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CHAPTER 12 Section 12.1 Solutions-------------------------------------------------------------------------------- 1. 1, 2,3,4 2. 1, 4,9,16 3. 1,3,5,7 4. 2 3 4, , ,x x x x
5. 1 2 3 4, , ,2 3 4 5
6. 3 4 52, , ,2 3 4
7.
4 8 162 1 3 2 1 4 3 2 1
4 8 162, , , ,2! 3! 4!
4 2which is equivalent to 2, 2, ,3 3
= = =⋅ ⋅ ⋅ ⋅ ⋅ ⋅
8. ! !
( 1)!n n
n=
+ ( 1) !n n+1 ,
11 1 1 1so the terms are , , ,2 3 4 5
n=
+
9. 2 3 4 5, , ,x x x x− − 10. 1, 4,9, 16− − 11.
1 1 1 1, , ,2 3 3 4 4 5 5 6
1 1 1 1which is equivalent to , , ,6 12 20 30
− −⋅ ⋅ ⋅ ⋅
− −
12.
14
1 4 90, , ,9 16 25
=
13. 91 1
2 512⎛ ⎞ =⎜ ⎟⎝ ⎠
14. ( )2
15 1525616
=
15. ( )( )
191 19! 1 121 ! 21 20 420
− − −= =
⋅ 16. ( ) ( )( )141 12 15 180
13 13−
=
17. 2 21 101 10,2011
100 100 10,000⎛ ⎞ ⎛ ⎞+ = =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
18. 2
1 99110 100
− =
19. 23log10 23log10 23= = 20. ln 49 49e = 21. 2 , 1na n n= ≥ 22. 3 , 1na n n= ≥
23. ( )
1 , 11na n
n n= ≥
+ 24. 1 , 1
2
n
na n⎛ ⎞= ≥⎜ ⎟⎝ ⎠
25. ( ) 2 ( 1) 21 , 13 3
n n nn
n na n−⎛ ⎞= − = ≥⎜ ⎟⎝ ⎠
26. 13 , 1
2
n
n na n−
= ≥
27. ( ) 11 , 1n n+− ≥ 28. ( ) 11 , 12
n n nn
+− ≥+
Chapter 12
1712
29. 9! 9 8 7!7!
⋅ ⋅=
7!72= 30. 4! 4!
6!=
6 5 4!⋅ ⋅1
30=
31. 29! 29 28 27!27!
⋅ ⋅=
27!812= 32. 32! 32 31 30!
30!⋅ ⋅
=30!
992=
33. 75! 75!77!
=77 76 75!⋅ ⋅
15852
= 34. 100! 100!103!
=103 102 101 100!⋅ ⋅ ⋅
11,061,106
=
35. 97! 97 96 95 94 93!93!
⋅ ⋅ ⋅ ⋅=
93!83,156,160= 36. 101! 101 100 99 98!
98!⋅ ⋅ ⋅
=98!
999,900=
37. ( )( )
( )1 !1 !1 !
nnn
−−=
+ ( ) ( )1 1 !n n n+ − ( )11n n
=+
38. ( )2 ! ( 2)( 1) !
!n n n n
n+ + +
=!n
( ) ( )2 1n n= + +
39.
( )( )
( )( ) ( )2 3 2 2 2 1 !2 3 !2 1 !
n n nnn
+ + ++=
+ ( )2 1 !n +
( )( )2 3 2 2n n= + +
40. ( )( )
( )( )( ) ( )2 2 2 1 2 2 1 !2 2 !2 1 !
n n n nnn
+ + −+=
− ( )2 1 !n −
( )( )( )2 2 2 1 2n n n= + +
41. 7,10,13,16 42. 2,3, 4,5 43. 1, 2,6,24 44. 2,6, 24,120 45.
253
50 25 2550 253 4 3 2 723 2 3
100 50100, , ,2! 3! 4!
25 25which is equivalent to 100, 50, , .3 72
== == = ⋅ ⋅ ⋅⋅
46.
2 2 2
55144
520 5 920, , , ,2 3 4
5 5which is equivalent to 20, 5, ,9 144
= =
47. 3 2 1 4 3 21, 2, 2, 4 ( ote that , )a a a a a a= ⋅ = ⋅N
48. 1 23 4
2 3
11, 2, , 4 (Note that , )2
a aa aa a
= =
49. ( ) ( ) ( ) ( )2 2 2 2
2 5
1, 1, 1 1 , 1 2
which is equivalent to 1, 1, 2,5=− =
⎡ ⎤ ⎡ ⎤− − + − − + −⎣ ⎦ ⎣ ⎦
− −
50. ( ) ( ) ( )
1 5
3 2 1 4 3 2
1, 1, 2 1 1, 3 1 2 1
which is equivalent to 1, 1, 1, 5(Note that 2 1 , 3 2 .)a a a a a a
=− =−
− − + − + −
− − −= + = +
51. 2 5 10⋅ = 52. 7 5 35⋅ = 53. 2 2 2 2 20 1 2 3 4 30+ + + + = 54. 1 1 1 251
2 3 4 12+ + + =
Section 12.1
1713
55. 6
1(2 1) 1 3 5 7 9 11 36
nn
=
− = + + + + + =∑
56.
( )6
11 2 3 4 5 6 7 27
nn
=
+ = + + + + + =∑
57.
( )4 4
0 01 1, for all . So, 1 1 5 1 5.n n
n nn
= =
= = = =∑ ∑
58. 0 1 2 3 42 2 2 2 2 1 2 4 8 16
31+ + + + = + + + +
=
59. 2 31 x x x− + −
60. 2 3 4x x x x− + − +
61. 0 1 2 3 4 5
4 8 16 322 1 3 2 4 3 2 5 4 3 2
2 2 2 2 2 20! 1! 2! 3! 4! 5!
4 2 41 2 23 3 15
4 109715 15
= = = =⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅
+ + + + + =
+ + + + + =
+ =
62. 1 1 1 1 1 10! 1! 2! 3! 4! 5!
1
− + − + − =
1−1 1 1 1 112 6 24 120 30
+ − + − =
63. 2 3 4
2 3 4
10! 1! 2! 3! 4!
12 6 24
x x x x
x x xx
+ + + + =
+ + + +
64. 2 3 4
3 42
1 (1) (2) (3) (4)0! 1! 2! 3! 4!
12 6
x x x x
x xx x
− − − − =
− − − −
65. ( )0
209
0.1 22 2.21 0.1 0.9
= = =−
66.
0
509
15105 5.51 0.91
10
⎛ ⎞⎜ ⎟⎝ ⎠⋅ = = =−
67. Not possible. (The result is infinite.) 68. Not possible. (The result is infinite.)
69. ( )6
0
112
nn
n=−∑ 70.
0
12n
n
∞
=∑
71. ( ) 1
11 n
nn
∞−
=
−∑ 72. 23
1nn
=∑
73. ( )( )
6 5
1 1
1 !( 1)
1 !(Remember that 0! 1! 1.)n n
nn n
n= =
+= +
−
= =
∑ ∑ 74.
( )1
1 0
2 21 ! !
n n
n nn n
−∞ ∞
= =
=−∑ ∑
Chapter 12
1714
75. ( ) ( ) ( )1
1
1 0
1 11 ! !
n nn n
n n
x xn n
−∞ ∞−
= =
− = −−∑ ∑ 76.
( )6
1 1 !
n
n
xn= −∑
77. 72
720.0620,000 1 $28,640.89.12
A ⎛ ⎞= + ≈⎜ ⎟⎝ ⎠
So, she has approximately $28,640.89 in her account after 6 years (or 72 months).
78. 12
120.057,000 1 $8,125.28
4A ⎛ ⎞= + ≈⎜ ⎟
⎝ ⎠
So, she has approximately $8,125.28 in her account after 3 years (or 36 months).
79. Let number of years experience. n = Then, the salary per hour is given by
20 2 , 0.na n n= + ≥
So, 20 20 2(20) 60.a = + = Thus, a paralegal with 20 years experience would make $60 per hour.
80. ( )
3
1
Here, 275,000 75,000 1 , 1.
So, salary for a 3 year career.
n
nn
a n n
a=
= + − ≥
=∑
81. Let number of years on the job. n =
0
1 1 1
raiseprevious year
Then, the salary is given recursively by30,000
0.03 1.03n n n n
aa a a a− − −
== + =
82. Let #20 minute periods.n =
0
11
36
37 11
Then, the number of E.coli cells is described by:2
2 2 , 1.After 12 hours, 36 20-minute periodshave passed and so, E.coli cells exist,
which equals 2 1.374 10 .After 48 hours, 144 2
nn n
a
a a n
a
+−
=
= ⋅ = ≥
≈ ×
144145 43
0-minute periodshave passed and E.coli cells exist,
which equals 2 4.46 10 .
a
≈ ×
83. Let number of years. n =
1
1
1
1
Then, the number of T-cells in body is given by:10001000 75 , 1.
We must find such that 200.To do so, note that the above formula can be expressed explicitly as 1000 75 .So, we must
n
n
n
aa n nn a
a n
+
+
+
== − ≥
≤
= −
solve 1000 75 200 :75 800
10.7
nnn
− ==≈
As such, after approximately 10.7 years, the person would have full blown AIDS.
84.
( )( )
( )
3
4
4
3
3.8 1.6 3 8.6 billion in sales
3.8 1.6 4 10.2 billion in sales
1 1So, 8.6 10.2 9.42 2
represents the average sales for the years 2003 and 2004.
nn
a
a
a=
= + =
= + =
= + =∑
Section 12.1
1715
85. Observe that ( )( )( )( )( )( )( )( )( )
1
22
33
44
3636
100,000 1.001 1 100
100,000 1.001 1 200.10
100,000 1.001 1 300.30
100,000 1.001 1 400.60
100,000 1.001 1 3663.72
A
A
A
A
A
= − =
= − =
= − ≈
= − ≈
= − ≈
86. Observe that ( )( )( )( )( )( )( )( )( )
1
22
33
44
4848
50,000 1.001 1 50
50,000 1.001 1 100.05
50,000 1.001 1 150.15
50,000 1.001 1 200.30
50,000 1.001 1 2457.27
A
A
A
A
A
= − =
= − =
= − ≈
= − ≈
= − ≈
87. At x = 2 1 1 1 x+ 3
2
21 xx+ + 5 2 3
2 3!1 x xx+ + + 6.33≈ 2 3 4
2 3! 4!1 x x xx+ + + + 7.0≈
Approx. of 2 7.38906e ≈
88. 6
6 195,000(1.035)239,704.7886
A ==
So, approximately $239,705.
89. At
x = 1.1 ( )1x − 0.1
( ) ( )2121 xx −− − 0.0950
( ) ( ) ( )2 31 12 31 x xx − −− − + 0.0953
( ) ( ) ( ) ( )2 3 41 1 12 3 41 x x xx − − −− − + − 0.09531
( ) ( ) ( ) ( ) ( )2 3 4 51 1 1 12 3 4 51 x x x xx − − − −− − + − +
0.095306 Approx of
ln(1.1)
90. (1.06) 150000.06
n
FV⎡ ⎤−
= ⎢ ⎥⎣ ⎦
n FV 1 5,000 2 10,300 3 15,918 4 21,873.08 5 28,185.4648
Chapter 12
1716
91. The mistake is that 6! 3!2!,≠ but rather 6! 6 5 4 3 2 1.= ⋅ ⋅ ⋅ ⋅ ⋅
92. The mistake is that ( ) ( )( )( )
( ) ( )( )( )
( )
2 2 ! 2 2 2 4 2 6 (2)The terms should be consecutive and decrease by 1 (not 2). Therefore, it should be2 2 ! 2 2 2 3 2 4 (1).
The same error is made when computing2 2 !.
n n n n
n n n n
n
− ≠ − − − ⋅ ⋅
− = − − − ⋅ ⋅
+
94. Same error as 93.
95. True
93.
( ) 1 1, 1,3,5,1
1, 2,4,6,So, the terms should all be the opposite sign.
n nn
+ =⎧− = ⎨− =⎩
……
96. True
97. False. Let 1 and 5.k ka b n= = =
5
1 1
5 5
1 1 1 1
Then, observe that
1 1 5 5
whereas
1 1 25.
n
k kk k
n n
k kk k k k
a b
a b
= =
= = = =
= = ⋅ =
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⋅ = ⋅ =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
∑ ∑
∑ ∑ ∑ ∑
98. False. Let 3, 2.a b= =
( )( ) ( )( ) ( )( )
( ) ( )
Then, observe that ! ! 3! 2! 3 2 1 2 1 12
whereas ! 2 3 ! 6! 720.
a b
ab
= = ⋅ ⋅ ⋅ =
= ⋅ = =
99.
( )( )( )
1
2
3
4
2
2 3
a Ca C D
a C D D C D
a C D D C D
=
= +
= + + = +
= + + = +
100.
( )( )( )
1
2
23
2 34
a Ca D C
a D DC D C
a D D C D C
=
=
= =
= =
101.
( ) ( )1
1 5 1 5 2 5 12 5 2 5
F+ − −
= = =
( ) ( )2 2
2 2
1 5 1 5
2 51 2 5 5 1 2 5 5 1
4 5
F+ − −
=
+ + − + −= =
102. n
1 , 2n na a n−= ≥ 1 7 2 1
27 7= 3 1
47 7= 4 1
87 7= 5 1
167 7= In general,
1127 , 2n
na n−= ≥ .
Section 12.1
1717
103. Using a calculator yields the following table of values:
n 11n
n⎛ ⎞+⎜ ⎟⎝ ⎠
100 2.704813 2.705≈ 1000 2.716923 2.717≈ 10000 2.718146 2.718≈ 100000 2.718268 1000000 2.718280 10000000 2.718281693
↓ ↓ ∞ e
104.
n na 1na + 1n
n
aa+
1 1 1 1 2 1 2 2 3 2 3 1.5 4 3 5 1.66 5 5 8 1.6 6 8 13 1.625 7 13 20 1.53846 8 20 33 1.65 9 33 53 1.60606 10 53 86 1.62264 11 86 139 1.61627912 139 228 1.61870513 225 364 1.61777 14 364 589 1.61813115 589 953 1.61799616 953 1542 1.61804817 1542 2495 1.61802818 2498 4037 1.61803619 4037
105. The calculator gives 10915 , which
agrees with Exercise 61. 106. The calculator gives 11
30 , which agrees with Exercise 62.
