y5 1 80u3y5 1 32

su3y5 1 2d5 5 su3y5d5 1 5su3y5d4s2d 1 10su3y5d3s2d2 1 10su3y5d2s2d3 1 5su3y5ds2d4 1 25

59.

5 3x2 1 3xh 1 h2, h Þ 0

5hs3x2 1 3xh 1 h2d

h

5x3 1 3x2h 1 3xh2 1 h3 2 x3

h

f sx 1 hd 2 f sxd

h5

sx 1 hd3 2 x3

h60.

5 4x3 1 6x2h 1 4xh2 1 h3, h Þ 0

5hs4x3 1 6x2h 1 4xh2 1 h3d

h

5x 4 1 4x3h 1 6x2h2 1 4xh3 1 h4 2 x 4

h

f sx 1 hd 2 f sxdh

5sx 1 hd4 2 x 4

h

61.

51

!x 1 h 1 !x, h Þ 0

5sx 1 hd 2 x

hs!x 1 h 1 !xd

5!x 1 h 2 !x

h?!x 1 h 1 !x

!x 1 h 1 !x

f sx 1 hd 2 f sxdh

5!x 1 h 2 !x

h62.

5 21

xsx 1 hd, h Þ 0

5

2h

xsx 1 hdh

5

x 2 sx 1 hdxsx 1 hd

h

fsx 1 hd 2 f sxdh

5

1

x 1 h2

1

x

h

63.

5 24

5 1 1 4i 2 6 2 4i 1 1

s1 1 id4 5 4C0s1d4 1 4C1s1d3i 1 4C2s1d2i 2 1 4C3s1di 3 1 4C4i4

64.

5 238 2 41i

5 32 2 80i 2 80 1 40i 1 10 2 i

s2 2 id5 5 5C025 2 5C12

4i 1 5C223i2 2 5C32

2i3 1 5C42i4 2 5C5i5

65.

5 2035 1 828i

5 64 2 576i 2 2160 1 4320i 1 4860 2 2916i 2 729

5 s1ds64d 2 s6ds32ds3id 1 15s16ds29d 2 20s8ds227id 1 15s4ds81d 2 6s2ds243id 1 s1ds2729d

s2 2 3id6 5 6C026 2 6C12

5s3id 1 6C224s3id2 2 6C32

3s3id3 1 6C422s3id4 2 6C52s3id5 1 6C6s3id6

66.

5 210 1 198i

5 125 1 225i 2 135 2 27i

5 53 1 3 ? 52s3id 1 3 ? 5s3id2 1 s3id3

s5 1 !29 d3 5 s5 1 3id3


67.

5 1

51

8 f21 1 3!3i 1 9 2 3!3ig

121

21

!3

2i2

3

51

8 fs21d3 1 3s21d2s!3id 1 3s21ds!3id2

1 s!3id3g

69.

5 1 1 0.16 1 0.0112 1 0.000448 1 . . . < 1.172

5 1 1 8s0.02d 1 28s0.02d2 1 56s0.02d3 1 70s0.02d4 1 56s0.02d5 1 28s0.02d6 1 8s0.02d7 1 s0.02d8

s1.02d8 5 s1 1 0.02d8

68.

5 184 2 440!3i

5 625 2 500!3i 2 450 1 60!3i 1 9

s5 2 !3id4 5 54 2 4 ? 53s!3id 1 6 ? 52s!3id2 2 4 ? 5s!3id3 1 s!3id4

70.

< 1049.890

5 1024 1 25.6 1 0.288 1 0.00192 1 0.0000084 1 . . .

1 10s2ds0.005d9 1 s0.005d10

1 252s2d5s0.005d5 1 210s2d4s0.005d6 1 120s2d3s0.005d7 1 45s2d2s0.005d8

s2.005d10 5 s2 1 0.005d10 5 210 1 10s2d9s0.005d 1 45s2d8s0.005d2 1 120s2d7s0.005d3 1 210s2d6s0.005d4

71.

< 531,441 2 21,257.64 1 389.7234 2 4.3303 1 0.0325 2 0.0002 1 . . . < 510,568.785

5 2 220s3d3s0.01d9 1 66s3d2s0.01d10 2 12s3ds0.01d11 1 s0.01d12

5 2 792s3d7s0.01d5 1 924s3d6s0.01d6 2 792s3d5s0.01d7 1 495s3d4s0.01d8

5 312 2 12s3d11s0.01d 1 66s3d10s0.01d2 2 220s3d9s0.01d3 1 495s3d8s0.01d4

s2.99d12 5 s3 2 0.01d12

72.

< 467.721

5 512 2 46.08 1 1.8432 2 0.043008 1 0.00064512 1 0.0000064512 1 . . .

2 126s2d4s0.02d5 1 84s2d3s0.02d6 2 36s2d2s0.02d7 1 9s2ds0.02d8 2 s0.02d9

s1.98d9 5 s2 2 0.02d9 5 29 2 9s2d8s0.02d 1 36s2d7s0.02d2 2 84s2d6s0.02d3 1 126s2d5s0.02d4

73.

The graph of g is the same as the graph of f shifted four units to the left.

5 x3 1 12x2 1 44x 1 48

5 x3 1 12x2 1 48x 1 64 2 4x 2 16

5 x3 1 3x2s4d 1 3xs4d2 1 s4d3 2 4x 2 16

5 sx 1 4d3 2 4sx 1 4d

gsxd 5 f sx 1 4d−8

−4

4

fg

4f sxd 5 x3 2 4x


74.

The graph of g is the same as the graph of f shifted three units to the right.

5 2x4 1 12x3 2 50x2 1 84x 2 46

5 2x4 1 12x3 2 54x2 1 108x 2 81 1 4x2 2 24x 1 36 2 1

5 2sx4 1 4x3s23d 1 6x2s23d2 1 4xs23d3 1 s23d4d 1 4sx2 2 6x 1 9d 2 1

5 2sx 2 3d4 1 4sx 2 3d2 2 1

gsxd 5 f sx 2 3d g

−4

−3

8

f

5f sxd 5 2x4 1 4x2 2 1, gsxd 5 f sx 2 3d

75. 7C411

224

11

223

57!

3!4!11

16211

82 5 351 1

16211

82 < 0.273 76. 10C311

42313

427

5 1201 1

64212187

16,3842 < 0.2503

77. 8C411

324

12

324

58!

4!4!11

812116

812 5 701 1

812116

812 < 0.171 78. 8C411

224

11

224

5 701 1

16211

162 < 0.273

79. (a)

(b)

(c)

5 0.0025t 3 1 0.06t2 1 1.33t 1 17.5

2 0.015st2 1 20t 1 100d 1 0.88st 1 10d 1 7.7

5 0.0025st 3 1 30t 2 1 300t 1 1000d

1 0.88st 1 10d 1 7.7

gstd 5 f st 1 10d 5 0.0025st 1 10d3 2 0.015st 1 10d2

0

0 13

24

f std < 0.0025t3 2 0.015t2 1 0.88t 1 7.7 (d)

(e) For 2008 use in and in

gallons

gallons

Both models yield the same answer.

(f ) The trend is for the per capita consumption of bottled

water to increase. This may be due to the increasing

concern with contaminants in tap water.

gs8d 5 33.26

f s18d 5 33.26

gstd.t 5 8f stdt 5 18

0

0 13

60

f

g

80.

(a) (b)

(c)

gstd: gs7d 5 2007

f std: f s17d 5 2007

5 0.031t 2 1 1.44t 1 17.4

5 0.031st 2 1 20t 1 100d 1 0.82st 1 10d 1 6.1

5 0.031st 1 10d2 1 0.82st 1 10d 1 6.1

0

0 20

60

f

g gstd 5 f st 1 10d

f std 5 0.031t 2 1 0.82t 1 6.1

81. True. The coefficients from the Binomial Theorem can be

used to find the numbers in Pascal’s Triangle.

82. False. Expanding binomials that represent differences is

just as accurate as expanding binomials that represent

sums, but for differences the coefficient signs are

alternating.

83. False.

The coefficient of the -term is

The coefficient of the -term is 12C5s3d5 5 192,456.x14

12C7s3d7 5 1,732,104.x10

84. The first and last numbers in each row are 1. Every other

number in each row is formed by adding the two numbers

immediately above the number.


85.

1 8 28 56 70 56 28 8 1

1 7 21 35 35 21 7 1

1 6 15 20 15 6 1

1 5 10 10 5 1

1 4 6 4 1

1 3 3

1 2

1

1

1

1

1

1 9 36 84 126 126 84 36 9 1

1 10 45 120 210 252 210 120 1045 1

86. termssn 1 1d 87. The signs of the terms in the expansion of

alternate from positive to negative.

sx 2 ydn

88. The functions and

(choices (a) and (d)) have identical graphs, because is

the expansion of

−4

−6 6

4

h

g

p k = f

f sxd.ksxd

ksxd 5 1 2 3x 1 3x2 1 x3f sxd 5 s1 2 xd3

90. 0 5 s1 2 1dn 5 nC0 2 nC1 1 nC2 2 nC3 1 . . . ± nCn

91.