Chapter 12
1718
Section 12.2 Solutions-------------------------------------------------------------------------------- 1. Yes, 3d = . 2. Yes, 3d =− . 3. No. 4. No. 5. Yes, 0.03d = − . 6. Yes, 0.5d = . 7. Yes, 2 / 3d = . 8. Yes, 1 3d = . 9. No. 10. No. 11. First four terms: 3,1, 1, 3− − So, yes and 2d =− .
12. First four terms: 7, 4, 1,2− − − So, yes and 3d = .
13. First four terms: 1,4,9,16 So, no.
14. First four terms are:
4 9 3 16 22 3 2 2 4 3 2 3
9 1642! 3! 4!
2
1, , ,= = =i i i
which is equivalent to 3 21,2, ,2 3
. So, no.
15. First four terms: 2,7,12,17 So, yes and 5d = .
16. First four terms: 1, 3, 7, 11− − − So, yes and 4d =− .
17. First four terms: 0,10,20,30 So, yes and 10d = .
18. First four terms: 4,12,20,28 So, yes and 8d = .
19. First four terms: 1,2, 3,4− − So, no.
20. First four terms: 2, 4,6, 8− − So, no.
21. ( )11 1 5 5 6na n n= + − = + 22. ( )5 1 11 11 6na n n= + − = −
23. ( )4 1 (2) 6 2na n n=− + − =− + 24. ( )( )2 1 4 4 6na n n= + − − =− +
25. ( ) 2 2 20 13 3 3na n n= + − = − 26. ( ) 3 3 11 1
4 4 4na n n− −⎛ ⎞=− + − = −⎜ ⎟⎝ ⎠
27. ( )0 1na n e en e= + − = − 28. ( ) ( )1.1 1 0.3 0.3 1.4na n n= + − − =− + 29. Here, 1 7, 13.a d= =
( )So, 7 1 13na n= + − 13 6n= − . Thus, 10 130 6 124a = − = .
30. Here, 1 7, 6a d= =− .
( )( )So, 7 1 6na n= + − − 6 13n=− +
Thus, ( )19 6 19 13 101a =− + =− . 31. Here, 1 9, 7.a d= =−
( )( )So, 9 1 7na n= + − − 7 16n=− + Thus, 100 700 16 684a =− + =− .
32. Here, 1 13, 6.a d= =
( )( )So, 13 1 6na n= + − 6 7n= +
Thus, ( )90 6 90 7 547a = + = .
33. 5 44a = Find 17 5: 12d a a d= + Find ( )1 5 1: 4 9a a a= + So, ( )8 1 9na n= + − 17 152a = 152 44 12d= + 144 36a= + 9 1n= − 9 d= 18 a=
Section 12.2
1719
34. 9 19a =− Find 21 9: 12d a a d= + Find ( )1 9 1: 8 3a a a= + − So, ( )( )5 1 3na n= + − − 21 55a =− 55 19 12d− =− + 119 24a− = − 3 8n=− + 36 12d− = 15 a= 3 d− = 35. 7 1a =− Find 17 7: 10d a a d= + Find ( )1 7 1: 6 4a a a= + − So, ( )( )23 1 4na n= + − − 17 41a =− 41 1 10d− =− + 11 24a− = − 4 27n=− + 40 10d− = 123 a= 4 d− = 36. 8 47a = Find 21 8: 13d a a d= + Find ( )1 8 1: 7 5a a a= + So, ( )( )12 1 5na n= + − 21 112a = 112 47 13d= + 147 35a= + 5 7n= + 65 13d= 112 a= 5d =
37. 4 3a = Find 22 4: 18d a a d= + Find 1 4 12: 33
a a a ⎛ ⎞= + ⎜ ⎟⎝ ⎠
So, ( ) 21 13na n= + −
22 15a = 15 3 18d= + 13 2a= + 2 13 3
n= +
12 18d= 11 a= 2
3 d=
38. 11 3a =− Find 31 1: 20d a a d= + Find 1 11 11: 102
a a a ⎛ ⎞= + −⎜ ⎟⎝ ⎠
So, ( ) 12 12na n −⎛ ⎞= + − ⎜ ⎟
⎝ ⎠
31 13a =− 13 3 20d− =− + 13 5a− = − 1 52 2
n=− +
10 20d− = 12 a= 1
2d =−
39. [ ]23 23
1 1
232 2 2 46 5522k k
k k= =
= = + =∑ ∑ 40. 20 20
0 1
205 5 [5 100] 10502k k
k k= =
= = + =↑
∑ ∑
Note that the 1st term when 0 is 0k = .
41. ( )30 30 30
1 1 1
2 5 2 5n n n
n n= = =
− + = − +∑ ∑ ∑
[ ]30 3 552
= − 930 150 780= − + =−
42. ( ) ( )17 17
0 1
3 10 10 3 10n n
n n= =
− =− + −∑ ∑
[ ]1710 7 412
= − + − +
( ) ( )17 187 3 10 17
2⎛ ⎞
= − + −⎜ ⎟⎝ ⎠
279=
Chapter 12
1720
43. 14 14 2
3 1 10.5 0.5 0.5
j j jj j j
= = =
⎡ ⎤= −⎢ ⎥⎣ ⎦
∑ ∑ ∑
[ ]14 30.5 72 2
= + − [ ]0.5 102 51= =
44. 33
1
33 1 33 140.254 2 4 4j
j=
⎡ ⎤= + =⎢ ⎥⎣ ⎦∑
45. arithmetic
2 7 12 62+ + + +… . We need to find such that 62nn a = . To this end, observe that
1 2a = and 5d = . Thus, since the sequence is arithmetic, ( )2 1 5 5 3na n n= + − = − . So, we need to solve 5 3 62na n= − = : 5 65, so that 13n n= = . Therefore, this sum
equals [ ]13 2 62 4162
+ = .
46. arithmetic
1 3 7 ... 75− − − − . We need to find such that 75nn a = − . To this end, observe that
1 1a = and 4d =− . Thus, since the sequence is arithmetic, ( ) ( )1 1 4na n= + − − 4 5n=− + . So, we need to solve 4 5 75na n= − + =− : 4 80, so that 20n n− =− = . Therefore, this
sum equals [ ]20 1 75 7402
− = − .
47. arithmetic
4 7 10 151+ + + +… . We need to find such that 151nn a = . To this end, observe
that 1 4a = and 3d = . Thus, since the sequence is arithmetic, ( )4 1 3na n= + − 3 1n= + . So, we need to solve 3 1 151na n= + = : 3 150, so that 50n n= = . Therefore, this sum
equals ( )50 4 151 38752
= + = .
48.
arithmetic
2 0 2 56+ − − −… . We need to find such that 56nn a =− . To this end, observe
that 1 2a = and 2d =− . Thus, since the sequence is arithmetic,
( ) ( )2 1 2na n= + − − 2 4n=− + . So, we need to solve 2 4 56na n= − + =− : 2 60, so that 30n n− =− = . Therefore, this
sum equals ( )30 2 56 8102
− =− .
Section 12.2
1721
49.
arithmetic
1 1 1 136 6 2 2− − − −… . We need to find 13
2such that nn a =− . To this end, observe
that 11 6a = and 1
3d =− . Thus, since the sequence is arithmetic,
( )1 116 3na n −⎛ ⎞= + − ⎜ ⎟
⎝ ⎠1 13 2
n=− + . So, we need to solve 1 1 133 2 2na n= − + =− :
1 7, so that 213
n n−=− = . Therefore, this sum equals 21 1 13 21 1 39 133
2 6 2 2 6 2−⎡ ⎤ ⎛ ⎞− = =−⎜ ⎟⎢ ⎥⎣ ⎦ ⎝ ⎠
.
50.
arithmetic
11 7 17 1412 6 12 3
+ + + +… . We need to find 143such that nn a = . To this end, observe
that 111 12a = and 7 11 3 1
6 12 12 4d = − = = . Thus, since the sequence is arithmetic,
( )11 1112 4na n ⎛ ⎞= + − ⎜ ⎟
⎝ ⎠1 24 3
n= + .
So, we need to solve 1 2 144 3 3na n= + = : 1 12 4, so that 16
4 3n n= = = .
Therefore, this sum equals 16 11 14 111 56 13482 12 3 12 3
−⎡ ⎤ ⎛ ⎞= + = =⎜ ⎟⎢ ⎥⎣ ⎦ ⎝ ⎠.
51. Here, 14 and 1d a= = . Hence,
1 ( 1)(4) 4 3na n n= + − = − . So,
( ) ( )12 2 1 4 3 (2 1)n nn nS a a n n n= + = + − = − .
As such, 18 18(35) 630.S = =
52. Here, 13 and 2d a= = . Hence, 2 ( 1)(3) 3 1na n n= + − = − . So,
( ) ( )12 2 22 3 1 (3 1)n n nn nS a a n n= + = + − = + .
As such, 2121 2 (64) 672.S = =
53. Here, 1
12 and 1d a= − = . Hence,
( ) 31 12 2 21 ( 1)na n n= + − − = − + . So,
( ) ( ) ( )3 51 112 2 2 2 2 2 21n n n
n nS a a n n= + = − + = − + .
As such, ( )4343 4
8175 432
S = − = − .
54. Here, 312 and 3d a= − = . Hence,
( )3 3 92 2 23 ( 1)na n n= + − − = − + . So,
( ) ( ) ( )3 9 3 1512 2 2 2 2 2 23n n n
n nS a a n n= + = − + = − + .
As such, ( )3737 4 15 3(37) 888.S = − = −
55. Here, 110 and 9d a= = − . Hence,
9 ( 1)(10) 10 19na n n= − + − = − . So, ( ) ( )12 2 9 10 19 (5 14)n n
n nS a a n n n= + = − + − = − . As such, 18 18(5(18) 14) 1368.S = − =
56. Here, 110 and 2d a= = − . Hence, 2 ( 1)(10) 10 12na n n= − + − = − . So, ( ) ( )12 2 22 10 12 (10 14)n n n
n nS a a n n= + = − + − = − . As such, 21 21(5(21) 7) 2058.S = − =
Chapter 12
1722
57. Colin Camden 1 28,000a = 1 25,000a = 1500d = 2000d = So ( )( )28,000 1 1500na n= + − So, ( )25,000 1 2000na n= + − 26,500 1500n= + 23,000 2000n= + Thus, 10 41,500a = . So, 10 43,000a = After 10 years, Colin will have After 10 years, Camden will have accumulated 10
2 (28,000 41,500)+ accumulated 102 (25,000 43,000)+
= $347,500. = $340,000. 58. Jasmine Megan 1 80,000a = 1 90,000a = 2000d = 5000d = So, ( )( )80,000 1 2000na n= + − So, ( )90,000 1 5000na n= + − 78,000 2000n= + 85,000 5000n= + Thus, 15 108,000a = . Thus, 15 160,000a = . After 15 years, Jasmine will have After 15 years, Megan will have accumulated 15
2 (80,000 108,000)+ accumulated 152 (90,000 160,000)+
= $1,410,000. = $1,875,000.
59. We are given that 1 22a = . We seek25
1i
ia
=∑ . Observe that 1d = . So,
( )( )22 1 1 21na n n= + − = + . Thus, 25 21 25 46a = + = . As such, using the formula
( )12n nnS a a= + , we conclude that
25
1i
ia
=∑ = [ ]25 22 46
2= + 850= .
60. We are given that 1 1a = . We seek20
1i
ia
=∑ . Observe that 1d = . So,
( )( )1 1 1na n n= + − = . Thus, 20 20a = . As such, using the formula ( )12n nnS a a= + , we
conclude that 20
1i
ia
=∑ = [ ]20 1 20 210
2= + = . So, there are 210 tulips in each delta.
61. We are given that 1 1a = and 56n= . We need to find 56 and .a d Well, note that
( ) ( )56
1 56 561
28
56 5630,856 12 2i
ia a a a
=
= = + = +∑ . Hence,
Section 12.2
1723
( )56 5630,82830,856 28 1 , so that 1101
28a a= + = = .
Now, since ( )1 1na a n d= + − , we can substitute 156, 1n a= = in to find d:
( )1101 1 56 1 , so that 1100 55 and hence, 20d d d= + − = = . So, there are 1101 glasses on the bottom row and each row had 20 fewer glasses than the one before. 62. From the information given in the problem, we infer that 1 1a = , 25 25a = , and 1d = .
Since the sequence is arithmetic, the sum must be [ ]25 1 252
+ 325 logs in the pile= .
63. We are given that 1 16, 32, 10a d n= = = . Hence,
( )( )16 1 32 16 32na n n= + − = − + .
10So, 16 320 304a = − + = . Now, observe
that [ ]10
1
10 16 3042i
ia
=
= +∑
1600 feet in 10 seconds=
64. We are given that 4.9ia = , 9.8d = , 10n = . Hence,
( )( )4.9 1 9.8 4.9 9.8na n n= + − = − + So, 10 93.1a = . Now, observe that
[ ]10
1
10 4.9 93.12i
ia
=
= +∑
490 meters in 10 seconds= 65. The sum is 20 + 19 + 18 + . . . + 1. This is arithmetic with 1 20, 1a d= = − . So, 20 ( 1)( 1) 21na n n= + − − = − . Thus,
2 2(20 21 ) (41 )n nnS n n= + − = − . So,
20 10(21) 210S = = . There are 210 oranges in the display.
66. The sum is 45000 + 46500 + … + na . This is arithmetic with
1 45000, 1500a d= = . So,
45000 ( 1)(1500) 1500 43500na n n= + − = + a. 35 96,000a = -- salary in year 35 b. Total earnings in 35 years = 35S
( )352 45000 1500(35) 43500 2, 467,500= + − =
67. 1 2 3 26 27 28 ...a a a+ + + + + + + na Here, 1d = , so that 1 23.a = a. There are 23 seats in the first row. b. 23 ( 1)(1) 22.na n n= + − = + So, 2 (23 22)n
nS n= + + , and hence,
30 1125S = seats in the theater.
68. Here 2 11and e ed a= = . Hence,
1 1 2( 1)(2)n e e ea n n= + − = − + . So,
( )2312 1 1442n e e eS = + = . Since there are 12
terms all told, the total is 1212 132 eS = + .
69. ( )1 11 , not na a n d a nd= + − + 70. Here, 2 d = − since consecutive terms decrease by 2
71. There are 11 terms, not 10. So, 11n = ,
and thus, ( )1111 1 21 1212
S = + = . 72. ( )12n n
nS a a= + , not ( )12n nnS a a= −
Chapter 12
1724
73. False. In a series you are adding terms, while in a sequence you are not.
74. False. For instance,
( )1
2 1i
i∞
=
+∑ doesn’t exist. All FINITE
arithmetic series can be computed, however.
75. True, since d must be constant. If a sequence alternates, the difference between consecutive terms would need to change sign.
76. False. If the terms are decreasing, d would be negative.
77.
( ) ( ) ( ) ( ) ( )( )1
1
1 21( 1)2 2
n
k
n a nbna a b a nb a k b a a nb+
=
+ +++ + + + + = + − = + + =∑…
78. First, observe that 30 30
29 29ln 29 28 1 0 1 30k
k ke k
=− =−
= = − − − − + + + +∑ ∑ … … .
Now, we simplify as follows: 29= − 28− 2− −… 1− 0
30 29+ 28+ 2+ +… 1+
30
So, the sum is 30. 79. As n gets larger, 2
1n
goes to zero. So,
( )2 21 1
k nv R= − gets closer to 2
Rk
, which in
this case is 27,419.5.
80. Here, 1 4, 6a d= − = . So, 4 ( 1)(6) 6 10na n n= − + − = − . Thus,
2 ( 4 6 10) (3 7)nnS n n n= − + − = − .
Solve (3 7) 570 :n n− = 2
383
3 7 570 0
7 49 4(3)(570)15 or
6
n n
n
− − =
± += = −
81. Compute 100
1i
i=∑ - you should get 5050. 82. Compute
50
1
2i
i=∑ - you should get
2550.
83. Compute ( )50
1
2 1i
i=
−∑ - you should get
2500.
84. Same answer.
85. 18,850 86. 12,320
Section 12.3
1725
Section 12.3 Solutions-------------------------------------------------------------------------------- 1. Yes, 3r = 2. Yes, 2r =
3. 9 16No, for instance4 9≠ 4.
1 19 16No, for instance1 14 9
≠
5. 1Yes, 2
r = 6. 1Yes, 2
r = −
7. Yes, 1.7r = 8. Yes, 2.2r = 9. 6,18,54,162,486 10. 17,34,68,136,272 11. 1, 4,16, 64,256− − 12. 3,6, 12,24, 48− − − 13.
10,000, 10,600, 11,236,11,910.16, 12,624.77
14. 10000, 8000, 6400, 5120, 4096
15. 2 1 1 1 1, , , ,3 3 6 12 24
16. 1 1 1 1 1, , , ,10 50 250 1250 6250
− −
17. ( ) 15 2 nna −= 18. ( ) 112 3 n
na −=
19. ( ) 11 3 nna −= − 20. ( ) 14 2 n
na −= − −
21. ( ) 11000 1.07 nna −= 22. ( ) 11000 0.5 n
na −=
23. 116 1
3 4
n
na−
⎛ ⎞= −⎜ ⎟⎝ ⎠
24. ( ) 11 5200
nna −=
25. Since 1 2, 2a r= − = − , we have
( ) 1 2 2 nna −= − − .
In particular, ( )7 17 2 2 128a −= − − = − .
26. Since 1 1, 5a r= = − , we have
( ) 15 nna −= −
In particular, ( )10 110 5 1,953,125a −= − = − .
27. Since 11 , 23
a r= = , we have
( ) 11 23
nna −=
In particular, ( )13 1 409613 3
1 23
a −= = .
28. Since 11100,5
a r= = , we have
111005
n
na−
⎛ ⎞= ⎜ ⎟⎝ ⎠
In particular, 9 1
49
1100 2.56 105
a−
−⎛ ⎞= = ×⎜ ⎟⎝ ⎠
.
Chapter 12
1726
29. Since 111000,20
a r= = , we have
11100020
n
na−
⎛ ⎞= ⎜ ⎟⎝ ⎠
In particular, 15 1
1615
11000 6.10 1020
a−
−⎛ ⎞= ≈ ×⎜ ⎟⎝ ⎠
.
30. Since 141000,5
a r= = − , we have
1410005
n
na−
⎛ ⎞= −⎜ ⎟⎝ ⎠
In particular, 8 1
841000 209.71525
a−
⎛ ⎞= − = −⎜ ⎟⎝ ⎠
.
31. Use the formula ( )
1
11
n
n
rS a
r−
=−
.
Since 11 , 13, 23
a n r= = = , the given sum
is13
131 1 2 81913 1 2 3
S⎛ ⎞−
= =⎜ ⎟−⎝ ⎠.
32. Use the formula ( )
1
11
n
n
rS a
r−
=−
.
Since 11 31, 11,a n r= = = , the given sum
is
11
11
1131 1.5113
S
⎛ ⎞⎛ ⎞−⎜ ⎟⎜ ⎟⎝ ⎠⎜ ⎟= ≈
⎜ ⎟−⎜ ⎟⎝ ⎠
.
33. Use the formula ( )
1
11
n
n
rS a
r−
=−
.
Since 1 2, 10, 3a n r= = = , the given sum is 10
101 32 59,0481 3
S⎛ ⎞−
= =⎜ ⎟−⎝ ⎠.
34. Use the formula ( )
1
11
n
n
rS a
r−
=−
.
Since 1 1, 4, 10a r n= = = , the given sum
is10
101 41 349,5251 4
S⎛ ⎞−
= =⎜ ⎟−⎝ ⎠.
35. Use the formula ( )
1
11
n
n
rS a
r−
=−
.
Since 1 2, 0.1, 11a r n= = = , the given sum is
( )10
11
1 0.12 2.2
1 0.1S
⎛ ⎞−= =⎜ ⎟
⎜ ⎟−⎝ ⎠.
(Note: The sum starts with 0n = , so there are 11 terms.)
36. Use the formula ( )
1
11
n
n
rS a
r−
=−
.
Since 1 3, 0.2, 12a r n= = = , the given sum is
( )12
12
1 0.23 3.75
1 0.2S
⎛ ⎞−= ≈⎜ ⎟
⎜ ⎟−⎝ ⎠
(Note: The sum starts with 0n = , so there are 12 terms.)
37. Use the formula ( )
1
11
n
n
rS a
r−
=−
.
Since 1 2, 3, 8a r n= = = , the given sum is 8
81 32 65601 3
S⎛ ⎞−
= =⎜ ⎟−⎝ ⎠.
38. Use the formula ( )
1
11
n
n
rS a
r−
=−
.
Since 12 , 5, 93
a r n= = = , the given sum
is 9
92 1 5 325,520.66673 1 5
S⎛ ⎞−
= ≈⎜ ⎟−⎝ ⎠.
Section 12.3
1727
39. Use the formula ( )
1
11
n
n
rS a
r−
=−
.
Since 1 1, 2, 14a r n= = = , the given sum
is14
141 21 16,3831 2
S⎛ ⎞−
= =⎜ ⎟−⎝ ⎠.
40. Use the formula ( )
1
11
n
n
rS a
r−
=−
.
Since 111, , 142
a r n= = = , the given sum
is14
14
1121 2.0112
S
⎛ ⎞⎛ ⎞−⎜ ⎟⎜ ⎟⎝ ⎠⎜ ⎟= ≈
⎜ ⎟−⎜ ⎟⎝ ⎠
.
41. Use the formula 1
1aS
r∞ = −.
Since 111,2
a r= = , the given sum is
1 2112
S∞ = =−
.
42. Use the formula 1
1aS
r∞ = −.
Since 11 1,3 3
a r= = , the given sum is
113
1 213
S∞ = =−
.
43. Use the formula 1
1aS
r∞ = −.
Since 11 1,3 3
a r= − = − , the given sum is
113
1 413
S∞
−= = −
⎛ ⎞− −⎜ ⎟⎝ ⎠
.
44. Use the formula 1
1aS
r∞ = −.
Since 111,2
a r= = − , the given sum is
1 21 312
S∞ = =⎛ ⎞− −⎜ ⎟⎝ ⎠
.
45. Not possible - diverges. 46. Not possible - diverges.
47. Use the formula 1
1aS
r∞ = −.
Since 119,3
a r= − = , the given sum is
9 271 213
S∞
−= = −
−
.
48. Use the formula 1
1aS
r∞ = −.
Since 118,2
a r= − = − , the given sum is
8 161 312
S∞
−= = −
⎛ ⎞− −⎜ ⎟⎝ ⎠
.
49. Use the formula 1
1aS
r∞ = −.
Since 1 10,000, 0.05a r= = , the given sum
is 10,000 10,5261 0.05
S∞ = ≈−
.
50. Use the formula 1
1aS
r∞ = −.
Since 1 200, 0.04a r= = , the given sum is 200 625
1 0.04 3S∞ = =
−.
Chapter 12
1728
51. The sum is 0.4 21 0.4 3
=−
. 52. The sum can be written as ( )110
13 n
n
∞
=∑ ,
which equals ( )110
110
3 11 3
=−
.
53. The sum is 0(0.99) 100
1 0.99=
−.
54. Diverges – infinite sum
55. Observe that 1 34,000, 1.025a r= = . We need to find 12a . Observe that the nth term is given by ( ) 11
1 34,000 1.025 nnna a r −−= = . Hence,
( )1112 34000 1.025
44,610.95a =
≈
So, the salary after 12 years is approximately $44,610.95.
56. Observe that 1 22,000, 1.15a r= = . We need to find 10a . Observe that the nth
term is given by ( ) 122,000 1.15 nna −= .
Hence, ( )9
10 22,000 1.1577,393.28
a =
≈
So, the salary after 10 years is approximately $77,393.28.
57. Since 1 2000, 0.5a r= = , we see that
( )Value of laptop after years
2000 0.5 .nn
n
a =
In particular, we have
( )
( )
44
Worth when graduating from college.
77
Worth when graduating from graduate school.
2000 0.5 125
2000 0.5 16
a
a
= =
= =
58.
( ) ( )( ) ( )
1 1
1 1
9 910 10
BMW Honda35,000 25,000
0.8 0.9
35,000 0.8 25,000 0.9
35,000 0.8 25,000 0.94697.62 9685.51
n nn n
a ar r
a a
a a
− −
= == =
= =
= =
= =
So, the Honda will be worth (much) more in 10 years.
59. Since 1 100, 0.7a r= = , we see that
( )100 0.7 nna = .
We need to find 5a and the value of n such that 0na = . Indeed, we see that
( )55 100 0.7 17 feeta = ≈ .
And to answer the second part, we note that 0na ≠ , for all values of n. As such, the object would never come to a rest, theoretically.
60. Since 1 200, 0.65a r= = , we see that
( ) 1200 0.65 nna −= .
We need to find 8a and the value of n such that 0na = . Indeed, we see that
( )78 200 0.65 9.80 feeta = ≈ .
And to answer the second part, we note that 0na ≠ , for all values of n. As such, the object would never come to a rest, theoretically.
Section 12.3
1729
61. Since
137,80036,000, 1.0536,000
a r= = = ,
we see that ( ) 136,000 1.05 nna −= . So, in
particular, ( )1011 36,000 1.05 58,640a = ≈ .
As such, in the year 2010, there will be approximately 58,640 students.
62. Since 1 20000, 1.05a r= = , we see
that ( ) 120,000 1.05 nna −= .
So, in particular, we have ( )51
52 20,000 1.05 240,815a = ≈ . So, there will be about 240,815 hits per week one year from now.
63. Since 1 1000, 0.90a r= = , we know
that ( ) 11000 0.9 nna −= .
First, find the value of 00 such that 1nn a ≤ .
Since 1 1
ln(0.001)ln(0.9)
1000(0.9) 1 (0.9) 0.001
67
n n
n
− −< ⇒ <
⇒ > ≈
We see that after 67 days, he will pay less than $1 per day. Next, the total amount of money he paid in January is given by:
( ) ( )31311
1
1 0.91000 0.9 1000 9618
1 0.9n
n
−
=
−= ≈
−∑
64. Since 1 0.01, 2a r= = , we see that
( ) 10.01 2 nna −= .
Hence, the total paid out in January is given by:
( )3131
1
1
1 20.01 2 0.011 2
$21,474,836.47
n
n
−
=
⎛ ⎞−= ⎜ ⎟−⎝ ⎠≅
∑
65. Use the formula 1ntrA P
n⎛ ⎞= +⎜ ⎟⎝ ⎠
. Note
that in the current problem, 0.05, 12,r n= = so that the formula
becomes:
( )12
120.051 1.004212
ttA P P⎛ ⎞= + =⎜ ⎟
⎝ ⎠
Let 12nt = , where n is the number of
months of investment. Also, let ( )1.0042 n
nA P= . First deposit of $100 gains interest for 36 months: ( )36
36 $100 1.0042A = Second deposit of $100 gains interest for 35 months: ( )35
35 $100 1.0042A =
Last deposit of $100 gains interest for 1 month: ( )11 $100 1.0042A = Now, sum the amount accrued for 36 deposits:
( )36
11 36
1
100 1.0042 n
n
A A −
=
+ + =∑…
Note that here 1 100, 1.0042a r= = . So,
( )36
36
1 1.0042So, 100
1 1.0042
3877.64
S⎛ ⎞−
= ⎜ ⎟⎜ ⎟−⎝ ⎠
≈
So, he saved $3,877.64 in 3 years.
Chapter 12
1730
66. Use the formula 1ntrA P
n⎛ ⎞= +⎜ ⎟⎝ ⎠
. Note
that in the current problem, 0.04,r = 52,n = so the formula becomes:
( )52
520.041 1.0007752
ttA P P⎛ ⎞= + =⎜ ⎟
⎝ ⎠
Let 52nt = , where n is number of weeks of
investment. Also, let ( )1.00077 nnA P= .
First deposit of $50 gains interest for 52 weeks: ( )52
52 50 1.00077A = Second deposit of $50 gains interest for 51 weeks: ( )51
51 50 1.00077A =
Last deposit of $50 gains interest for 1 week: ( )11 50 1.00077A = Now, sum the amount accrued for 52 weeks:
( )52
11 52
1
50 1.00077 n
n
A A −
=
+ + =∑…
Note that here 1 50, 1.00077a r= = . So,
( )52
52
1 1.00077So, 50
1 1.00077
2651.71
S⎛ ⎞−
= ⎜ ⎟⎜ ⎟−⎝ ⎠
≈
So, he saved $2,651.71 in one year.