5 n11Cr

5sn 1 1d!

fsn 1 1d 2 rg!r!

5n!fn 1 1g

r!sn 2 r 1 1d!

5n!sn 2 rd!fsn 2 r 1 1d 1 rg

sn 2 rd!r!sn 2 r 1 1d!

5n!sr 2 1d!fsn 2 r 1 1d! 1 rsn 2 rd!g

sn 2 rd!r!sn 2 r 1 1d!sr 2 1d!

5n!fsn 2 r 1 1d!sr 2 1d! 1 r!sn 2 rd!g


5n!sn 2 r 1 1d!sr 2 1d! 1 n!sn 2 rd!r!


nCr 1 nCr21 5n!

sn 2 rd!r!1

n!

sn 2 r 1 1d!sr 2 1d!92. nC0 1 nC1 1 nC2 1 nC3 1 . . . 1 nCn 5 s1 1 1dn 5 2n

89.

5 nCr

5n!

sn 2 rd!r!

5n!

r!sn 2 rd!

nCn2r 5n!

sn 2 sn 2 rdd!sn 2 rd!

93. The graph of is

shifted three units to the

right. Thus, gsxd 5 sx 2 3d2.

x

4 6

4

8

2

6

2

−2

−2−4

yf sxd 5 x2 94. The graph of has

been reflected in the x–axis,

shifted two units to the left,

and shifted three units

upward. Thus,

gsxd 5 2sx 1 2d2 1 3.x

4 53

5

1−1−2−3−5

3

−2

4

2

−3

−4

−5

yf sxd 5 x2

Section 9.6 Counting Principles 877

Section 9.6 Counting Principles

n You should know The Fundamental Counting Principle.

n is the number of permutations of n elements taken r at a time.

n Given a set of n objects that has of one kind, of a second kind, and so on, the number of distinguishable

permutations is

n is the number of combinations of n elements taken r at a time.nCr 5n!

sn 2 rd!r!

n!

n1!n2! . . . nk!.

n2n1

nPr 5n!

sn 2 rd!

Vocabulary Check

1. Fundamental Counting Principle 2. permutation

3. 4. distinguishable permutations

5. combinations

nPr 5n!

sn 2 rd!

2. Even integers: 2, 4, 6, 8, 10, 12

6 ways

1. Odd integers: 1, 3, 5, 7, 9, 11

6 ways

3. Prime integers: 2, 3, 5, 7, 11

5 ways

95. The graph of is

shifted two units to the left

and shifted one unit upward.

Thus, gsxd 5 !x 1 2 1 1.

x

1

3

5

4

−3 2 3−1−2

−1

1

2

yf sxd 5 !x 96. The graph of has

been reflected in the x–axis,

shifted one unit to the left,

and shifted two units

downward. Thus,

gsxd 5 2!x 1 1 2 2.

x

1 5

3

2

1

2 3 4−1−2−3

−4

−5

yf sxd 5 !x

97. A21 51

s26ds4d 2 s5ds25d34

5

25

264 5 34

5

25

264 98.

A21 5 320

10

11.5

64

1.9R2 1 R1 → 31

0

0

1

:

:

20

10

11.5

64 5 fI : A21g

5R2 →

31

0

21.9

1

:

:

1

10

0.1

64

2R1 1 R2 →

31

0

21.9

0.2

:

:

1

2

0.1

1.24

0.1R2 1 R1 → 3 1

22

21.9

4

:

:

1

0

0.1

14

fA : Ig 5 3 1.2

22

22.3

4

:

:

1

0

0

14


15.

distinct license plate numbers

26 ? 26 ? 26 ? 10 ? 10 ? 10 ? 10 5 175,760,000 16.

distinct license plates

24 ? 24 ? 10 ? 10 ? 10 ? 10 5 5,760,000

17. (a)

(b)

(c)

(d) 6 ? 10 ? 10 5 600

9 ? 10 ? 2 5 180

9 ? 9 ? 8 5 648

9 ? 10 ? 10 5 900

19. 403 5 64,000

18. (a) numbers

(b) numbers

(c) numbers

(d) numbers9 ? 10 ? 10 ? 5 5 4500

4 ? 10 ? 10 ? 10 5 4000

9 ? 9 ? 8 ? 7 5 4536

9 ? 10 ? 10 ? 10 5 9000

20. combinations503 5 125,000

21. (a)

(b) 8 ? 1 ? 6 ? 1 ? 4 ? 1 ? 2 ? 1 5 384

8 ? 7 ? 6 ? 5 ? 4 ? 3 ? 2 ? 1 5 40,320 22. (a) orders

(b) orders4!4! 5 576

8! 5 40,320

23.

So, 4P4 54!

0!5 4! 5 24.

nPr 5n!

sn 2 rd!24.

5P5 55!

s5 2 5d!5

5!

0!5 120

nPr 5n!

sn 2 rd!25. 8P3 5

8!

5!5 8 ? 7 ? 6 5 336

13. 26 5 64 14. 212 5 4096 ways

12. 1st position: 2

2nd position: 1

3rd position: 6

4th position: 5

5th position: 4

6th position: 3

7th position: 2

8th position: 1

Total: 2!6! 5 1440 ways

4. Greater than 9: 10, 11, 12

3 ways

6. Divisible by 3: 3, 6, 9, 12

4 ways

8. Two distinct integers whose sum is 8:

6 ways

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

10. Chemist: 5 choices

Statistician: 3 choices

Total: 5 ? 3 5 15 ways

5. Divisible by 4: 4, 8, 12

3 ways

7. Sum is 9:

8 ways

6 1 3, 7 1 2, 8 1 1

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

9. Amplifiers: 3 choices

Compact disc players: 2 choices

Speakers: 5 choices

Total: 3 ? 2 ? 5 5 30 ways

11. Math courses: 2

Science courses: 3

Social sciences and

humanities courses: 5

Total: 2 ? 3 ? 5 5 30 schedules

26. 20P2 520!

18!5 20 ? 19 5 380 27. 5P4 5

5!

1!5 120 28. 7P4 5

7!

3!5 7 ? 6 ? 5 ? 4 5 840


30. Note: for this to be defined.

n 5 9 or n 5 10

sn 2 9dsn 2 10d 5 0

n2 2 19n 1 90 5 0

n2 2 n 5 18n 2 90

nsn 2 1dsn 2 2dsn 2 3dsn 2 4d 5 18sn 2 2dsn 2 3dsn 2 4dsn 2 5d

n!

sn 2 5d!5 181sn 2 2d!

sn 2 6d!2n ≥ 6nP5 5 18 ? n22P4

We can divide by

since and n Þ 4.n Þ 3,n Þ 2,sn 2 4dsn 2 2d, sn 2 3d,1 2

31. 20P5 5 1,860,480 33. 100P3 5 970,20032. 100P5 5 9,034,502,400

39. 12P4 512!

8!5 12 ? 11 ? 10 ? 9 5 11,880 ways 40. 4! 5 24 orders

41.7!

2!1!3!1!5

7!

2!3!5 420 42.

8!

3!5!5 56

43.7!

2!1!1!1!1!1!5

7!

2!5 7 ? 6 ? 5 ? 4 ? 3 5 2520 44.

11!

1!4!4!2!5

11!

4!4!2!5 34,650

45. ABCD BACD CABD DABC

ABDC BADC CADB DACB

ACBD BCAD CBAD DBAC

ACDB BCDA CBDA DBCA

ADBC BDAC CDAB DCAB

ADCB BDCA CDBA DCBA

46. ABCD

ACBD

DBCA

DCBA

47.

different batting orders

15 P9 515!

6!5 1,816,214,400 49. ways40C12 5

40!

28!12!5 5,586,853,48048. 6 P3 5

6!

3!5 120

34. 10P8 5 1,814,400 35. 20C5 5 15,504 36. 10C7 5 120

37. ways5! 5 120 38. 6! 5 720 ways

29. Note: for this to be defined.

(We can divide here by since )

n 5 5 or n 5 6

0 5 sn 2 5dsn 2 6d

0 5 n2 2 11n 1 30

14n 2 28 5 n2 1 3n 1 2

14sn 2 2d 5 sn 1 2dsn 1 1dn Þ 0, n Þ 1.nsn 2 1d14nsn 2 1dsn 2 2d 5 sn 1 2dsn 1 1dnsn 2 1d

141 n!

sn 2 3d!2 5sn 1 2d!sn 2 2d!

n ≥ 314 ? nP3 5 n12P4


50.

5 4.42 3 1016

5100!

86!14!

100C14 5100!

s100 2 14d!14!51.

The 15 ways are listed below.

AB, AC, AD, AE, AF, BC, BD, BE,

BF, CD, CE, CF, DE, DF, EF

6C2 5 15

52. 20C5 5 15,504 groups 53. 35C5 535!