67. Use the formula 1ntrA P
n⎛ ⎞= +⎜ ⎟⎝ ⎠
. Note that in the current problem,
0.06, 52,r n= = so that the formula becomes: ( )52
520.061 1.001252
ttA P P⎛ ⎞= + =⎜ ⎟
⎝ ⎠
Let 52nt = , where n is the number of weeks of investment. Also, let ( )1.0012 n
nA P= .
26 week investment: First deposit of $500 gains interest for 26 weeks: ( )26
26 500 1.0012A = Second deposit of $500 gains interest for 25 weeks: ( )25
25 500 1.0012A =
Last deposit of $500 gains interest for 1 week: ( )11 500 1.0012A = Now, the amount accrued for 26 weeks:
( )26
11 26
1500 1.0012 n
nA A −
=
+ + =∑…
Note that here 1 500, 1.0012a r= = . So,
( )26
26
1 1.0012500 13,196.88
1 1.0012S
⎛ ⎞−= ≈⎜ ⎟
⎜ ⎟−⎝ ⎠.
52 week investment: First deposit of $500 gains interest for 52 weeks: ( )52
52 500 1.0012A = Second deposit of $500 gains interest for 51 weeks: ( )51
51 500 1.0012A =
Last deposit of $500 gains interest for 1 week: ( )11 500 1.0012A = Now, the amount accrued for 52 weeks:
( )52
11 52
1500 1.0012 n
nA A −
=
+ + =∑…
Note that here 1 500, 1.0012a r= = . So,
( )52
52
1 1.0012500 26,811.75
1 1.0012S
⎛ ⎞−= ≈⎜ ⎟
⎜ ⎟−⎝ ⎠.
Section 12.3
1731
68. Use the formula 1ntrA P
n⎛ ⎞= +⎜ ⎟⎝ ⎠
. Note
that in the current problem, 0.05, 12,r n= = so that the formula becomes:
( )12
120.051 1.004212
ttA P P⎛ ⎞= + =⎜ ⎟
⎝ ⎠
Let 12nt = , where n is the number of months
of investment. Also, let ( )1.0042 nnA P= .
First deposit of $300 gains interest for 60 months: ( )60
36 $300 1.0042A = Second deposit of $300 gains interest for 59 months: ( )59
35 $300 1.0042A =
Last deposit of $300 gains interest for 1 month: ( )11 $300 1.0042A = Now, sum the amount accrued for 36 deposits:
( )60
11 60
1
300 1.0042 n
n
A A −
=
+ + =∑…
Note that here 1 300, 1.0042a r= = . So,
( )60
60
1 1.0042So, 300
1 1.0042
20, 422.66
S⎛ ⎞−
= ⎜ ⎟⎜ ⎟−⎝ ⎠
≈
So, he saved $20,422.66 in 5 years.
69. Here, 1 195,000a = and r = 1 + 0.065 = 1.065. So, 1195,000(1.065) , 1n
na n−= ≥ . So, its worth after 15 years is
15 $501,509a = .
70. Here, 11 39 and a r= = . Each time
the ball bounces, it travels up and down by the same distance. So, the total distance traveled is:
( ) ( )131
3 11 3
189 2 9 9 18 ft.
1n
n
∞
=
+ ⋅ = + =−∑
71. The sum is 12
12
11
=−
. 72. The finite sum is
30301 21 2 1 1,073,741,823
1 2−
⋅ = − =−
.
This is clearly the better deal than a one-time check of $10 million.
73. 1Should be 3
r = − 74. 1Should be 2a = , so that you have
1
12 2
nk
k
−
=
⋅∑ .
75. 1
Should use 3 all the way throughthe calculation. Also, 12(not 4).
ra
= −= −
76. Formula for only applies if 1.S r∞ <
77.
1
False. 1 1 1 11, , , is geometric with 1,2 4 8 2
a r− − = = −
78. False. Finite geometric series can always be computed, but infinite geometric series only exist if 1.r <
Chapter 12
1732
79. True. We could have either of the following:
1 1 11, , , (Here .)2 4 2
1 1 1 11, , , , (Here .)2 4 8 2
r
r
=
− − = −
80. False. Common ratio must be 1 in absolute value in order for an infinite geometric series to exist.
<
81. 1
1
11
This sum exists iff b 1. In such case, the
sum is . (Here, , , )1 1
n n
n
a a b a b ab
aa a a r b Sb r
∞−
=
∞
+ ⋅ + + ⋅ + =
<
= = =− −
∑… …
82. Since 1 1, 2, 21a r n= = = , and we can simplify the given series as:
20 202
0 0log10 2
k k
k k= =
=∑ ∑
we have 21
211 21 2,097,1511 2
S⎛ ⎞−
= =⎜ ⎟−⎝ ⎠.
83. This decimal can be represented as
( )47
1001100 1
1 100
47471 99
n
n
∞
=
= =−∑ .
84. If 21 x−
is the sum of a geometric
series, then 1 2a = and r = x. Hence, the
series is 1
12 n
nx
∞−
=∑ .
a. First five terms: 2 3 42 2 2 2 2x x x x+ + + +
b. This series converges for 1x < . 85. Since 1 1, 2, 50a r n= = − = , we have
( ) ( )( )
50501
1
1 21 2 1
1 2
375,299,968,947,541
k
k
−
=
⎛ ⎞− −− = ⎜ ⎟
⎜ ⎟− −⎝ ⎠= −
∑
86. Yes, you should get:
11 1 31, and S .13 21
3
a r ∞= = = =−
87. We expect that 0
11
n
n
xx
∞
=
=−∑ since as
n →∞ , the graph of 1 nny x= +…+ gets
closer to the graph of 11 x−
.
Notes on the graph: Solid curve: 2 3 4
1 1y x x x x= + + + +
Dashed curve: 21
1y
x=
−, assuming |x| < 1.
Section 12.4
1733
88. The plot of the two functions is:
The series will sum to 1
1 x+ assuming |x| <
1.
89. The plot of the two functions is:
The series will sum to 1
1 2x−, assuming
that |x| < ½
Section 12.4 Solutions –----------------------------------------------------------------------------- 1. Claim: 2 3, for all 1n n n≤ ≥ . Proof. Step 1: Show the statement is true for 1n = .
2 31 1≤ is true since 2 31 1 1= = . Step 2: Assume the statement is true for n k= : 2 3k k≤ Show the statement is true for 1n k= + : 2 3( 1) ( 1)k k+ ≤ +
2 2
3
3
3 2 2
3
( 1) 2 12 1 (by assumption)3 1 (since 2 3 , for 0)3 3 1 (since 3 0)
( 1)
k k kk kk k k k kk k k kk
+ = + +
≤ + +
≤ + + ≤ >
≤ + + + >
= +
This completes the proof. ■ 2. Claim: If 0 1, then 0 1, for all 1nx x n< < < < ≥ . Proof. Step 1: Show the statement is true for 1n = .
The statement If 0 1, then 0 1x x< < < < tautologically true. Step 2: Assume the statement is true for n k= : If 0 1, then 0 1kx x< < < < Show the statement is true for 1n k= + : 1If 0 1, then 0 1kx x +< < < < Starting from 0 1,kx< < multiply through the inequality by x (which is fine since 0x > . This yields the inequality
1
0 1k
k
xx
x x x+ ==
< ⋅ < ⋅ , which simplifies to
Chapter 12
1734
10 kx x+< < (1). Since from Step 1 we know that 1x < , we can use this in (1) to further conclude that 1 10 1, so that 0 1, as desired.k kx x x+ +< < < < < This completes the proof. ■ 3. Claim: 2 2 , for all 1nn n≤ ≥ . Proof. Step 1: Show the statement is true for 1n = .
12(1) 2≤ is true since both terms equal 2. Step 2: Assume the statement is true for n k= : 2 2kk ≤ Show the statement is true for 1n k= + : 12( 1) 2kk ++ ≤
1
1
2( 1) 2 2 2 2 (by assumption)2 2 (since 2 2 , for 0)2(2 ) 2
k
k k k
k k
k kk
+
+ = + ≤ +
≤ + ≤ >
= =
This completes the proof. ■ 4. Claim: 15 5 , for all 1n n n+< ≥ . Proof. Step 1: Show the statement is true for 1n = .
1 1 1 25 5 5 25+< = = is clearly true. Step 2: Assume the statement is true for n k= : 15 5k k+< Show the statement is true for 1n k= + : 1 25 5k k+ +<
1 1 25 5 5 5 5 (by assumption) 5k k k k+ + += ⋅ < ⋅ = This completes the proof. ■ 5. Claim: ! 2 , for all 4nn n> ≥ . Proof. Step 1: Show the statement is true for 4n = .
44! 2> since 4! 4 3 2 1 24= ⋅ ⋅ ⋅ = and 42 16= . Step 2: Assume the statement is true for n k= : ! 2kk > Show the statement is true for 1n k= + : 1( 1)! 2kk ++ >
1
( 1)! ( 1) !( 1) 2 (by assumption)2 2 (since 1 2 for 4)2
k
k
k
k k kk
k k+
+ = + ⋅
> + ⋅
> ⋅ + > ≥
=
This completes the proof. ■
Section 12.4
1735
6. Claim: For any 1c ≥ , (1 ) , for all 1nc nc n+ ≥ ≥ . Proof. Step 1: Show the statement is true for 1n = .
1(1 ) 1c c+ ≥ ⋅ is clearly true. Step 2: Assume the statement is true for n k= : (1 )kc kc+ ≥ Show the statement is true for 1n k= + : 1(1 ) ( 1)kc k c++ ≥ +
1
(1 )(1 ) (1 )(1 ) (1 )2(1 )(1 )(1 ) (since 1)(1 )
k
k
k k
k
k
k
ck c c cc cc c
cc c cc +
+ ≤ + +
≤ + + +
≤ + + +
= +
≤ + + ≥
= +
This completes the proof. ■ 7. Claim: ( 1)( 1) is divisible by 3, for all 1n n n n+ − ≥ . Proof. Step 1: Show the statement is true for 1n = .
(1)(1 1)(1 1) 1(2)(0) 0+ − = = , which is clearly divisible by 3. Step 2: Assume the statement is true for n k= : ( 1)( 1) is divisible by 3k k k+ − Show the statement is true for 1n k= + : ( 1)( 2)( ) is divisible by 3k k k+ + First, note that since by assumption ( 1)( 1) is divisible by 3k k k+ − , we know that there exists an integer m such that ( 1)( 1) 3k k k m+ − = (1) . Now, observe that
2 3 2( 1)( 2)( ) ( 3 2)( ) 3 2k k k k k k k k k+ + = + + = + + (At this point, write 2 3 and group the terms as shown.)k k k= −
3 2
2
2
( ) (3 3 )3 3( ) (by )3( )
k k k km k km k k
= − + +
= + +
= + +
(1)
Now, choose 2p m k k= + + to see that you have expressed ( 1)( 2)( )k k k+ + as 3p , thereby showing ( 1)( 2)( )k k k+ + is divisible by 3. This completes the proof. ■
Chapter 12
1736
8. Claim: 3 is divisible by 3, for all 1n n n− ≥ . Proof. Step 1: Show the statement is true for 1n = .
31 1 0− = , which is clearly divisible by 3. Step 2: Assume the statement is true for n k= : 3 is divisible by 3k k− Show the statement is true for 1n k= + : 3( 1) ( 1) is divisible by 3k k+ − + First, note that since by assumption 3 is divisible by 3k k− , we know that there exists an integer m such that 3 3k k m− = (1) . Now, observe that
( ) ( )3 3 2 3 2( 1) ( 1) 3 3 1 1 3 2k k k k k k k k k+ − + = + + + − + = + + (At this point, write 2 3 and group the terms as shown.)k k k= −
3 2
2
2
( ) (3 3 )3 3( ) (by )3( )
k k k km k km k k
= − + +
= + +
= + +
(1)
Now, choose 2p m k k= + + to see that you have expressed 3( 1) ( 1)k k+ − + as3p , thereby showing 3( 1) ( 1)k k+ − + is divisible by 3. This completes the proof. ■ 9. Claim: 2 3 is divisible by 2, for all 1n n n+ ≥ . Proof. Step 1: Show the statement is true for 1n = .
21 3(1) 4 2(2)+ = = , which is clearly divisible by 2. Step 2: Assume the statement is true for n k= : 2 3 is divisible by 2k k+ Show the statement is true for 1n k= + : 2( 1) 3( 1) is divisible by 2k k+ + + First, note that since by assumption 2 3 is divisible by 2k k+ , we know that there exists an integer m such that 2 3 2k k m+ = (1) . Now, observe that
( ) ( )2 2 2( 1) 3( 1) 2 1 3 1 5 4k k k k k k k+ + + = + + + + = + + (At this point, write 5 3 2 and group the terms as shown.)k k k= +
2( 3 ) (2 4)2 2( 2) (by )2( 2)
k k km km k
= + + += + += + +
(1)
Now, choose 2p m k= + + to see that you have expressed 2( 1) 3( 1)k k+ + + as 2 p , thereby showing 2( 1) 3( 1)k k+ + + is divisible by 2. This completes the proof. ■
Section 12.4
1737
10. Claim: ( 1)( 2) is divisible by 6, for all 1n n n n+ + ≥ . Proof. Step 1: Show the statement is true for 1n = .
(1)(1 1)(1 2) 1(2)(3) 6+ + = = , which is clearly divisible by 6. Step 2: Assume the statement is true for n k= : ( 1)( 2) is divisible by 6k k k+ + Show the statement is true for 1n k= + : ( 1)( 2)( 3) is divisible by 6k k k+ + + First, note that since by assumption ( 1)( 2) is divisible by 6k k k+ + , we know that there exists an integer m such that ( 1)( 2) 6k k k m+ + = (1) . Now, observe that
2
3 2 2
3 2
( 1)( 2)( 3) ( 1)( 5 6)5 6 5 66 11 6
k k k k k kk k k k kk k k
+ + + = + + +
= + + + + +
= + + +
2 2 2(At this point, write 6 3 3 and
11 2 9 and group the terms as shown.)k k k
k k k= +
= +
3 2 2
2
For any , one of theseconsecutive integers mustbe even. So, the productmust be divisible by 2.
( 3 2 ) (3 9 6)6 3( 3 2) (by )6 3 ( 2)( 1)
6 3(2 ), for some integer 6( )
k
k k k k km k km k k
m s sm s
= + + + + +
= + + += + + +
= += +
(1)
Now, choose p m s= + to see that you’ve expressed ( 1)( 2)( 3)k k k+ + + as 6 p , thereby showing ( 1)( 2)( 3)k k k+ + + is divisible by 6. This completes the proof. ■
11. Claim: 2 4 ... 2 ( 1), for all 1n n n n+ + + = + ≥ . Proof. Step 1: Show the statement is true for 1n = .
2 1(2)= , which is clearly true. Step 2: Assume the statement is true for n k= : 2 4 ... 2 ( 1)k k k+ + + = + Show the statement is true for 1n k= + : 2 4 ... 2( 1) ( 1)( 2)k k k+ + + + = + + Observe that
( )2 4 ... 2 2( 1) 2 4 ... 2 2( 1)( 1) 2( 1) (by assumption)
( 2)( 1)
k k k kk k kk k
+ + + + + = + + + + +
= + + += + +
This completes the proof. ■
Chapter 12
1738
12. Claim: 21 3 ... (2 1) , for all 1n n n+ + + − = ≥ Proof. Step 1: Show the statement is true for 1n = .
21 1= , which is clearly true. Step 2: Assume the statement is true for n k= : 21 3 ... (2 1)k k+ + + − = Show the statement is true for 1n k= + : 21 3 ... (2( 1) 1) ( 1)k k+ + + + − = +
( )2 1
2
2
1 3 ... (2 1) (2( 1) 1) 1 3 ... (2 1) (2 1)
(2 1) (by assumption)( 1) (factor)
k
k k k k
k kk
+
+ + + − + + − = + + + − + +
= + +
= +
This completes the proof. ■ 13. Claim: 13 1
21 3 ... 3 , for all 1nn n+ −+ + + = ≥ Proof. Step 1: Show the statement is true for 1n = .
21 3 121 3 4−+ = = is clearly true.
Step 2: Assume the statement is true for n k= : 13 121 3 ... 3 kk + −+ + + =
Show the statement is true for 1n k= + : 21 3 121 3 ... 3 kk ++ −+ + + =
( )1
1 1
1 2
1 1
13 12
3 1 2(3 )2
3(3 ) 1 3 12 2
1 3 ... 3 3 1 3 ... 3 3
3 (by assumption)k
k k
k k
k k k k
k+
+ +
+ +
+ +
+−
− +
− −
+ + + + = + + + +
= +
=
= =
This completes the proof. ■ 14. Claim: 12 4 ... 2 2 2, for all 1n n n++ + + = − ≥ Proof. Step 1: Show the statement is true for 1n = .
1 12 2 2+= − is clearly true. Step 2: Assume the statement is true for n k= : 12 4 ... 2 2 2k k++ + + = − Show the statement is true for 1n k= + : 1 22 4 ... 2 2 2k k+ ++ + + = −
( )( )
1 1
1 1
1
2
2 4 ... 2 2 2 4 ... 2 2
2 2 2 (by assumption)
2(2 ) 22 2
k k k k
k k
k
k
+ +
+ +
+
+
+ + + + = + + + +
= − +
= −
= −
This completes the proof. ■
Section 12.4
1739
15. Claim: ( 1)(2 1)2 261 ... , for all 1n n nn n+ ++ + = ≥
Proof. Step 1: Show the statement is true for 1n = .
1(1 1)(2(1) 1)2 66 61 1+ += = = is clearly true.
Step 2: Assume the statement is true for n k= : ( 1)(2 1)2 261 ... k k kk + ++ + =
Show the statement is true for 1n k= + : 2 3
( 1)( 2)(2( 1) 1)2 261 ... ( 1)
k
k k kk+
+ + + ++ + + =
( )
2
3 2 2
3 2
2 2 2 2 2 2
( 1)(2 1) 26
( 1)(2 1) 6( 1)6
2 3 6 12 66
2 9 13 66
( 1)( 2)(2 3)6
1 ... ( 1) 1 ... ( 1)
( 1) (by assumption)
=
k k k
k k k k
k k k k k
k k k
k k k
k k k k
k+ +
+ + + +
+ + + + +
+ + +
+ + +
+ + + + = + + + +
= + +
=
=
=
This completes the proof. ■ 16. Claim:
2 2( 1)3 341 ... , for all 1n nn n++ + = ≥
Proof. Step 1: Show the statement is true for 1n = .
2 21 (1 1)341 1+= = is clearly true.
Step 2: Assume the statement is true for n k= : 2 2( 1)3 3
41 ... k kk ++ + =
Show the statement is true for 1n k= + : 2 2( 1) ( 2)3 341 ... ( 1) k kk + ++ + + =
( )
( )
2 2
2 2 3
2 2
2 2
3 3 3 3 3 3
( 1) 34
( 1) 4( 1)4
( 1) 4( 1)
4
( 1) ( 2)4
1 ... ( 1) 1 ... ( 1)
( 1) (by assumption)k k
k k k
k k k
k k
k k k k
k+
+ + +
+ + +
+ +
+ + + + = + + + +
= + +
=
=
=
This completes the proof. ■
Chapter 12
1740
17. Claim: 1 1 11 2 2 3 ( 1) 1... , for all 1n
n n n n⋅ ⋅ + ++ + + = ≥ Proof. Step 1: Show the statement is true for 1n = .
1 1 11 2 1 1 2⋅ += = , which is clearly true.
Step 2: Assume the statement is true for n k= : 1 1 11 2 2 3 ( 1) 1... k
k k k⋅ ⋅ + ++ + + = Show the statement is true for 1n k= + : 11 1 1
1 2 2 3 ( 1)( 2) 2... kk k k
+⋅ ⋅ + + ++ + + =
( )
2
2
1 1 1 1 1 1 1 11 2 2 3 ( 1) ( 1)( 2) 1 2 2 3 ( 1) ( 1)( 2)
11 ( 1)( 2)
( 2) 1( 1)( 2)
2 1( 1)( 2)
( 1)( 1)( 2)
1( 2)
... ...
(by assumption)k k k k k k k k
kk k k
k kk k
k kk k
kk k
kk
⋅ ⋅ + + + ⋅ ⋅ + + +
+ + +
+ ++ +
+ ++ +
++ +
++
+ + + + = + + + +
= +
=
=
=
=
This completes the proof. ■ 18. 1 1
2 3 ( 1)( 2) 2( 2)... , for all 1nn n n n⋅ + + ++ + = ≥
Proof. Step 1: Show the statement is true for 1n = .
1 1 12 3 2(1 2) 2(3)⋅ += = , which is clearly true.
Step 2: Assume the statement is true for n k= : 1 12 3 ( 1)( 2) 2( 2)... k
k k k⋅ + + ++ + = Show the statement is true for 1n k= + : 11 1
2 3 ( 2)( 3) 2( 3)... kk k k
+⋅ + + ++ + =
( )
2
1 1 1 1 1 12 3 ( 1)( 2) ( 2)( 3) 2 3 ( 1)( 2) ( 2)( 3)
12( 2) ( 2)( 3)
( 3) 22( 2)( 3)
3 22( 2)( 3)
( 2)
... ...
(by assumption)k k k k k k k k
kk k k
k kk k
k kk k
k
⋅ + + + + ⋅ + + + +
+ + +
+ ++ +
+ ++ +
+
+ + + = + + +
= +
=
=
= ( 1)2 ( 2)
kk
+
+ ( 3)
12( 3)
k
kk
+
++=
This completes the proof. ■
Section 12.4
1741
19. Claim: ( 1)( 2)3(1 2) (2 3) ... ( 1) , for all 1n n nn n n+ +⋅ + ⋅ + + + = ≥ .
Proof. Step 1: Show the statement is true for 1n = .
1(1 1)(1 2) 1(2)(3)3 3(1 2) 2+ +⋅ = = = , which is clearly true.
Step 2: Assume the statement is true for n k= : ( 1)( 2)
3(1 2) (2 3) ... ( 1) k k kk k + +⋅ + ⋅ + + + = Show the statement is true for 1n k= + :
( 1)( 2)( 3)3(1 2) (2 3) ... ( 1)( 2) k k kk k + + +⋅ + ⋅ + + + + =
( )( 1)( 2)
3
( 1)( 2) 3( 1)( 2)3
( 1)( 2)( 3)3
(1 2) (2 3) ... ( 1) ( 1)( 2) (1 2) (2 3) ... ( 1) ( 1)( 2)
( 1)( 2) (by assumption)
(factor out ( 1)( 2))
k k k
k k k k k
k k k
k k k k k k k k
k k
k k
+ +
+ + + + +
+ + +
⋅ + ⋅ + + + + + + = ⋅ + ⋅ + + + + + +
= + + +
=
= + +
This completes the proof. ■ 20. Claim: ( 1)(2 7)
6(1 3) (2 4) ... ( 2) , for all 1n n nn n n+ +⋅ + ⋅ + + + = ≥ . Proof. Step 1: Show the statement is true for 1n = .
1(1 1)(2(1) 7) 1(2)(9)6 6(1 3) 3+ +⋅ = = = , which is clearly true.
Step 2: Assume the statement is true for n k= : ( 1)(2 7)
6(1 3) (2 4) ... ( 2) k k kk k + +⋅ + ⋅ + + + = Show the statement is true for 1n k= + :
2 9
( 1)( 2)(2( 1) 7)6(1 3) (2 4) ... ( 1)( 3)
k
k k kk k+
+ + + +⋅ + ⋅ + + + + = ( )
[ ]
2
( 1)(2 7)6
( 1)(2 7) 6( 1)( 3)6
( 1) (2 7) 6( 3)6
( 1) 2 13 18
6(
(1 3) (2 4) ... ( 2) ( 1)( 3) (1 3) (2 4) ... ( 2) ( 1)( 3)
( 1)( 3) (by assumption)
(factor out ( 1))
k k k
k k k k k
k k k k
k k k
k k k k k k k k
k k
k
+ +
+ + + + +
+ + + +
⎡ ⎤+ + +⎣ ⎦
⋅ + ⋅ + + + + + + = ⋅ + ⋅ + + + + + +
= + + +
=
= +
=
= 1)( 2)(2 9)6
k k k+ + +
This completes the proof. ■
Chapter 12
1742
21. Claim: 1 111 ... , for all 1nn x
xx x n− −−+ + + = ≥ .
Proof. Step 1: Show the statement is true for 1n = .
1111 x
x−−= , which is clearly true.
Step 2: Assume the statement is true for n k= : 1 111 ... kk x
xx x − −−+ + + =
Show the statement is true for 1n k= + : 1111 ... kk x
xx x +−−+ + + =
( )1 1
11
1 (1 )1
1
1 ... 1 ...k
k k
k
k k k k
kxx
x x xx
x
x x x x
x
− −
−−
− + −−
−
+ + + = + + +
= +
=
=kx+
1
1
11
k
k
x xx
xx+
−−
−−=
This completes the proof. ■ 22. Claim: 1 1 1 1
2 4 2 2... 1 , for all 1n n n+ + + = − ≥ .
Proof. Step 1: Show the statement is true for 1n = .
11 12 2
1= − , which is clearly true.
Step 2: Assume the statement is true for n k= : 1 1 1 12 4 2 2
... 1k k+ + + = −
Show the statement is true for 1n k= + : 1 11 1 1 12 4 2 2
... 1k k+ ++ + + = −
( )( )
1 1
1
1
1
1
1
1
1 1 1 1 1 1 1 12 4 2 42 2 2 2
1 12 2
2 2 12
2 12
12
... ...
1
1
k k k k
k k
k
k
k
k
k
+ +
+
+
+
+
+
+
− +
−
+ + + + = + + + +
= − +
=
=
= −
This completes the proof. ■
Section 12.4
1743
23. Claim: [ ]1 1 1 12( ) ... ( ( 1) ) 2 ( 1) , for all 1na a d a n d a n d n+ + + + + − = + − ≥ . Proof. Step 1: Show the statement is true for 1n = .
[ ] [ ]1 11 1 12 22 (1 1) 2a a d a= + − = , which is clearly true.
Step 2: Assume the statement is true for n k= : [ ]1 1 1 12( ) ... ( ( 1) ) 2 ( 1)ka a d a k d a k d+ + + + + − = + −
Show the statement is true for 1n k= + :
11 1 1 12( ) ... ( (( 1) 1) ) 2 ( 1 1)
k
ka a d a k d a k d+⎡ ⎤⎢ ⎥+ + + + + + − = + + −⎢ ⎥⎣ ⎦
Observe that
( )[ ]
[ ]
1 1 1 1
1 1 1 1
1 12
( 1)1 12
11 2
11 2
( 1) 112 2
( 1)12
( ) ... ( ( 1) ) ( )( ) ... ( ( 1) ) ( )
2 ( 1) ( )
( 1) ( 1)( 1) ( )
2 ( )
2
k
k k
k
k
k k
k
a a d a k d a kda a d a k d a kd
a k d a kd
ka d a kdk a kdk a kd
a kd
a kd
−
−
+
+ +
+
+ + + + + − + +
= + + + + + − + +
= + − + +
= + + +
= + + +
= + +
= +
= +
This completes the proof. ■ 24. Claim: ( )1 1
1 1 1 11... , for all 1nn rra ra r a a n− −
−+ + + = ≥ .
Proof. Step 1: Show the statement is true for 1n = .
( )111 11
1
,rra a−
−
=
= which is clearly true.
Step 2: Assume the statement is true for n k= : ( )1 11 1 1 11... kk r
ra ra r a a− −−+ + + =
Show the statement is true for 1n k= + : ( )111 1 1 11... kk r
ra ra r a a+−−+ + + =
Observe that ( )( )
1 11 1
1 11 1 1 1 1 1 1 1
11 11
1(1 ) (1 )1
... ...k
kk k
k k k k
krr
a a ra r r a rr
a ra r a r a a ra r a r a
a r a
− −
−−
−− + −−
+ + + + = + + + +
= +
= = 1kr a+ ( )1 11 1
11 1
k kr a rr r a
+ +− −− −=
This completes the proof. ■
Chapter 12
1744
25. Label the disks 1, 2, and 3 (smallest = 1 and largest = 3), and label the posts A, B, and C. The following are the moves on would take to solve the problem in the fewest number of step. (Note: The manner in which the disks are stacked (from top to bottom) on each peg form the contents of each cell; a blank cell means that no disk is on that peg in that particular move.)