30!5!5 324,632 ways 54. 40C6 5 3,838,380 ways

55. There are 7 good units and 3 defective units.

(a)

(b)

(c)

5 203 ways

5 35 1 35 ? 3 1 21 ? 3

7C4 1 7C3 ? 3C1 1 7C2 ? 3C2 57!

3!4!1

7!

4!3!?

3!

2!1!1

7!

5!2!?

3!

1!2!

7C2 ? 3C2 57!

5!2!?

3!

1!2!5 21 ? 3 5 63 ways

7C4 57!

3!4!5 35 ways

58. (a)

(b) 3C2 ? 5C2 53!

s3 2 2d!2!?

5!

s5 2 2d!2!5 3 ? 10 5 30 ways

8C4 58!

s8 2 4d!4!5

8!

4!4!5

8 ? 7 ? 6 ? 5

4 ? 3 ? 25 70 ways

59. 7C1 ? 12C3 ? 20C2 57!

s7 2 1d!1!?

12!

s12 2 3d!3!?

20!

s20 2 2d!2!5 292,600

56. (a) relationships

(b) relationships8C2 58!

2!6!5

8 ? 7

25 28

3C2 53!

2!1!5 3 (c) relationships

(d) relationships20C2 520!

2!18!5

20 ? 19

25 190

12C2 512!

2!10!5

12 ? 11

25 66

57. (a) Select type of card for three of a kind:

Select three of four cards for three of a kind:

Select type of card for pair:

Select two of four cards for pair:

(b) Select two jacks:

Select three aces:

4C2 ? 4C3 54!

s4 2 2d!2!?

4!

s4 2 3d!3!5 24

4C3

4C2

13C1 ? 4C3 ? 12C1 ? 4C2 513!

s13 2 1d!1!?

4!

s4 2 3d!3!?

12!

s12 2 1d!1!?

4!

s4 2 2d!2!5 3744

4C2

12C1

4C3

13C1

60. (a) different faces

(b) different facess89ds105ds74d 5 691,530

s195ds99ds89ds105ds74d < 1.335 3 1010


61. 5C2 2 5 5 10 2 5 5 5 diagonals 62. diagonals6C2 2 6 5 15 2 6 5 9

64. diagonals10C2 2 10 5 45 2 10 5 3563. 8C2 2 8 5 28 2 8 5 20 diagonals

65. (a)

(b) 1. If the jackpot is won, then there is only one

winning number.

(c) There are 22,957,480 possible winning numbers in the

state lottery, which is less than the possible number of

winning Powerball numbers.

53C5 ? s42d 5 120,526,770 66. (a) Permutation because order matters

(b) Combination because order does not matter

(c) Permutation because order matters

(d) Combination because order does not matter

67. False.

It is an example of a combination.

68. True by the definition of the Fundamental

Counting Principle

69. They are the same.nCr 5 nCn2r70.

Changing the order of any of the six elements selected

results in a different permutation but the same combination.

10P6 > 10C6

71. nPn215n!

sn 2 sn 2 1dd!5

n!

1!5

n!

0!5 nPn 72. nCn 5

n!

sn 2 nd!n!5

n!

0!n!5

n!

n!0!5

n!

sn 2 0d!0!5 nC0

73.

5n!

sn 2 1d!1!5 nC1

nCn21 5n!

sn 2 sn 2 1dd!sn 2 1d!5

n!

s1d!sn 2 1d! 74.

5 nPr

r!

5nsn 2 1dsn 2 2d . . . sn 2 r 1 1d

r!

5nsn 2 1dsn 2 2d . . . sn 2 r 1 1dsn 2 rd!

sn 2 rd!r!

nCr 5n!

sn 2 rd!r!

75.

This number is too large for some calculators to evaluate.

100P80 < 3.836 3 10139 76. The symbol denotes the number of ways to choose

and order r elements out of a collection of n elements.nPr

77.

(a)

(b)

(c) f s25d 5 3s25d2 1 8 5 83

f s0d 5 3s0d2 1 8 5 8

f s3d 5 3s3d2 1 8 5 35

f sxd 5 3x2 1 8 78.

(a)

(b)

(c) gsx 1 1d 5 !x 1 1 2 3 1 2 5 !x 2 2 1 2

gs7d 5 !7 2 3 1 2 5 4

gs3d 5 !3 2 3 1 2 5 2

gsxd 5 !x 2 3 1 2

80.

(a)

(b)

(c) f s220d 5 s220d2 2 2s220d 1 5 5 445

f s21d 5 2s21d2 2 2 5 23

f s24d 5 s24d2 2 2s24d 1 5 5 29

f sxd 5 5x2 2 2x

2x2

1

2

5,

2,

x

x

≤

>

24

2479.

(a)

(b)

(c) f s11d 5 2|11 2 5| 1 6 5 26 1 6 5 0

f s21d 5 2|21 2 5| 1 6 5 26 1 6 5 0

f s25d 5 2|25 2 5| 1 6 5 210 1 6 5 24

f sxd 5 2|x 2 5| 1 6


82.

5.5 5 t

8 1 3 5 2t

4

ts2td 1

3

2ts2td 5 1s2td

4

t1

3

2t5 1

84.

x 5 3 ln 16 < 8.32

x

35 ln 16

exy3 5 16

81.

By the Quadratic Formula we have:

is extraneous.

The only valid solution is x 513 1 !13

2< 8.30.

x 513 2 !13

2

x 513 ±!13

2

0 5 x2 2 13x 1 39

x 2 3 5 x2 2 12x 1 36

s!x 2 3d25 sx 2 6d2

!x 2 3 5 x 2 6

83.

x 5 35

x 2 3 5 32

x 2 3 5 25

log2sx 2 3d 5 5

Section 9.7 Probability

You should know the following basic principles of probability.

n If an event E has equally likely outcomes and its sample space has equally likely outcomes, then

the probability of event E is

where

n If A and B are mutually exclusive events, then

If A and B are not mutually exclusive events, then

n If A and B are independent events, then the probability that both A and B will occur is

n The complement of an event A is denoted by and its probability is PsA9d 5 1 2 PsAd.A9

PsAdPsBd.

PsA < Bd 5 PsAd 1 PsBd 2 PsA > Bd.

PsA < Bd 5 PsAd 1 PsBd.

0 ≤ PsEd ≤ 1.PsEd 5nsEdnsSd

,

nsSdnsEd

Vocabulary Check

1. experiment; outcomes 2. sample space

3. probability 4. impossible; certain

5. mutually exclusive 6. independent

7. complement 8. (a) iii (b) i (c) iv (d) ii

1.

HsT, 1d, sT, 2d, sT, 3d, sT, 4d, sT, 5d, sT, 6dJ

HsH, 1d, sH, 2d, sH, 3d, sH, 4d, sH, 5d, sH, 6d, 2. {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

3. HABC, ACB, BAC, BCA, CAB, CBAJ

4. Hsred, redd, sred, blued, sred, yellowd, sblue, blued, sblue, yellowdJ

Section 9.7 Probability 883

5. HAB, AC, AD, AE, BC, BD, BE, CD, CE, DEJ 6. {SSS, SSF, SFS, FSS, SFF, FFS, FSF, FFF}

7.

PsEd 5nsEdnsSd

53

8

E 5 HHHT, HTH, THHJ 8.

PsEd 5nsEdnsSd

54

85

1

2

E 5 HHHH, HHT, HTH, HTTJ

9.

PsEd 5nsEdnsSd

57

8

E 5 HHHH, HHT, HTH, HTT, THH, THT, TTHJ 10.

PsEd 5nsEdnsSd

54

85

1

2

E 5 HHHH, HHT, HTH, THHJ

11.

PsEd 5nsEdnsSd

512

525

3

13

E 5 HK¨, K©, Kª, K«, Q¨, Q©, Qª, Q«, J¨, J©, Jª, J«J 12. The probability that the card is not a face card is the

complement of getting a face card. (See Exercise 11.)

PsE9 d 5 1 2 PsEd 5 1 23

135

10

13

13.

PsEd 5nsEdnsSd

56

525

3

26

E 5 HK©, Kª, Q©, Qª, J©, JªJ 14. There are six possible cards in each of 4 suits:

PsEd 5nsEdnsSd

524

525

6

13

6 ? 4 5 24

15.

PsEd 5nsEdnsSd

53

365

1

12

E 5 Hs1, 3d, s2, 2d, s3, 1dJ 16. {(1, 6), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 3),

(4, 4), (4, 5), (4, 6), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),

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

PsEd 5nsEdnsSd

521

365

7

12

E 5

17. Use the complement.

PsEd 5 1 2 PsE9d 5 1 21

125

11

12

PsE9d 5nsE9dnsSd 5

3

365

1

12

E9 5 Hs5, 6d, s6, 5d, s6, 6dJ

18.

PsEd 5nsEdnsSd

54

365

1

9

E 5 Hs1, 1), s1, 2d, s2, 1d, s6, 6dJ

19.