Post A Post B Post C Initial placement 1
2 3
Move 1 2 3
1
Move 2 3 1 2 Move 3 3 1
2 Move 4 3 1
2 Move 5 1 3 2 Move 6 1 2
3
Move 7 1 2 3
So, the puzzle can be solved in as few as 7 steps. An argument as to why this is actually the fewest number of steps is beyond the scope of the text. (Note: Alternatively, we could have initially placed disk 1 on post C, and proceeded in a similar manner.) 26. Label the disks 1, 2, 3, and 4 (smallest = 1 and largest = 4), and label the posts A, B, and C. The following are the moves on would take to solve the problem in the fewest number of step. (Note: The manner in which the disks are stacked (from top to bottom) on each peg form the contents of each cell; a blank cell means that no disk is on that peg in that particular move.) (This solution continues onto the next page.)
Post A Post B Post C Initial placement 1
2 3 4
Move 1 2 3 4
1
Section 12.4
1745
Move 2 3 4
1 2
Move 3 3 4
1 2
Move 4 4 3 1 2
Move 5 1 4
3
2
Move 6 1 4
2 3
Move 7 4
1 2 3
Move 8 1 2 3
4
Move 9 2 3
1 4
Move 10 2
3 1 4
Move 11 1 2
3 4
Move 12 1 2
3 4
Move 13 2 1 3 4
Move 14 1 2 3 4
Move 15 1 2 3 4
So, the puzzle can be solved in as few as 15 steps. An argument as to why this is actually the fewest number of steps is beyond the scope of the text. (Note: Alternatively, we could have initially placed disk 1 on post C, and proceeded in a similar manner.) 27. Using the strategy of Problem 25 (and 26), this puzzle can be solved in as few as 31 steps. Have fun trying it!! There are many classical references that discuss this problem, as well as several internet sites.
Chapter 12
1746
28. What follows is not a formal proof, but rather an intuitive discussion that hints at what the proof ought to be. To establish a more formal argument, one needs to know how to work with recurrence relations, a topic not addressed in the text. But, see if you can follow the spirit of the argument as presented below. First, note that the general strategy used to solve the problem with n disks is to:
- Move the topmost 1n − disks from the left peg to one of the other two, - Move the bottommost disk from left to right, - Move the 1n − disks from its current location to the right.
And, this process is replicated inductively until you are down to one disks. As such, since this amounts to two steps applied in succession for each of the n disks, this would imply that the number of steps would be 2n . However, 1 is subtracted since you start off with the problem with an initial configuration. So, truly, the problem can be solved in as few as 2 1n − moves. 29. If n = 2, then the number of wires needed is 2(2 1)
2 1− = . Assume the formula holds for k cities. Then, if k cities are to be connected directly to each other, the number of wires needed is ( 1)
2k k− . Now, if one more city is added, then
you must have k additional wires to connect the telephone to this additional city. So, the total wires in such case is 2( 1) ( 1)2
2 2 2k k k kk k kk − ++ −+ = = . Thus, the statement holds for
k+1 cities. Hence, we have proven the statement by induction. 30. For n = 3, the formula becomes (3-2)(180)=180. The formula is true when n =3 (a triangle), since it is known that the sum of the interior angles of a triangle is 180 degrees. Assume the formula is true for k sided regular polygon. Then, the sum of the interior angles is (k-2)(180). Looking at a few particular examples (for particular values of k) we observe that you can divide a k-sided polygon into (k-2) triangles, and each time we increase the number of sides by one, one additional triangle is formed. As such, the interior angle sum increases by 1. Hence, adding one side to a k-side polygon then has interior angle sum 180 + (k-2)(180) = 180(k-1), thereby showing the statement holds for k+1. Hence, we have proven the statement by induction. 31. False. You first need to show that 1S is true.
32. False. You must show that if kS is true, then 1kS + is true, for all values of k, not just the first pair of consecutive values of k.
Section 12.4
1747
33. Claim: 2( 1)(2 1)(3 3 1)4
301
, for all 1n
n n n n n
kk n+ + + −
=
= ≥∑ .
Proof. Step 1: Show the statement is true for 1n = .
21(1 1)(2(1) 1)(3(1) 3(1) 1)4301 1+ + + −= = , which is clearly true.
Step 2: Assume the statement is true for n p= : 2( 1)(2 1)(3 3 1)4
301
pp p p p p
k
k + + + −
=
=∑
Show the statement is true for 1n p= + : 2
1( 1)( 2)(2( 1) 1)(3( 1) 3( 1) 1)4
301
pp p p p p
kk
++ + + + + + + −
=
=∑
Observe that
2
2 4
2 3
4 3 2
14 4 4
1 1
( 1)(2 1)(3 3 1) 430
( 1)(2 1)(3 3 1) 30( 1)30
( 1) (2 1)(3 3 1) 30( 1)
30
( 1) 6 39 91 89 30
30
( 1)
( 1) (by assumption)
p p
k k
p p p p p
p p p p p p
p p p p p p
p p p p p
k k p
p
+
= =
+ + + −
+ + + − + +
⎡ ⎤+ + + − + +⎣ ⎦
⎡ ⎤+ + + + +⎣ ⎦
= + +
= + +
=
=
=
∑ ∑
(1)
Next, note that multiplying out these terms yields ( )( )( )
( )( )
2 22
3 2 2
5 4 3 2
3 2 2 3 3 6 3 3 3 1( 1)( 2)(2( 1) 1)(3( 1) 3( 1) 1)30 30
2 9 13 6 3 9 5
30
6 45 130 180 119 3030
p p p p p pp p p p p
p p p p p
p p p p p
+ + + + + + + −+ + + + + + + −
+ + + + +
+ + + + +
=
=
= (2)
Comparing (1) and (2), we see they are, in fact, equal. This is precisely what we needed to show to establish the claim. This completes the proof. ■
Chapter 12
1748
34. Claim: 2 2 2( 1) (2 2 1)5
121
, for all 1n
n n n n
kk n+ + −
=
= ≥∑ .
Proof. Step 1: Show the statement is true for 1n = .
2 2 21 (1 1) (2(1) 2(1) 1)5121 1+ + −= = , which is clearly true.
Step 2: Assume the statement is true for n p= : 2 2 2( 1) (2 2 1)5
121
pp p p p
k
k + + −
=
=∑
Show the statement is true for 1n p= + :
22 6 3
2 2 21
( 1) (( 1) 1) (2( 1) 2( 1) 1)512
1
p pp
p p p p
kk
= + ++
+ + + + + + −
=
=∑
Observe that
2 2 2
2 2 2 5
2 2 2 3
4 3 2
15 5 5
1 1
( 1) (2 2 1) 512
( 1) (2 2 1) 12( 1)12
( 1) (2 2 1) 12( 1)
12
( 1) 2 14 35 36 12
12
( 1)
( 1) (by assumption)
p p
k k
p p p p
p p p p p
p p p p p
p p p p p
k k p
p
+
= =
+ + −
+ + − + +
⎡ ⎤+ + − + +⎣ ⎦
⎡ ⎤+ + + + +⎣ ⎦
= + +
= + +
=
=
=
∑ ∑
(1)
Next, note that multiplying out these terms yields 2 2 2 2 2 2
2 2 2
2 4 3 2
( 1) (( 1) 1) (2( 1) 2( 1) 1) ( 1) ( 2) (2 6 3)12 12
( 1) ( 4 4)(2 6 3)12
( 1) (2 14 35 36 12)12
p p p p p p p p
p p p p p
p p p p p
+ + + + + + − + + + +
+ + + + +
+ + + + +
=
=
= (2)
Comparing (1) and (2), we see they are, in fact, equal. This is precisely what we needed to show to establish the claim. This completes the proof. ■
Section 12.4
1749
35. Claim: ( ) ( ) ( )1 1 11 21 1 ... 1 1, for all 1n n n+ ⋅ + ⋅ ⋅ + = + ≥ .
Proof. Step 1: Show the statement is true for 1n = .
( )111 1 1 2+ = + = , which is clearly true.
Step 2: Assume the statement is true for n k= : ( ) ( ) ( )1 1 11 21 1 ... 1 1k k+ ⋅ + ⋅ ⋅ + = +
Show the statement is true for 1n k= + : ( ) ( ) ( )1 1 11 2 1
2
1 1 ... 1 ( 1) 1k
k
k+
= +
+ ⋅ + ⋅ ⋅ + = + +
Observe that ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
( )
1 1 1 1 1 1 1 11 2 1 1 2 1
11
1 1 ... 1 1 1 1 ... 1 1
( 1) 1 (by assumption)
( 1)
k k k k
kk
k
+ +
+
⎡ ⎤+ ⋅ + ⋅ ⋅ + ⋅ + = + ⋅ + ⋅ ⋅ + ⋅ +⎣ ⎦= + ⋅ +
= + 1 11
kk+ ++( )
2k= +
This completes the proof. ■ 36. Claim: 2 2 is a factor of , for all 1n nx y x y n+ − ≥ . Proof. Step 1: Show the statement is true for 1n = .
2 2 ( )( )x y x y x y− = + − , so that x y+ is a factor. Step 2: Assume the statement is true for n k= :
2 2 is a factor of k kx y x y+ − Show the statement is true for 1n k= + :
2( 1) 2( 1) is a factor of k kx y x y+ ++ − Observe that
( ) ( )( )
2( 1) 2( 1) 2 2 2 2
2 2 2 2 2 2 2 2
2 2 2 2 2 2
2 2 2 2
2 2
(Add zero.)
( )( )
( ) ( , ) ( )( ) (by assumption)
k k k k
k k k k
k k k
k k k
k
x y x x y yx x y x y x y y
x x y y x y
x x y y x y x y
x x y p x y y x y x y
+ +− = −
= − + −
= − + −
= − + + −
= + + + − (1)
where ( , )p x y is some polynomial in x and y. Hence, since ( )x y+ can be factored out of both terms in (1), we conclude that, in fact, ( )x y+ is indeed a factor of 2( 1) 2( 1)k kx y+ +− . This completes the proof. ■
Chapter 12
1750
37. Claim: ( )1 2 1ln ... ln( ) ... ln( ), for all 1n nc c c c c n⋅ ⋅ ⋅ = + + ≥ . Proof. Step 1: Show the statement is true for 1n = .
( )1 1ln ln( )c c= is clearly true. Step 2: Assume the statement is true for n k= :
( )1 2 1ln ... ln( ) ... ln( )k kc c c c c⋅ ⋅ ⋅ = + + Show the statement is true for 1n k= + :
( )1 2 1 1 1ln ... ln( ) ... ln( )k kc c c c c+ +⋅ ⋅ ⋅ = + + Observe that
1 2 1 1 1
Treat as a single quantity
1 1
ln ... ln( ... ) ln( )
ln( ) ... ln( ) ln( ) (by assumption)
k k k k
k k
c c c c c c c
c c c
+ +
+
⎛ ⎞⎜ ⎟
⋅ ⋅ ⋅ ⋅ = ⋅ ⋅ +⎜ ⎟⎜ ⎟⎝ ⎠
= + + +
This completes the proof. ■ 38. 2,706,800. Yes. 39. 255
256 . Yes. Section 12.5 Solutions -------------------------------------------------------------------------------
1. ( )
7 7! 7! 7 6 5 4!3 7 3 !3! 4!3!⎛ ⎞ ⋅ ⋅ ⋅
= = =⎜ ⎟ −⎝ ⎠ 4! ( )35
3 2 1=
⋅ ⋅
2. ( )
8 8! 8! 8 7 6!2 8 2 !2! 6!2!⎛ ⎞ ⋅ ⋅
= = =⎜ ⎟ −⎝ ⎠ 6! ( )28
2 1=
⋅
3. ( )
10 10! 10! 10 9 8!8 10 8 !8! 2!8!
⎛ ⎞ ⋅ ⋅= = =⎜ ⎟ −⎝ ⎠ 8! ( )
452 1
=⋅
4. ( )
23 23! 23! 23 22 21!21 23 21 !21! 2!21!⎛ ⎞ ⋅ ⋅
= = =⎜ ⎟ −⎝ ⎠ 21! ( )253
2 1=
⋅
5. ( )
17 17! 17!0 17 0 !0!
⎛ ⎞= =⎜ ⎟ −⎝ ⎠ 17!
1 10!0!
= =
6. ( )
100 100! 100!0 100 0 !0!
⎛ ⎞= =⎜ ⎟ −⎝ ⎠ 100!
1 10!0!
= =
7. ( )
99 99! 99!99 99 99 !99!⎛ ⎞
= =⎜ ⎟ −⎝ ⎠ 0! 99!1 10!
= =
Section 12.5
1751
8. ( )
52 52! 52!52 52 52 !52!⎛ ⎞
= =⎜ ⎟ −⎝ ⎠ 0! 52!1 10!
= =
9. ( )
48 48! 48! 48 47 46 45!45 48 45 !45! 3!45!⎛ ⎞ ⋅ ⋅ ⋅
= = =⎜ ⎟ −⎝ ⎠ 45! ( )17,296
3 2 1=
⋅ ⋅
10. ( )
29 29! 29! 29 28 27 26!26 29 26 !26! 3!26!⎛ ⎞ ⋅ ⋅ ⋅
= = =⎜ ⎟ −⎝ ⎠ 26! ( )3654
3 2 1=
⋅ ⋅
11. 4
4 4
0
4 0 3 1 2 2 1 3 0 4
4 0 3 1 2 2 1 3 0 44! 4! 4! 4! 4!4!0! 3!1! 2!2! 1!3! 0!4!
1 4 6 4 1
4 3 2
4( 2) 2
4 4 4 4 42 2 2 2 2
0 1 2 3 4
2 2 2 2 2
8 24 32 16
k k
k
x xk
x x x x x
x x x x x
x x x x
−
=
= = = = =
⎛ ⎞+ = ⎜ ⎟
⎝ ⎠⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
= + + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
= + + + +
= + + + +
∑
12. 5
5 5
0
5 0 4 1 3 2 2 3 1 4 0 5
5 0 4 1 3 2 2 3 1 4 0 55! 5! 5! 5! 5! 5!5!0! 4!1! 3!2! 2!3! 1!4! 0!5!
1 5 10 10 5 1
5 4 3
5( 3) 3
5 5 5 5 5 53 3 3 3 3 3
0 1 2 3 4 5
3 3 3 3 3 3
15 90 27
k k
k
x xk
x x x x x x
x x x x x x
x x x
−
=
= = = = = =
⎛ ⎞+ = ⎜ ⎟
⎝ ⎠⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
= + + + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
= + + + + +
= + + +
∑
20 405 243x x+ +
13. 5
5 5
0
5 0 4 1 3 2 2 3 1 4 0 5
5 0 4 1 3 2 2 35! 5! 5! 5! 5!5!0! 4!1! 3!2! 2!3! 1!4!