PsEd 5nsEdnsSd

512

365

1

3

nsEd 5 2 1 4 1 6 5 12

E 5 E3 < E5 < E7

E7 5 Hs1, 6d, s2, 5d, s3, 4d, s4, 3d, s5, 2d, s6, 1dJ, nsE7d 5 6

E5 5 Hs1, 4d, s2, 3d, s3, 2d, s4, 1dJ, nsE5d 5 4

E3 5 Hs1, 2d, s2, 1dJ, nsE3d 5 2 20. {(1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5),

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

(5, 6), (6, 1), (6, 3), (6, 5)}

PsEd 5nsEdnsSd

519

36

E 5

21. PsEd 53C2

6C2

53

155

1

522. PsEd 5

2C2

6C2

51

1523. PsEd 5

4C2

6C2

56

155

2

5

24.

52 1 3 1 6

155

11

15

PsEd 51C1 ? 2C1 1 1C1 ? 3C1 1 2C1 ? 3C1

6C2

25. PsE9d 5 1 2 PsEd 5 1 2 0.7 5 0.3


32. 1 2 PsE9d 5 1 261100 5

39100 33. (a)

(b)

(c)2

500 5 0.004 5 0.4%

478500 5 0.956 5 95.6%

290500 5 0.58 5 58% 34. (a)

(b)

(c)23100 5 0.23 5 23%

45100 5 0.45 5 45%

34100 5 0.34 5 34%

35. (a) adults

(b)

(c) 52% 1 12% 5 64% 516

25

2% 51

50

0.24s1011d < 243 36. (a)

(b)

(c) 1 213

1005

87

100

6% 1 11% 5 17% 517

100

59% 559

10037. (a)

(b)

(c)672 2 124

12545

548

12545

274

627

582

12545

97

209

672

12545

112

209

38. (a)

(b)

(c)24

2025

12

101

1 262

1015

39

101

71 1 53

2025

124

2025

62

10140. 1 2 0.37 2 0.44 5 0.19 5 19%39.

Taylor:

Moore:

Jenkins: 0.25 51

4

0.25 51

4

0.50 51

2

p 5 0.25

p 1 p 1 2p 5 1

41. (a) (b)

(c) 15C9 ? 5C1

20C10

115C10

20C10

525,025 1 3003

184,7565

28,028

184,7565

49

323< 0.152

15C8 ? 5C2

20C10

564,350

184,7565

225

646< 0.34815C10

20C10

53003

184,7565

21

1292< 0.016

42. Total ways to insert paychecks: ways

5 correct: 1 way

4 correct: not possible

3 correct: ways (because once you choose

the three envelopes that will contain the

correct paychecks, there is only one way

to insert the paychecks so that the other

two are wrong)

2 correct: ways (because once you

choose the two envelopes that will contain

the correct paychecks, there are two ways

to fill the next envelope incorrectly, then

only one incorrect way to insert the

remaining paychecks)

(a) (b)45 1 20 1 10 1 1

1205

19

30

45

1205

3

8

5C3 ? 2 5 20

5C3 5 10

5! 5 120

1 correct: ways (five ways to choose

which envelope is paired with the correct

paycheck, three ways to fill the next

envelope incorrectly, then three ways

to fill the envelope whose correct paycheck

was placed in the second envelope, and only

one way to fill the remaining two envelopes

such that both are incorrect)

0 correct: ways120 2 1 2 10 2 20 2 45 5 44

5 ? 3 ? 3 5 45

26. PsE9d 5 1 2 PsEd 5 1 2 0.36 5 0.64 27. PsE9d 5 1 2 PsEd 5 1 214 5

34 28. 1 2 PsEd 5 1 2

23 5

13

29.

5 1 2 0.14 5 0.86

PsEd 5 1 2 PsE9d 30. 1 2 PsE9 d 5 1 2 0.92 5 0.08 31. PsEd 5 1 2 PsE9 d 5 1 21735 5

1835


49. s0.78d3 < 0.4746 50. s0.32d2 5 0.1024

51. (a)

(b)

(c) PsFFd 5 s0.015d2 < 0.0002

PsSd 5 1 2 PsFFd 5 1 2 s0.015d2 < 0.9998

PsSSd 5 s0.985d2 < 0.9702 52. (a)

(b)

(c) PsAd 5 1 2 PsNNd 5 1 2 0.01 5 0.99

PsNNd 5 s0.10d2 5 0.01

PsAAd 5 s0.90d2 5 0.81

53. (a)

(b)

(c)

5 1 2 PsGGGGd 5 1 21

16 51516

Psat least one boyd 5 1 2 Psno boysd

PsBBBBd 1 PsGGGGd 5 s12d4

1 s12d4

518

PsBBBBd 5 s12d4

51

16 54. (a)

(b)

(c)

(d)

(e)

(f) a.

b.

c.

d.

e.

The probabilities are better for European roulette.

1837 ?

1837 ?

1837 5

583250,653

137 ?

137 5

11369

137 1

1837 5

1937

1837

137

1838 ?

1838 ?

1838 5

583254,872 5

7296859

138 ?

138 5

11444

238 1

1838 5

2038 5

1019

1838 5

919

138

43. (a)

(b)1

4P4

51

24

1

5P5

51

12044. (a)

(b) 8C2 ? 25C2 ? 25C3

108C7

5

8!

6!2!?

25!

23!2!?

25!

22!3!

108!

101!7!

5 0.00069

8C2 ? 100C5

108C7

5

8!

6!2!?

100!

95!5!

108!

101!7!

5 0.076

45. (a)

(b)

(c)16

525

4

13

26

525

1

2

20

525

5

1346.

56

4165

53744

2,598,960

13C1 ? 4C3 ? 12C1 ? 4C2

52C5

513 ? 4 ? 12 ? 6

2,598,960

47. (a) (4 good units)

(b) (2 good units)

(c) (3 good units)

At least 2 good units:12

551

28

551

14

555

54

55

9C3 ? 3C1

12C4

5252

4955

28

55

9C2 ? 3C2

12C4

5108

4955

12

55

9C4

12C4

5126

4955

14

5548. (a)

(b)

(c)

(d) PsN1N1d 540

40?

1

405

1

40

PsN1 < 30, N2 < 30d 529

40?

29

405

841

1600

PsEO or OEd 5 2120

402120

402 51

2

PsEEd 520

40?

20

405

1

4


55. 1 2s45d2

s60d25 1 2 145

6022

5 1 2 13

422

5 1 29

165

7

16

56. (a) If the center of the coin falls within the circle of radius around a vertex, the coin will cover the vertex.

(b) Experimental results will vary.

5

n3p1d

222

4nd2

5p

4

Pscoin covers a vertexd 5

Area in which coin may fallso that it covers a vertex

Total area

dy2

57. True. Two events are independent if the occurance of one

has no effect on the occurance of the other.

58. False. The complement of the event is to roll a number

greater than or equal to 3 and its probability is 2y3.

59. (a) As you consider successive people with distinct birthdays, the probabilities must decrease to take

into account the birth dates already used. Because the birth dates of people are independent events,

multiply the respective probabilities of distinct birthdays.

(b)

(c)

Pn 5365

365?

364

365?

363

365? . . . ?

365 2 sn 2 1d365

5365 2 sn 2 1d

365Pn21

P3 5365

365?

364

365?

363

3655

363

365 P2 5

365 2 s3 2 1d365

P2

P2 5365

365?

364

3655

364

365 P1 5

365 2 s2 2 1d365

P1

P1 5365

3655 1

365

365?

364

365?

363

365?

362

365

(d) is the probability that the birthdays are not distinct which is equivalent to at least two people

having the same birthday.

(e)

(f) 23, see the chart above.

Qn

n 10 15 20 23 30 40 50

0.970.890.710.510.410.250.12Qn

0.030.110.290.490.590.750.88Pn

60. If a weather forecast indicates that the probability of rain

is 40%, this means the meteorological records indicate

that over an extended period of time with similar weather

conditions it will rain 40% of the time.

61.

No real solution

x2 5 243

6x2 5 28

6x2 1 8 5 0

62.

523 ± !57

4

x 52b ± !b2 2 4ac

2a5

23 ± !32 2 4s2ds26d2s2d

2x2 1 3x 2 6 5 0

4x2 1 6x 2 12 5 0 63.

x 51 ± !1 2 4s1ds23d

2s1d 51 ± !13

2

x 5 0 or x2 2 x 2 3 5 0

xsx2 2 x 2 3d 5 0

x3 2 x2 2 3x 5 0


64.

x 5 0, ±1

x2 2 1 5 0 ⇒ x 5 ±1

x 5 0

xsx2 1 2dsx2 2 1d 5 0

xsx 4 1 x2 2 2d 5 0

x5 1 x3 2 2x 5 0 65.

24 5 x

12 5 23x

12

x5 23 66.

±4 5 x

16 5 x2

32 5 2x2

32

x5 2x

67.

11

25 x

22 5 4x

2 5 4x 2 20

2 5 4sx 2 5d

2

x 2 55 4 68.

x 5 21

8x 5 28

4 5 8x 1 12

4 5 4s2x 1 3d

4

2x 1 35 4

3

2x 1 31

1

2x 1 35 4

3

2x 1 32 4 5

21

2x 1 3

69.

x 5 210

x 1 6 5 24

x2 1 x 1 6 5 x2 2 4

3x 1 6 1 x2 2 2x 5 x2 2 4

3sx 1 2d 1 xsx 2 2d 5 1sx 2 2dsx 1 2d

3

x 2 21

x

x 1 25 1 70.

x 5 3

3x 5 9

24 2 3x 5 213

2x 2 4 2 5x 5 213

2sx 2 2d 2 5x

x2 2 2x5

213

x2 2 2x

2

x2

5

x 2 25

213

x2 2 2x

71. y

x

−4 −2 2

12

10

8

4

2

864 12

5 y ≥ 23

x ≥ 21

2x 2 y ≥ 28

72. y

x

−8 −6 −4 −2

−4

−6

−8

4

2

4 6 8

73. y

x−8 −6 −4 −2

−14

−12

−8

2

4 6 8

5x2 1 y

y

≥

≥

22

x 2 4

74. y

x

−4 −3

−4

−3

1

3

4

1 3 4


Review Exercises for Chapter 9

1.

a5 5 2 16

55

16

5

a4 5 2 16

45

7

2

a3 5 2 16

35 4

a2 5 2 16

25 5

a1 5 2 16

15 8

an 5 2 16

n2.

a5 5s21d55s5d2s5d 2 1

5 225

9

a4 5s21d45s4d2s4d 2 1

520

7

a3 5s21d35s3d2s3d 2 1

5 23

a2 5s21d25s2d2s2d 2 1

510

3

a1 5s21d15s1d2s1d 2 1

5 25

an 5s21dn5n

2n 2 13.

a5 572

5!5

3

5

a4 572

4!5 3

a3 572

3!5 12

a2 572

2!5 36

a1 572

1!5 72

an 572

n!

4.

a5 5 5s5 2 1d 5 20

a4 5 4s4 2 1d 5 12

a3 5 3s3 2 1d 5 6

a2 5 2s2 2 1d 5 2

a1 5 1s1 2 1d 5 0

an 5 nsn 2 1d 5.

an 5 2s21dn

22, 2, 22, 2, 22, . . .

6.

n: 1 2 3 4 5 . . . n

Terms: 2 7 14 23 . . .

Apparent pattern: Each term is 2 less than the square of n,

which implies that an 5 n22 2.

an21

21, 2, 7, 14, 23, . . . 7.

an 54

n

4, 2, 43, 1,

45, . . .

8.

n: 1 2 3 4 5 . . . n

Terms: 1 . . .

Apparent pattern: Each term is times the

reciprocal of n, which implies that an 5s21dn11

n.

s21dn11

an

1

52

1

4

1

32

1

2

1, 21

2,

1

3, 2

1

4,

1

5, . . . 9. 5! 5 5 ? 4 ? 3 ? 2 ? 1 5 120

10. 3! ? 2! 5 s3 ? 2 ? 1d ? s2 ? 1d 5 12 11.3! 5!

6!5

s3 ? 2 ? 1d5!

6 ? 5!5 1 12.

7! ? 6!

6! ? 8!5

7! ? 6!

6!s8 ? 7!d 51

8

13. o6

i51

5 5 6s5d 5 30 14.

5 8 1 12 1 16 1 20 5 56

o5

k52

4k 5 4s2d 1 4s3d 1 4s4d 1 4s5d

Review Exercises for Chapter 9 889

21. o`

i51

5

10i5 0.5 1 0.05 1 0.005 1 0.0005 1 . . . 5 0.5555 . . . 5

5

9

22. o`

i51

3

10i5 o

`

i51

31 1

10i2 5

310

1 21

10

51

3

23. o`

k51

2

100k5 0.02 1 0.0002 1 0.000002 1 . . . 5 0.020202 . . . 5

2

99

24. o`

k52

9

10k5 o

`

k52

91 1

10k2 5

9100

1 21

10

51

10

25.

(a)

(b) A120 < $22,196.40

A10 < $10,687.03

A9 < $10,616.25

A8 < $10,545.95

A7 < $10,476.10

A6 < $10,406.73

A5 < $10,337.81

A4 < $10,269.35

A3 < $10,201.34

A2 < $10,133.78

A1 < $10,066.67

An 5 10,00011 10.08

12 2n

26.

a12 5 920.28

a11 5 889.57

a10 5 861.00

a9 5 834.57

a8 5 810.28

a7 5 788.13

a6 5 768.12

a5 5 750.25

0

4 13

1000a4 5 734.52

16.

51

21

2

31

3

41

4

51

5

61

6

71

7

81

8

9< 6.17

o8

i51

i

i 1 15

1

1 1 11

2

2 1 11

3

3 1 11

4

4 1 11

5

5 1 11

6

6 1 11

7

7 1 11

8

8 1 1

17. o10

k51

2k35 2s1d3

1 2s2d31 2s3d3

1 . . . 1 2s10d35 6050

18.

5 1 1 2 1 5 1 10 1 17 5 35

o4

j50

s j21 1d 5 s02

1 1d 1 s121 1d 1 s22

1 1d 1 s321 1d 1 s42

1 1d

19.1

2s1d1

1

2s2d1

1

2s3d1 . . . 1

1

2s20d5 o

20

k51

1

2k20.

1

21

2

31

3

41 . . . 1

9

105 o

9

k51

k

k 1 1

15. o4

j51

6

j25

6

121

6

221

6

321

6

425 6 1

3

21

2

31

3

85

205

24


27.


5, 3, 1, 21, 23, . . . 28. 0, 1, 3, 6, 10, . . .


29.

Arithmetic sequence, d 512

12, 1,

32, 2,

52, . . .

30.


99,

89,

79,

69,

59, . . . 31.

a5 5 13 1 3 5 16

a4 5 10 1 3 5 13

a3 5 7 1 3 5 10

a2 5 4 1 3 5 7

a1 5 4

a1 5 4, d 5 3 32.

a5 5 0 2 2 5 22

a4 5 2 2 2 5 0

a3 5 4 2 2 5 2

a2 5 6 2 2 5 4

a1 5 6

a1 5 6, d 5 22

33.

a5 5 34 1 3 5 37

a4 5 31 1 3 5 34

a3 5 28 1 3 5 31

a2 5 25 1 3 5 28

a1 5 25

a1 5 25, ak11 5 ak 1 3 34.

a5 5 5.4 1 0.4 5 5.8

a4 5 5.0 1 0.4 5 5.4

a3 5 4.6 1 0.4 5 5.0

a2 5 4.2 1 0.4 5 4.6

a1 5 4.2

a1 5 4.2, ak11 5 ak 1 0.4 35.

5 12n 2 5

5 7 1 12n 2 12

an 5 7 1 sn 2 1d12

a1 5 7, d 5 12

36.

So, an 5 23n 1 28.

c 5 a1 2 d 5 25 2 s23d 5 28

an 5 23n 1 c

an 5 dn 1 c

a1 5 25, d 5 23 37.

5 3ny 2 2y

5 y 1 3ny 2 3y

an 5 y 1 sn 2 1d3y

a1 5 y, d 5 3y 38.

So, an 5 xn 2 3x.

c 5 a1 2 d 5 22x 2 x 5 23x

an 5 xn 1 c

an 5 dn 1 c

a1 5 22x, d 5 x

39.

an 5 a1 1 sn 2 1dd 5 100 1 sn 2 1ds27d 5 27n 1 107

a1 5 a2 2 d ⇒ a1 5 93 2 s27d 5 100

a6 5 a2 1 4d ⇒ 65 5 93 1 4d ⇒ 228 5 4d ⇒ d 5 27

a2 5 93, a6 5 65

40.

an 5 a1 1 sn 2 1dd ⇒ an 5 10 1 sn 2 1ds213d ⇒ an 5 2

13n 1

313

a1 5 a7 2 6d ⇒ a1 5 8 2 6s213d ⇒ a1 5 10

a13 5 a7 1 6d ⇒ 6 5 8 1 6d ⇒ d 5 213

a7 5 8, a13 5 6

41. o10

j51

s2j 2 3d is arithmetic. Therefore, a1 5 21, a10 5 17, S10 5102 f21 1 17g 5 80.

42. o8

j51

s20 2 3jd 5 o8

j51

20 2 3o 8

j51

j 5 8s20d 2 33s8ds9d2 4 5 52

43. o11

k51

s23k 1 4d is arithmetic. Therefore, a1 5

143 , a11 5

343 , S11 5

112 f14

3 1343 g 5 88.


51.

The sequence is geometric, r 5 22

13, 2

23,

43, 2

83, . . . 52.


14,

25,

36,

47, . . .