1 5 10 10 5
5( 3) ( 3)
5 5 5 5 5 5( 3) ( 3) ( 3) ( 3) ( 3) ( 3)
0 1 2 3 4 5
( 3) ( 3) ( 3) ( 3)
k k
k
y yk
y y y y y y
y y y y
−
=
= = = = =
⎛ ⎞− = −⎜ ⎟
⎝ ⎠⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
= − + − + − + − + − + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
= − + − + − + − +
∑
1 4 0 55!0!5!
1
5 4 3 2
( 3) ( 3)
15 90 270 405 243
y y
y y y y y=
− + −
= − + − + −
Chapter 12
1752
14. 4
4 4
0
4 0 3 1 2 2 1 3 0 4
4 0 3 1 2 2 1 3 0 44! 4! 4! 4! 4!4!0! 4!0! 4!0! 4!0! 4!0!
1 4 6 4 1
4 3
4( 4) ( 4)
4 4 4 4 4( 4) ( 4) ( 4) ( 4) ( 4)
0 1 2 3 4
( 4) ( 4) ( 4) ( 4) ( 4)
16 96
k k
ky y
k
y y y y y
y y y y y
y y
−
=
= = = = =
⎛ ⎞− = −⎜ ⎟
⎝ ⎠⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
= − + − + − + − + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
= − + − + − + − + −
= − +
∑
2 256 256y y− +
15. 5
5 5
0
5 0 4 1 3 2 2 3 1 4 0 5
5 0 4 1 3 2 2 3 1 4 0 55! 5! 5! 5! 5! 5!5!0! 4!1! 3!2! 2!3! 1!4! 0!5!
1 5 10 10 5 1
5 4 3 2
5( )
5 5 5 5 5 50 1 2 3 4 5
5 10
k k
k
x y x yk
x y x y x y x y x y x y
x y x y x y x y x y x y
x x y x y
−
=
= = = = = =
⎛ ⎞+ = ⎜ ⎟
⎝ ⎠⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
= + + + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
= + + + + +
= + + +
∑
2 3 4 510 5x y xy y+ +
16. 6
6 6
0
6 0 5 1 4 2 3 3
2 4 1 5 0 6
6 0 5 1 4 26! 6! 6! 6!6!0! 5!1! 4!2! 3!3!
1 6 15 20
6( ) ( )
6 6 6 6( ) ( ) ( ) ( )
0 1 2 3
6 6 6( ) ( ) ( )
4 5 6
( ) ( ) ( )
k k
k
x y x yk
x y x y x y x y
x y x y x y
x y x y x y
−
=
= = = =
⎛ ⎞− = −⎜ ⎟
⎝ ⎠⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
= − + − + − + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ − + − + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
= − + − + − +
∑
3 3
2 4 1 5 0 66! 6! 6!2!4! 1!5! 0!6!
15 6 1
6 5 4 2 3 3 2 4 5 6
( )
( ) ( ) ( )
6 15 20 15 6
x y
x y x y x y
x x y x y x y x y xy y= = =
−
+ − + − + −
= − + − + − +
Section 12.5
1753
17. 3
3 3
0
3 0 2 1 1 2 0 3
3 0 2 1 1 2 0 33! 3! 3! 3!3!0! 2!1! 1!2! 0!3!
1 3 3 1
3 2 2 3
3( 3 ) (3 )
3 3 3 3(3 ) (3 ) (3 ) (3 )
0 1 2 3
(3 ) (3 ) (3 ) (3 )
9 27 27
k k
kx y x y
k
x y x y x y x y
x y x y x y x y
x x y xy y
−
=
= = = =
⎛ ⎞+ = ⎜ ⎟
⎝ ⎠⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
= + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
= + + +
= + + +
∑
18. 3
3 3
0
3 0 2 1 1 2 0 3
3 0 2 1 1 2 0 33! 3! 3! 3!3!0! 2!1! 1!2! 0!3!
1 3 3 1
3 2 2
3(2 ) (2 ) ( )
3 3 3 3(2 ) ( ) (2 ) ( ) (2 ) ( ) (2 ) ( )
0 1 2 3
(2 ) ( ) (2 ) ( ) (2 ) ( ) (2 ) ( )
8 12 6
k k
k
x y x yk
x y x y x y x y
x y x y x y x y
x x y xy
−
=
= = = =
⎛ ⎞− = −⎜ ⎟
⎝ ⎠⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
= − + − + − + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
= − + − + − + −
= − +
∑
3y−
19. 3
3 3
0
3 0 2 1 1 2 0 3
3 0 2 1 1 2 0 33! 3! 3! 3!3!0! 2!1! 1!2! 0!3!
1 3 3 1
3 2
3(5 2) (5 ) ( 2)
3 3 3 3(5 ) ( 2) (5 ) ( 2) (5 ) ( 2) (5 ) ( 2)
0 1 2 3
(5 ) ( 2) (5 ) ( 2) (5 ) ( 2) (5 ) ( 2)
125 150 60
k k
k
x xk
x x x x
x x x x
x x
−
=
= = = =
⎛ ⎞− = −⎜ ⎟
⎝ ⎠⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
= − + − + − + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
= − + − + − + −
= − +
∑
8x −
20.
33 3
0
3 0 2 1 1 2 0 3
3 0 2 1 1 2 0 33! 3! 3! 3!3!0! 2!1! 1!2! 0!3!
1 3 3 1
3 2
3( 7 ) ( ) ( 7 )
3 3 3 3( ) ( 7 ) ( ) ( 7 ) ( ) ( 7 ) ( ) ( 7 )
0 1 2 3
( ) ( 7 ) ( ) ( 7 ) ( ) ( 7 ) ( ) ( 7 )
21 147
k k
k
a b a bk
a b a b a b a b
a b a b a b a b
a a b ab
−
=
= = = =
⎛ ⎞− = −⎜ ⎟
⎝ ⎠⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
= − + − + − + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
= − + − + − + −
= − +
∑
2 3343b−
Chapter 12
1754
21.
( )
( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( )
4441 1
0
4 3 2 1 00 1 2 3 41 1 1 1 1
4 3 2 10 1 2 34! 4! 4! 4! 4!1 1 1 1 14!0! 3!1! 2!2! 1!3! 0!4!
1 4 6 4 1
4( 5 ) (5 ) , 0
4 4 4 4 4(5 ) (5 ) (5 ) (5 ) (5 )
0 1 2 3 4
(5 ) (5 ) (5 ) (5 )
k kx x
k
x x x x x
x x x x x
y y xk
y y y y y
y y y y
−
=
= = = = =
⎛ ⎞+ = ≠⎜ ⎟
⎝ ⎠⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
= + + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
= + + + +
∑
2 3
4 3 2
0 4
41
(5 )
20 150 500 625y y yxx x x
y
y= + + + +
22.
( )
( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( )
4443 3
0
4 3 2 1 00 1 2 3 43 3 3 3 3
4 3 2 10 1 2 34! 3 4! 3 4! 3 4! 3 4!4!0! 3!1! 2!2! 1!3! 0!4!
1 4 6 4 1
4(2 ) 2 ( ) , 0
4 4 4 4 42 ( ) 2 ( ) 2 ( ) 2 ( ) 2 ( )
0 1 2 3 4
2 ( ) 2 ( ) 2 ( ) 2 ( ) 2
k ky y
k
y y y y y
y y y y
x x yk
x x x x x
x x x x x
−
=
= = = = =
⎛ ⎞+ = ≠⎜ ⎟
⎝ ⎠⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
= + + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
= + + + +
∑
3 2
2 3 4
0 43
4 1
( )
16 96 216 216 81
y
x x xy y y y
x= + + + +
23.
( )
( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( )
4 42 2 4 2 2
0
4 3 2 1 02 2 0 2 2 1 2 2 2 2 2 3 2 2 4
4 3 2 1 02 2 0 2 2 1 2 2 2 2 2 3 2 24! 4! 4! 4! 4!4!0! 3!1! 2!2! 1!3! 0!4!
1 4 6 4 1
4( ) ( )
4 4 4 4 4( ) ( ) ( ) ( ) ( )
0 1 2 3 4
( ) ( ) ( ) ( ) (
k k
kx y x y
k
x y x y x y x y x y
x y x y x y x y x y
−
=
= = = = =
⎛ ⎞+ = ⎜ ⎟
⎝ ⎠⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
= + + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
= + + + +
∑
4
8 6 2 4 4 2 6 8
)
4 6 4x x y x y x y y= + + + +
24. 3
3 3 3 3 3 3
0
3 3 3 0 3 2 3 1 3 1 3 2 3 0 3 3
3 3 3 0 3 2 3 1 3 1 3 2 3 0 3 33! 3! 3! 3!3!0! 2!1! 1!2! 0!3!
1 3 3 1
9
3( ) ( ) ( )
3 3 3 3( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
0 1 2 3
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
3
k k
k
r s r sk
r s r s r s r s
r s r s r s r s
r
−
=
= = = =
⎛ ⎞− = −⎜ ⎟
⎝ ⎠⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
= − + − + − + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
= − + − + − + −
= −
∑
6 3 3 6 93r s r s s+ −
Section 12.5
1755
25. 5
5 5
0
5 0 4 1 3 2 2 3
1 4 0 5
5 0 4 15! 5! 5!5!0! 4!1! 3!2!
1 5 10
5( ) ( ) ( )
5 5 5 5( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
0 1 2 3
5 5( ) ( ) ( ) ( )
4 5
( ) ( ) ( ) ( ) ( )
k k
k
ax by ax byk
ax by ax by ax by ax by
ax by ax by
ax by ax by ax
−
=
= = =
⎛ ⎞+ = ⎜ ⎟
⎝ ⎠⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
= + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
= + +
∑
3 2 2 35!2!3!
10
1 4 0 55! 5!1!4! 0!5!
5 1
5 5 4 4 3 2 3 2 2 2 2 3 4 4 5 5
( ) ( ) ( )
( ) ( ) ( ) ( )
5 10 10 5
by ax by
ax by ax by
a x a bx y a b x y a b x y ab xy b y
=
= =
+
+ +
= + + + + +
26. 5
5 5
0
5 0 4 1 3 2 2 3
1 4 0 5
5 0 4 15! 5! 5!5!0! 4!1! 3!
1 5
5( ) ( ) ( )
5 5 5 5( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
0 1 2 3
5 5( ) ( ) ( ) ( )
4 5
( ) ( ) ( ) ( )
k k
k
ax by ax byk
ax by ax by ax by ax by
ax by ax by
ax by ax by
−
=
= =
⎛ ⎞− = −⎜ ⎟
⎝ ⎠⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
= − + − + − + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⎛ ⎞ ⎛ ⎞+ − + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
= − + − +
∑
3 2 2 35!2! 2!3!10 10
1 4 0 55! 5!1!4! 0!5!
5 1
5 5 4 4 3 2 3 2 2 2 2 3 4 4 5 5
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
5 10 10 5
ax by ax by
ax by ax by
a x a bx y a b x y a b x y ab xy b y
= =
= =
− + −
+ − + −
= − + − + −
Chapter 12
1756
27.
( )
( ) ( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( ) ( )
( ) ( )
6 66
0
6 5 4 30 1 2 3
2 1 04 5 6
6 5 4 30 1 2 36! 6! 6! 6!
6!0! 5!1! 4!2! 3!3!1 6 15 20
2 146! 6!
2!4! 1!5!
15 6
6( 2) 2 , 0
6 6 6 62 2 2 2
0 1 2 3
6 6 62 2 2
4 5 6
2 2 2 2
2
kk
k
x x xk
x x x x
x x x
x x x x
x x
−
=
= = = =
= =
⎛ ⎞+ = ≥⎜ ⎟
⎝ ⎠⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
= + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
= + + +
+ +
∑
( )5 3 1
2 2 2
05 66!
0!6!
1
3 2
2 2
12 60 160 240 192 64
x
x x x x x x=
+
= + + + + + +
28.
( )
( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( )31
2 2
44 4
0
0 1 2 3 44 3 2 1 0
0 1 2 3 44 3 2 1 04! 4! 4! 4! 4!
4!0! 3!1! 2!2! 1!3! 0!4!
1 4 6 4 1
2
4(3 ) 3 , 0
4 4 4 4 43 3 3 3 3
0 1 2 3 4
3 3 3 3 3
81 108 54 12
kk
ky y y
k
y y y y y
y y y y y
y y y y
−
=
= = = = =
⎛ ⎞+ = ≥⎜ ⎟
⎝ ⎠⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
= + + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
= + + + +
= + + + +
∑
29.
( ) ( )
( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( )
( ) ( )
3 31 14 4 4 4
3 3 31 1 14 4 4 4 4 4
3 31 14 4 4 4
3 3 31 1 14 4 4 4 4 4
3 14 4
4 44
0
4 3 20 1 2
1 03 4
4 3 20 1 24! 4! 4!
4!0! 3!1! 2!2!1 4 6
1 34! 4!
1!3! 0!4!
4
4( ) , , 0
4 4 40 1 2
4 43 4
k k
k
a b a b a bk
a b a b a b
a b a b
a b a b a b
a b
−
=
= = =
=
⎛ ⎞+ = ≥⎜ ⎟
⎝ ⎠⎛ ⎞ ⎛ ⎞ ⎛ ⎞
= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
= + +
+ +
∑
( ) ( )3 14 4
9 3 3 31 14 4 2 2 4 4
0 4
1
3 4 6 4
a b
a a b a b a b b=
= + + + +
Section 12.5
1757
30.
( ) ( )
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
2 1 2 13 3 3 3
2 1 2 1 2 1 2 13 3 3 3 3 3 3 3
2 1 2 1 2 1 2 13 3 3 3 3 3 3 3
4 1 2 23 3 3 3
3 33
0
3 0 2 1 1 2 0 3
3 0 2 1 1 2 0 33! 3! 3! 3!
3!0! 2!1! 1!2! 0!3!1 3 3 1
2
3( )
3 3 3 30 1 2 3
3 3
k k
k
x y x yk
x y x y x y x y
x y x y x y x y
x x y x y y
−
=
= = = =
⎛ ⎞+ = ⎜ ⎟
⎝ ⎠⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
= + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
= + + +
= + + +
∑
31.