53.

a5 5 2116s21

4d 5164

a4 514s21

4d 5 2116

a3 5 21s214d 5

14

a2 5 4s214d 5 21

a1 5 4

a1 5 4, r 5 214

54.

a5 5 16s2d 5 32

a4 5 8s2d 5 16

a3 5 4s2d 5 8

a2 5 2s2d 5 4

a1 5 2

a1 5 2, r 5 2

55.

or

a5 5 283s22

3d 5169a5 5

83s2

3d 5169

a4 5 4s223d 5 2

83a4 5 4s2

3d 583

a3 5 26s223d 5 4a3 5 6s2

3d 5 4

a2 5 9s223d 5 26a2 5 9s2

3d 5 6

a1 5 9a1 5 9

49 5 r 2 ⇒ r 5 ±

23

4 5 9r2

a3 5 a1r2

a1 5 9, a3 5 4 56.

or

a5 5 212!6s2!6d 5 72a5 5 12!6s!6 d 5 72

a4 5 12s2!6d 5 212!6a4 5 12s!6d 5 12!6

a3 5 22!6s2!6d 5 12a3 5 2!6s!6d 5 12

a2 5 2s2!6d 5 22!6a2 5 2s!6d 5 2!6

a1 5 2a1 5 2

±!6 5 r

6 5 r2

12 5 2r2

a3 5 a1r2

a1 5 2, a3 5 12

44. o25

k5113k 1 1

4 2 53

4 o25

k51 k 1 o

25

k51

1

45

3

43s25ds26d

2 4 1 2511

42 5 250

45. o100

k51 5k is arithmetic. Therefore, a1 5 5, a100 5 500, S500 5

1002 s5 1 500d 5 25,250.

46. o80

n520

n 5 o80

n51

n 2 o19

n51

n 5s80ds81d

22

s19ds20d2

5 3050

47.

(a)

(b) S5 552s34,000 1 43,000d 5 $192,500

a5 5 34,000 1 4(2250d 5 $43,000

an 5 34,000 1 sn 2 1ds2250d 48.

S8 582s123 1 46d 5 676

a8 5 123 1 7s211d 5 46

n 5 8

a1 5 123, d 5 112 2 123 5 211

49.

The sequence is geometric, r 5 2

5, 10, 20, 40, . . . 50.

Geometric sequence, r 5 21854 5 2

13

54, 218, 6, 22, . . .


57.

a20 5 16s212d19 < 23.052 3 1025

an 5 16s212dn21

a2 5 a1r ⇒ 28 5 16r ⇒ r 5 212

a1 5 16, a2 5 28 58.

a20 5 216s16d19

5 3.545 3 10213

an 5 216s16dn21

a1 5 216

6 5 a1s 136d

6 5 a1s16d2

a3 5 a1r2

r 516

6r 5 1

a3r 5 a4

a3 5 6, a4 5 1

59.

a20 5 100s1.05d19 < 252.695

an 5 100s1.05dn21

a1 5 100, r 5 1.05 60.

a20 5 5s0.2d19 < 2.62 3 10213

an 5 5s0.2dn21

a1 5 5, r 5 0.2

61. o7

i51 2

i215

1 2 27

1 2 25 127 63. o

4

i5111

22i

51

21

1

41

1

81

1

165

15

16

65. o5

i51

s2di215 1 1 2 1 4 1 8 1 16 5 31

62. o5

i51

3i215 111 2 35

1 2 3 2 5 121

64. 51 2

1729

1 213

5364

243o6

i5111

32i21

5 11 2 s13d6

1 2132

66. o4

i51

6s3di5 6s3d11 2 34

1 2 3 2 5 720 67. o10

i51 1013

52i21

< 24.85 68. o15

i51

20s0.2di215 25

69. o25

i51

100s1.06di21 < 5486.45 70. o20

i51

816

52i21

5 1493.50 71. o`

i5117

82i21

51

1 278

5 8

72. o`

i5111

32i21

51

1 213

53

273. o

`

i51

s0.1di215

1

1 2 0.15

10

974. o

`

i51

s0.5di215

1

1 2 0.55 2

75. o`

k51 412

32k21

54

1 223

5 12 76. o`

k51

1.31 1

102k21

51.3

1 2110

513

977. (a)

(b)

5 $20,168.40

a5 5 120,000s0.7d5

at 5 120,000s0.7dt

78. Monthly:

Continuously:

5200e0.06y12ses0.06ds10d

21de0.06y12

2 15 $32,967.03

A 5Pery12sert

2 1dery12

2 1

5 $32,939.75

5 200311 10.06

12 212?10

2 1411 112

0.062

A 5 P311 1r

12212t

2 1411 112

r 2


80. 1. When

2.

Thus, the formula holds for all positive integers n.

5k 1 1

4fsk 1 1d 1 3g.

5sk 1 1dsk 1 4d

4

5k2

1 5k 1 4

4

5ksk 1 3d 1 2sk 1 2d

4

5k

4sk 1 3d 1

1

2sk 1 2d

Sk11 5 Sk 1 ak11 5 11 13

21 2 1

5

21 . . . 1

1

2sk 1 1d2 1

1

2sk 1 2d

Assume that Sk 5 1 13

21 2 1

5

21 . . . 1

1

2sk 1 1d 5

k

4sk 1 3d. Then,

n 5 1, S1 5 1 51

4s1 1 3d 5 1.

81. 1. When

2. Assume that

Then,

Therefore, by mathematical induction, the formula is valid for all positive integer values of n.

5as1 2 rk

1 rk2 rk11d

1 2 r5

as1 2 rk11d1 2 r

.

Sk11 5 ok

i50

ari5 1o

k21

i50

ari2 1 ark5

as1 2 rkd1 2 r

1 ark

Sk 5 ok21

i50

ari5

as1 2 rkd1 2 r

.

n 5 1, a 5 a11 2 r

1 2 r2.

79. 1. When

2. Assume that

Therefore, by mathematical induction, the formula is valid for all positive integer values of n.

5 sk 1 1dfsk 1 1d 1 2g.

5 sk 1 1dsk 1 3d

5 k21 4k 1 3

5 ksk 1 2d 1 2k 1 3

Then, Sk11 5 3 1 5 1 7 1. . .

1 s2k 1 1d 1 f2sk 1 1d 1 1g 5 Sk 1 s2k 1 3d

Sk 5 3 1 5 1 7 1 . . . 1 s2k 1 1d 5 ksk 1 2d.

n 5 1, 3 5 1s1 1 2d.


83.

Sn 5 ns2n 1 7d

S4 5 9 1 13 1 17 1 21 5 60 5 4s15d 5 4f2s4d 1 7g

S3 5 9 1 13 1 17 5 39 5 3s13d 5 3f2s3d 1 7g

S2 5 9 1 13 5 22 5 2s11d 5 2f2s2d 1 7g

S1 5 9 5 1s9d 5 1f2s1d 1 7g 84.

Sn 5 4ns18 2 nd

S4 5 68 1 60 1 52 1 44 5 224 5 16 ? 14

S3 5 68 1 60 1 52 5 180 5 12 ? 15

S2 5 68 1 60 5 128 5 8 ? 16

S1 5 68 5 4 ? 17

82. 1. When

2. Assume that Then,

Thus, the formula holds for all positive integers n.

52ia 1 isi 2 1dd 1 2a 1 2id

25

2asi 1 1d 1 idsi 1 1d2

5 1i 1 1

2 2f2a 1 idg.

oi1121

k50

sa 1 kdd 5i

2f2a 1 si 2 1ddg 1 fa 1 idg

Sk11 5 Sk 1 ak11

Sk 5 oi21

k50

sa 1 kdd 5i

2f2a 1 si 2 1ddg.

n 5 1, S1 5 a 1 0 ? d 5 a 51

2f2a 1 s1 2 1ddg 5 a.

85.

Since the series is geometric,

Sn 51 2 s3

5dn1 2

35

55

231 2 13

52n

4.

S4 5 1 13

51

9

251

27

1255

272

125

S3 5 1 13

51

9

255

49

25

S2 5 1 13

55

8

5

S1 5 1 86.

Since the series is geometric,

Sn 5 121 2 s2 1

12dn

f1 2 s2 112dg

5144

13 31 2 121

122n

4.

S4 5 12 2 1 11

122

1

1445

1595

144

S3 5 12 2 1 11

125

133

12

S2 5 12 2 1 5 11

S1 5 12

87. o30

n51

n 530s31d

25 465 88. o

10

n51

n25

10s10 1 1ds2 ? 10 1 1d6

510s11ds21d

65 385

89.

5 4676 2 28 5 4648

5s7ds8ds15ds167d

302

s7ds8d2

5s7ds8ds15dfs3ds49d 1 21 2 1g

302

s7ds8d2

o7

n51

sn42 nd 5 o

7

n51

n42 o

7

n51

n

90.

5s6d2s7d2s83d 2 2s6ds7ds13d

125 12,110

5s6d2s7d2s83d

122

6s7ds13d6

5s6d2s6 1 1d2f2s6d2

1 2s6d 2 1g12

26s6 1 1df2s6d 1 1g

6

o6

n51

sn52 n2d 5 o

6

n51

n52 o

6

n51

n2


91.

n: 1 2 3 4 5

First differences:

Second differences:

The sequence has a linear model.

an :

a5 5 20 1 5 5 25

a4 5 15 1 5 5 20

a3 5 10 1 5 5 15

a2 5 5 1 5 5 10

a1 5 5

a1 5 f s1d 5 5, an 5 an21 1 5

5 10 15 20 25

5 5 5 5

0 0 0

92.

23 27 213 221 231

First differences: 24 26 28 210

Second differences: 22 22 22

Since the second differences are all the same,

the sequence has a quadratic model.

an:

a5 5 a4 2 2s5d 5 221 2 10 5 231

a4 5 a3 2 2s4d 5 213 2 8 5 221

a3 5 a2 2 2s3d 5 27 2 6 5 213

a2 5 a1 2 2s2d 5 23 2 4 5 27

a1 5 23

an 5 an21 2 2n

a1 5 23

93.

n: 1 2 3 4 5

First differences:

Second differences:

The sequence has a linear model.

an :

a5 5 13 2 1 5 12

a4 5 14 2 1 5 13

a3 5 15 2 1 5 14

a2 5 16 2 1 5 15

a1 5 16

a1 5 f s1d 5 16, an 5 an21 2 1

16 15 14 13 12

21 21 21 21

0 0 0

94.

0 1 1 2 2

First differences: 1 0 1 0

Second differences: 21 1 21

Since neither the first differences nor the second

differences are equal, the sequence does not have

a linear or a quadratic model.

an:

a4 5 4 2 a3 5 4 2 2 5 2

a3 5 3 2 a2 5 3 2 1 5 2

a2 5 2 2 a1 5 2 2 1 5 1

a1 5 1 2 a0 5 1 2 0 5 1

a0 5 0

a0 5 0, an 5 n 2 an21

95. 6C4 56!

2!4!5 15

97. 8C5 58!

3!5!5 56

99.

the 5th entry in the 7th row17

32 5 35,

1

111

11

1 11510105

1 6 15 20 15 6 1172135352171

4643 3 1

12

17

32 5 35

96.

510 ? 9 ? 8

3 ? 2 ? 15 120

10C7 510!

7!3!5

10 ? 9 ? 8 ? 7!

7!3!

98. 12C3 512!

3!9!5

12 ? 11 ? 10 ? 9!

3! ? 9!5

12 ? 11 ? 10

3 ? 2 ? 15 220

100.


42 5 126,

111

11

1 115101051

1 6 15 20 15 6 11

11

82856705628811 9 36 84 126 126 84 36 9

72135352171

4643 3 1

12


101.


62 5 28,

111

11

1 115101051

1 6 15 20 15 6 11

1828567056288172135352171

4643 3 1

12

18

62 5 28 102.


32 5 10,

111

11

1 115101051

4643 3 1

12

103.

5 x41 16x3

1 96x21 256x 1 256

sx 1 4d45 x4

1 4x3s4d 1 6x2s4d21 4xs4d3

1 44

104.

5 x62 18x5

1 135x42 540x3

1 1215x22 1458x 1 729

1 6C3sxd3s23d31 6C4sxd2s23d4

1 6C5sxds23d51 6C6sxd0s23d6 sx 2 3d6

5 6C0sxd6s23d01 6C1sxd5s23d 1 6C2sxd4s23d2

105.

5 a52 15a4b 1 90a3b2

2 270a2b31 405ab4

2 243b5

sa 2 3bd55 a5

2 5a4s3bd 1 10a3s3bd22 10a2s3bd3

1 5as3bd42 s3bd5

106.

5 2187x71 5103x6y2

1 5103x5y41 2835x4y6

1 945x3y81 189x2y10

1 21xy121 y14

5 s3xd71 7s3xd6y2

1 21s3xd5sy2d21 35s3xd4sy2d3

1 35s3xd3sy2d41 21s3xd2sy2d5

1 7s3xdsy2d61 sy2d7

1 7C6s3xdsy2d61 7C7sy2d7

s3x 1 y2d75 7C0s3xd7

1 7C1s3xd6sy2d 1 7C2s3xd5sy2d21 7C3s3xd4sy2d3

1 7C4s3xd3sy2d41 7C5s3xd2sy2d5

107.

5 625 1 1000i 2 600 2 160i 1 16 5 41 1 840i

5 625 1 1000i 1 600i2 1 160i3 1 16i4

s5 1 2id45 s5d4

1 4s5d3s2id 1 6s5d2s2id21 4s5ds2id3

1 s2id4

108.

5 2236 2 115i

5 64 2 240i 2 300 1 125i

5 432 3s4d2s5id 1 3s4ds5id2

2 s5id3

s4 2 5id35 3C0s43d 1 3C1s42ds25id 1 3C2s4ds25id2

1 3C3s25id3

109. First number: 1 2 3 4 5 6 7 8 9 10 11

Second number: 11 10 9 8 7 6 5 4 3 2 1

From this list, you can see that a total of 12 occurs 11 different ways.

110. 6C1 ? 5C1 ? 6C1 5 6 ? 5 ? 6 5 180

112. 3C1 ? 4C1 ? 6C1 5 3 ? 4 ? 6 5 72

111. different telephone numberss10ds10ds10ds10d 5 10,000

113.

5 10 ? 9 ? 8 5 720 different ways

10P3 510!

7!5

10 ? 9 ? 8 ? 7!

7!


120. (a)

(b)

(c)37500 5 0.074 or 7.4%

400500 5 0.8 or 80%

208500 5 0.416 or 41.6% 121. s1

6ds16ds1

6d 51

216

122. 16

6215

6214

6213

6212

6211

62 56!

665

720

46,6565

5

324123. 1 2

13

525 1 2

1

45

3

4

124. 1 2 PsHHHHHd 5 1 2 11

225

531

32125. True.

sn 1 2d!n!

5sn 1 2dsn 1 1dn!

n!5 sn 1 2dsn 1 1d

126. True by Properties of Sums 127. True. by the Properties of Sums.o8

k51

3k 5 3 o8

k51

k

128. True because 211 22

1 231 24

1 251 26

5 23221 2422

1 25221 2622

1 27221 2822

129. False. If or then nPr 5 nCr.r 5 1,r 5 0 130. The domain of an infinite sequence is the set of

natural numbers.

131. (a) Odd-numbered terms are negative.

(b) Even-numbered terms are negative.

132. (a) Arithmetic. There is a constant difference between

consecutive terms.

(b) Geometric. Each term is a constant multiple of the

previous term. In this case the common ratio is

greater than 1.

133. Each term of the sequence is defined in terms

of preceding terms.

134. Increased powers of real numbers between 0 and 1

approach zero.

135.

The sequence is geometric and

is decreasing.

Matches graph (d).

a1 5 4, a2 5 2, a10 51

128

an 5 4s12dn21

136.

and fluctuates from

positive to negative.

Matches graph (a).

ana1 5 4

an 5 4s212dn21

137.

and as

Matches graph (b).

n → `an → 8a1 5 4

an 5 on

k51

4s12dk21

114. 32C12 532!

20!12!5 225,792,840 115. 8C3 5

8!

5!3!5 56

116. Breads:

Meats:

Cheese:

Vegetables:

5 ? 128 ? 8 ? 64 5 327,680

5 1 1 6 1 15 1 20 1 15 1 6 1 1 5 646C0 1 6C1 1 6C2 1 6C3 1 6C4 1 6C5 1 6C6

5 1 1 3 1 3 1 1 5 83C0 1 3C1 1 3C2 1 3C3

5 1 1 7 1 21 1 35 1 35 1 21 1 7 1 1 5 1287C0 1 7C1 1 7C2 1 7C3 1 7C4 1 7C5 1 7C6 1 7C7

5C1 5 5

117. s1d11

92 51

9118. PsEd 5

nsEdnsSd

51

5!5

1

120119. (a)

(b) 100% 2 18% 5 82%

25% 1 18% 5 43%


Problem Solving for Chapter 9

1.

Conjecture: xn →!2 as n →`

x6 < x7 < x8 < x9 < 1.414213562

x5 51

2x4 1

1

x4

< 1.414213562

x4 51

21577

4082 11

577/408< 1.414213562

x3 51

2117

122 11

17/125

577

408< 1.414215686

x2 51

213

22 11

3/25

17

125 1.416

x1 51

2s1d 1

1

15

3

25 1.5

x0 5 1

x0 5 1 and xn 51

2 xn21 1

1

xn21

, n 5 1, 2, . . . 2.

(a)

(b)

(c)

(d) an → 0 as n →`

an → 0 as n →`

0

0 10

2

an 5n 1 1

n2 1 1

n 1 10 100 1000 10,000

110,001

100,000,0011001

1,000,001101

10,00111101an

3.

(a)

(b)

(c)

(d) As oscillates between 2 and 4 and does not

approach a fixed value.

n →`, an

an 5 52,

4,

if n is odd

if n is even

0

0 10

8

an 5 3 1 s21dn

n 1 10 101 1000 10,001

2 4 2 4 2an

4. Let an arithmetic sequence with a common

difference of d.

(a) If C is added to each term, then the resulting

sequence, is still

arithmetic with a common difference of d.

(b) If each term is multiplied by a nonzero constant C,

then the resulting sequence,

is still arithmetic. The

common difference is Cd.

(c) If each term is squared, the resulting sequence,

is not arithmetic.bn 5 an2 5 sdn 1 cd2

bn 5 Csdn 1 cd 5 Cdn 1 Cc

bn 5 an 1 C 5 dn 1 c 1 C

an 5 dn 1 c,

138 .

Matches graph (c).

a1 5 4 and an → 83 as n → `.

an 5 on

k51

4s212dk21

139.

S10 5 S9 1 S8 1 S7 5 1490 1 810 1 440 5 2740

S9 5 S8 1 S7 1 S6 5 810 1 440 1 240 5 1490

S8 5 S7 1 S6 1 S5 5 440 1 240 1 130 5 810

S7 5 S6 1 S5 1 S4 5 240 1 130 1 70 5 440

S6 5 S5 1 S4 1 S3 5 130 1 70 1 40 5 240

140. closed interval0 ≤ p ≤ 1,

Problem Solving for Chapter 9 899

(e) 1 16 81 256 625 1296 2401 4096 6561

First differences: 15 65 175 369 671 1105 1695 2465

Second differences: 50 110 194 302 434 590 770

Third differences: 60 84 108 132 156 180

In general, for the third differences. dn 5 24n 1 36

6. Distance:

Time:

In two seconds, both Achilles and the tortoise will be

40 feet away from Achilles starting point.

o`

n5111

22n21

51

1 212

5 2

o`

n51

2011

22n21

520

1 212

5 40 7. Side lengths:

for

Areas:

An 5!3

4 311

22n21

42

5!3

4 11

222n22

5!3

4Sn

2

!3

4, !3

4 11

222

, !3

4 11

422

, !3

4 11

822

, . . .

n ≥ 1Sn 5 11

22n21

1, 1

2,

1

4,

1

8, . . .

8.

(a)

a20 522 5 1a10 5

402 5 20

a19 542 5 2a9 5 3s13d 1 1 5 40

a18 5 3s1d 1 1 5 4a8 5262 5 13

a17 522 5 1a7 5

522 5 26

a16 542 5 2a6 5 3s17d 1 1 5 52

a15 582 5 4a5 5

342 5 17

a14 5162 5 8a4 5 3s11d 1 1 5 34

a13 5 3s5d 1 1 5 16a3 5222 5 11

a12 5102 5 5a2 5 3s7d 1 1 5 22

a11 5202 5 10a1 5 7

3an21 1 1, if an21 is odd

an21

2 , if an21 is even

an 5 5(b)

Eventually the terms repeat; 4, 2, 1 if is a positive

integer and if is a negative integer.a122, 21

a1

a10 5 22a10 5 4a10 5 4

a9 5 21a9 5 1a9 5 1

a8 5 22a8 5 2a8 5 2

a7 5 21a7 5 4a7 5 4

a6 5 22a6 5 1a6 5 1

a5 5 21a5 5 2a5 5 2

a4 5 22a4 5 4a4 5 4

a3 5 24a3 5 8a3 5 1

a2 5 28a2 5 16a2 5 2

a1 5 23a1 5 5a1 5 4

5. (a) 1 4 9 16 25 36 49 64 81


In general, for the first differences.

(b) Find the second differences of the perfect cubes.

(c) 1 8 27 64 125 216 343 512 729


Second differences: 12 18 24 30 36 42 48

In general, for the second differences.

(d) Find the third differences of the perfect fourth powers.

cn 5 6sn 1 1d 5 6n 1 6

bn 5 2n 1 1


10. (a) If is true and implies then is true for integers

(b) If are all true, then you can draw no conclusion about in general other than it is

true for

(c) If and are all true, but the truth of does not imply that is true, then is false for some

values of You can only conclude that it is true for and .

(d) If is true and implies then is true for all integers n ≥ 1.P2nP2k12,P2kP2

P3P2,P1,n ≥ 4.

PnPk11PkP3P2,P1,

1 ≤ n ≤ 50.

PnP50. . . ,P3,P2,P1,

n ≥ 3.PnPk11,PkP3

11. (a) The Fibonacci sequence is defined as follows: for

By this definition

1. For

2. Assume

Then,

Therefore, by mathematical induction, the formula is valid for all integers

(b) S20 5 f22 2 1 5 17,711 2 1 5 17,710

n ≥ 2.

5 fsk11d1221.5 fk13 2 15 s fk12 1 fk11d 2 1f1 1 f2 1 f3 1 . . . 1 fk 1 fk115fk12 2 1 1 fk11

f1 1 f2 1 . . . 1 fk 5 fk12 2 1.

n 5 2: f1 1 f2 5 2 and f4 2 1 5 2

f3 5 f1 1 f2 5 2, f4 5 f2 1 f3 5 3, f5 5 f4 1 f3 5 5, f6 5 f5 1 f4 5 8, . . .

n ≥ 3.f1 5 1, f2 5 1, fn 5 fn22 1 fn21

12. (a)

(b)

(c)

—CONTINUED—

PsEd 5odds in favor of E

odds in favor of E 1 1

PsEd 5nsEdnsSd 5

nsEdnsEd 1 nsE9d 5

nsEdynsE9dnsEdynsE9d 1 nsE9dynsE9d

Odds against choosing a blue marble 5number of yellow marbles

number of blue marbles5

7

3

Odds in favor of choosing a blue marble 5number of blue marbles

number of yellow marbles5

3

7

Total marbles 5 6 1 24 5 30

24 5 x (number of non-red marbles)

4

15

x

6

Odds against choosing a red marble 5number of non-red marbles

number of red marbles

9. The numbers 1, 5, 12, 22, 35, 51, . . . can be written recursively as Show that

1. For

2. Assume

Then,

Therefore, by mathematical induction, the formula is valid for all integers n ≥ 1.

5sk 1 1df3sk 1 1d 2 1g

2.

5sk 1 1ds3k 1 2d

25

3k2 1 5k 1 2

2

5ks3k 2 1d 1 2s3k 1 1d

25

ks3k 2 1d2

1 s3k 1 1d

Pk11 5 Pk 1 f3sk 1 1d 2 2g

Pk 5ks3k 2 1d

2.

n 5 1: 1 51s3 2 1d

2

Pn 5 ns3n 2 1dy2.Pn 5 Pn21 1 s3n 2 2d.

Problem Solving for Chapter 9 901

12. —CONTINUED—

(d)

Odds in favor of event E 5nsEdnsE9d 5

nsSdPsEdnsSdPsE9d 5

PsEdPsE9d

nsSdPsE9d 5 nsE9d nsSdPsEd 5 nsEd

PsE9d 5nsE9dnsSd PsEd 5

nsEdnsSd

13.1

3 14.

5 68.2%

< 0.682

1 2Area of triangle

Area of circle5 1 2

12s12ds6d

ps6d25 1 2

1

p

15. (a)

(b)

60

2.53< 24 turns

V 51

36s1d 1

1

36s4d 1

1

36s9d 1

1

36s16d 1

1

36s25d 1

1

36s36d 1

30

36s0d < 2.53

< 2$0.71

V 5 1 1

47C5s27d2s12,000,000d 1 11 21

47C5s27d2s21d


Chapter 9 Practice Test

1. Write out the first five terms of the sequence an 52n

sn 1 2d!.

2. Write an expression for the nth term of the sequence 43,

59,

627,

781,

8243, . . . .

3. Find the sum o6

i51

s2i 2 1d.

4. Write out the first five terms of the arithmetic sequence where and d 5 22.a1 5 23

5. Find for the arithmetic sequence with a1 5 12, d 5 3, and n 5 50.an

6. Find the sum of the first 200 positive integers.

7. Write out the first five terms of the geometric sequence with a1 5 7 and r 5 2.

8. Evaluate o10

n51

612

32n21

. 9. Evaluate o`

n50

s0.03dn.

10. Use mathematical induction to prove that 1 1 2 1 3 1 4 1 . . . 1 n 5nsn 1 1d

2.

11. Use mathematical induction to prove that n! > 2n, n ≥ 4.

12. Evaluate 13C4. 13. Expand sx 1 3d5.

14. Find the term involving x7 in sx 2 2d12. 15. Evaluate 30P4.

16. How many ways can six people sit at a table with six chairs?

17. Twelve cars run in a race. How many different ways can they come in first,

second, and third place? (Assume that there are no ties.)

18. Two six-sided dice are tossed. Find the probability that the total of the two dice

is less than 5.

19. Two cards are selected at random form a deck of 52 playing cards without

replacement. Find the probability that the first card is a King and the second

card is a black ten.

20. A manufacturer has determined that for every 1000 units it produces, 3 will be

faulty. What is the probability that an order of 50 units will have one or more

faulty units?

Documents

C H A P T E R 9 Sequences, Series, and Probability · C H A P T E R 9 Sequences, Series, and Probability Section 9.1 Sequences and Series 819 Vocabulary Check 1. infinite sequence