( ) ( )
( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
1 14 4
1 1 14 4 4
1 14 4
1 1 14 4 4
1 14 4
4 44
0
0 1 24 3 2
3 41 0
0 1 24 3 24! 4! 4!
4!0! 3!1! 2!2!1 4 6
3 41 04! 4!
1!3! 0!4!4 1
4( 2 ) 2 , , 0
4 4 42 2 2
0 1 2
4 42 2
3 4
2 2 2
2 2
kk
k
x y x y x yk
x y x y x y
x y x y
x y x y x y
x y x y
x
−
=
= = =
= =
⎛ ⎞+ = ≥⎜ ⎟
⎝ ⎠⎛ ⎞ ⎛ ⎞ ⎛ ⎞
= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
= + +
+ +
=
∑
3 31 1 14 2 2 4 2 28 24 32 16x y x y x y y+ + + +
Chapter 12
1758
32.
( ) ( )
( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( )
1 14 4
1 1 14 4 4
1 1 14 4 4
1 1 14 4 4
8 88
0
8 7 60 1 2
5 4 33 4 5
2 1 06 7 8
8( 3 ) 3 , , 0
8 8 83 3 3
0 1 2
8 8 83 3 3
3 4 5
8 8 83 3 3
6 7 8
k k
k
x y x y x yk
x y x y x y
x y x y x y
x y x y x y
−
=
⎛ ⎞− = − ≥⎜ ⎟
⎝ ⎠⎛ ⎞ ⎛ ⎞ ⎛ ⎞
= − + − + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎛ ⎞ ⎛ ⎞ ⎛ ⎞
+ − + − + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎛ ⎞ ⎛ ⎞ ⎛ ⎞
+ − + − + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
∑
( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( )
1 1 14 4 4
1 1 14 4 4
1 1 14 4 4
8 7 60 1 28! 8! 8!
8!0! 7!1! 6!2!
1 8 28
5 4 33 4 58! 8! 8!
5!3! 4!4! 3!5!56 70 56
2 1 06 7 88! 8! 8!
2!6! 1!7! 0!8!
28 8 1
3 3 3
3 3 3
3 3 3
x y x y x y
x y x y x y
x y x y x y
= = =
= = =
= = =
= − + − + −
+ − + − + −
+ − + − + −
7 5 31 12 4 2 2 4
3 5 3 712 4 2 2 4
4 3 2
2
24 252 1512 5670
13,608 20,412 17,496 6561
x x y x y x y x y
x y xy x y y
= − + − +
− + − +
33.
4 4 3 2 2 3 4
4 3 2 2 3 4
( ) 4 ( ) 6 ( ) 4 ( ) ( )4 6 4
r s r r s r s r s sr r s r s rs s
− = + − + − + − + −
= − + − +
34.
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )
7 7 6 1 5 2 4 3 3 42 2 2 2 2 2 2 2 2 2 2
2 5 1 6 72 2 2 2 2
14 12 2 10 4 8 6 6 8 4 10 2 12 14
7 21 35 35
21 7
7 21 35 35 21 7
x y x x y x y x y x y
x y x y y
x x y x y x y x y x y x y y
+ = + + + +
+ + +
= + + + + + + +
35.
( )6 6 5 4 2 3 3
2 4 1 5 6
6 6 5 5 4 2 4 2 3 3 3 3 2 4 2 4
5 5 6 6
( ) 6( ) ( ) 15( ) ( ) 20( ) ( )
15( ) ( ) 6( ) ( ) ( )6 15 20 156
ax by ax ax by ax by ax by
ax by ax by bya x a bx y a b x y a b x y a b x y
ab xy b y
+ = + + +
+ + +
= + + + +
+ +
Section 12.5
1759
36. 4 4 3 1 2 2 1 3 4
4 3 2 2 3 4
( 3 ) 4 (3 ) 6 (3 ) 4 (3 ) (3 )12 54 108 81
x y x x y x y x y yx x y x y xy y
+ = + + + +
= + + + +
37. This term is 10 4 4 610 10! 10 9 8 7 6!(2) (16)4 4!6!
x x−⎛ ⎞ ⋅ ⋅ ⋅ ⋅= =⎜ ⎟
⎝ ⎠ (4 3 2 1) 6!⋅ ⋅ ⋅6 63360x x=
So, the desired coefficient is 3360.
38. This term is 9 4 4 59 9! 9 8 7 6 5!(3) (81)4 5!4!
y y−⎛ ⎞ ⋅ ⋅ ⋅ ⋅= =⎜ ⎟
⎝ ⎠ (4 3 2 1) 5!⋅ ⋅ ⋅5 510,206y y=
So, the desired coefficient is 10,206.
39. This term is 8 4 4 48 8! 8 7 6 5 4!( 3) (81)4 4!4!
y y−⎛ ⎞ ⋅ ⋅ ⋅ ⋅− = =⎜ ⎟
⎝ ⎠ (4 3 2 1) 4!⋅ ⋅ ⋅4 45670y y=
So, the desired coefficient is 5670.
40. This term is 12 7 7 512 12! 12 11 10 9 8 7!( 1) ( 1)7 7!5!
x x−⎛ ⎞ ⋅ ⋅ ⋅ ⋅ ⋅− = − =⎜ ⎟
⎝ ⎠ (5 4 3 2 1) 7!⋅ ⋅ ⋅ ⋅5 5( 1) 792x x− = −
So, the desired coefficient is 792− . 41. This term is
7 4 4 3 47 7! 7 6 5 4!(2 ) (3 ) (8 81)4 4!3!
x y x y−⎛ ⎞ ⋅ ⋅ ⋅= ⋅ =⎜ ⎟
⎝ ⎠ (3 2 1) 4!⋅ ⋅3 4 3 4(648) 22,680x y x y=
So, the desired coefficient is 22,680. 42. This term is
9 7 7 2 79 9! 9 8 7!(3 ) ( 5 ) (9)( 78,125)7 7!2!
x y x y−⎛ ⎞ ⋅ ⋅− = − =⎜ ⎟
⎝ ⎠ (2 1) 7!⋅2 7 2 7( 703,125) ( 25,312,500)x y x y− = −
So, the desired coefficient is 25,312,500− .
43. This term is 2 8 4 4 8 48 8! 8 7 6 5 4!( ) ( )4 4!4!
x y x y−⎛ ⎞ ⋅ ⋅ ⋅ ⋅= =⎜ ⎟
⎝ ⎠ (4 3 2 1) 4!⋅ ⋅ ⋅8 4 8 470x y x y=
So, the desired coefficient is 70.
44. This term is 10 4 2 4 6 810 10! 10 9 8 7 6!( )4 4!6!
r s r s−⎛ ⎞ ⋅ ⋅ ⋅ ⋅− = =⎜ ⎟
⎝ ⎠ (4 3 2 1) 6!⋅ ⋅ ⋅6 8 6 8210r s r s=
So, the desired coefficient is 210.
45. Since 40 40! 40 39 38 37 36 35 34!6 6!34!
⎛ ⎞ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅= =⎜ ⎟
⎝ ⎠ 6! 34!3,838,380= , there are 3,838,380
such 6-number lottery numbers.
46. Since 60 60! 60 59 58 57 56 55 54!6 6!54!
⎛ ⎞ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅= =⎜ ⎟
⎝ ⎠ 6! 54!50,063,860= , there are
50,063,860 such 6-number lottery numbers.
Chapter 12
1760
47. Since 52 52! 52 51 50 49 48 47!5 5!47!
⎛ ⎞ ⋅ ⋅ ⋅ ⋅ ⋅= =⎜ ⎟
⎝ ⎠ 5! 47!2,598,960= , there are 2,598,960
possible 5-card poker hands.
48. Since 108 108! 108 107 ... 98 97!11 97!11!
⎛ ⎞ ⋅ ⋅ ⋅ ⋅= =⎜ ⎟
⎝ ⎠ 11! 97!344,985,116,800,000= , there are
344,985,116,800,000 different 11-card Canasta hands.
49. 7 7!5 5!⎛ ⎞
≠⎜ ⎟⎝ ⎠
, 7 7!5 5!2!⎛ ⎞
=⎜ ⎟⎝ ⎠
50. Should replace powers of y by powers of 2y . More precisely, the expansion should be
4 4 3 2 2 3 4
4 3 2 2 3 4
( 2 ) 4 (2 ) 6 (2 ) 4 (2 ) (2 )8 24 32 16
x y x x y x y x y yx x y x y xy y
+ = + + + +
= + + + +
51. False, there are 11 terms. 52. True 53. True
54. False. This is only defined for 0n = , and the result in such case is 0
10⎛ ⎞
=⎜ ⎟⎝ ⎠
.
55. Claim: , 0 , for any 1n n
k n nk n k⎛ ⎞ ⎛ ⎞
= ≤ ≤ ≥⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠.
Proof. Observe that
This is written in a different form.
! ! !!( )! ( )! ! ( )! ( ( )) !
k
n nn n nk n kk n k n k k n k n n k⎛ ⎞ ⎛ ⎞
= = = =⎜ ⎟ ⎜ ⎟−− − − − −⎝ ⎠ ⎝ ⎠ ■
56. Observe that
0 0 1 1 0
2 (1 1)
1 1 1 1 . . . 1 1 . . . 1 10 1
. . .0 1
n n
n n n k k nn n n nk n
n n nn
− − −
= +
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + + + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎛ ⎞ ⎛ ⎞ ⎛ ⎞
= + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
since all powers of 1 equal 1.
Section 12.5
1761
57. The binomial expansion of 3(1 )x− is 2 31 3 3x x x− + − .
Notes on the Graph: Heavy Solid Curve: 2 3
1 1 3 3y x x x= − + − Heavy Dashed Curve: 2 3
2 1 3 3y x x x= − + − + Heavy Dotted Curve: 3
3 (1 )y x= − The graphs of 1y and 3y are the same, so you only see two graphs.
58. The binomial expansion of 4( 3)x + is
4 3 212 54 108 81x x x x+ + + + . Notes on the Graph: Heavy Solid Curve:
4 3 23 12 54 108 81y x x x x= + + + +
Heavy Dashed Curve: 4 3 2
2 4 6 4 1y x x x x= + + + + Heavy Dotted Curve: 4
1 ( 3)y x= + The graphs of 1y and 3y are the same, so you only see two graphs.
59. We see from the graph that as each term is added, the graphs of the respective functions get closer to the graph of
34 (1 )y x= − when 1 2x< < . However, when
1x > , this is no longer true. Notes on the graph: Heavy solid curve: 4y Heavy dashed curve: 1y Heavy dotted curve: 2y Thin dashed curve: 3y
Chapter 12
1762
60. We see from the graph that as each term is added, the graphs of the respective functions get closer to the graph of
314 (1 )xy = − when 1 2x< < . However, when
0 1x< < , this is no longer true. Notes on the graph: Heavy solid curve: 4y Heavy dashed curve: 1y Heavy dotted curve: 2y Thin dashed curve: 3y 61. As each new term is added, the corresponding graph of the curve is a better approximation to the graph of ( )311 xy = + , for 1 2x< < . The series of functions does not get closer to this graph if 0 1x< < .
62. As each new term is added, the corresponding graph of the curve is a better approximation to the graph of
xy e= , for 1 1x− < < . The series of functions does not get closer to this graph if 1 2x< < .
Section 12.6 Solutions-------------------------------------------------------------------------------- 1.
( )6 46! 6! 6 5 4 3 2!
6 4 ! 2!P ⋅ ⋅ ⋅ ⋅= = =
− 2!360=
2. ( )7 3
7! 7! 7 6 5 4!7 3 ! 4!
P ⋅ ⋅ ⋅= = =
− 4!210=
3.
( )9 59! 9! 9 8 7 6 5 4!
9 5 ! 4!P ⋅ ⋅ ⋅ ⋅ ⋅= = =
− 4!15,120=
4.
( )9 49! 9! 9 8 7 6 5!
9 4 ! 5!P ⋅ ⋅ ⋅ ⋅= = =
− 5!3024=
5. ( )8 8
8! 8! 8! 40,3208 8 ! 0!
P = = = =−
6. ( )6 6
6! 6! 7206 6 ! 0!
P = = =−
7. ( )13 3
13! 13! 13 12 11 10!13 3 ! 10!
P ⋅ ⋅ ⋅= = =
− 10!1716=
8. ( )20 3
20! 20! 20 19 18 17!20 3 ! 17!
P ⋅ ⋅ ⋅= = =
− 17!6840=
Section 12.6
1763
9. ( )10 5
10! 10! 10 9 8 7 6 5!10 5 !5! 5!5!
C ⋅ ⋅ ⋅ ⋅ ⋅= = =
− 5! ( )252
5 4 3 2 1=
⋅ ⋅ ⋅ ⋅
10. ( )9 4
9! 9! 9 8 7 6 5!9 4 !4! 5!4!
C ⋅ ⋅ ⋅ ⋅= = =
− 5! ( )126
4 3 2 1=
⋅ ⋅ ⋅
11. ( )50 6
50! 50 49 48 47 46 45 44!50 6 !6!
C ⋅ ⋅ ⋅ ⋅ ⋅ ⋅= =
− 44! ( )15,890,700
6 5 4 3 2 1=
⋅ ⋅ ⋅ ⋅ ⋅
12.
( )50 1050! 50!
50 10 !10! 40!10!
50 49 48 47 46 45 44 43 42 41 40!
C = =−
⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅=
40! ( )10, 272, 278,170
10 1=
…
13. ( )7 7
7! 7! 17 7 !7! 0!7!
C = = =−
14. 8 8 1C =
15. ( )30 4
30! 30 29 28 27 26!30 4 !4!
C ⋅ ⋅ ⋅ ⋅= =
− 26! ( )27,405
4 1=
…
16. ( )13 5
13! 13 12 11 10 9 8!13 5 !5!
C ⋅ ⋅ ⋅ ⋅ ⋅= =
− 8! ( )1287
5 1=
…
17. ( )45 8
45! 45 44 43 42 41 40 39 38 37!45 8 !8!
C ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅= =
− 37! ( )215,553,195
8 7 1=
⋅ …
18. ( )30 4
30! 30! 30 29 28 27 26!30 4 !4! 26!4!
C ⋅ ⋅ ⋅ ⋅= = =
− 26! ( )27, 405
4 1=
…
19. 4 3 2 24⋅ ⋅ = different system configurations
20. 4 5 3 2 120⋅ ⋅ ⋅ = different houses to choose from
21. 2 3 2 12color envelopeswriting
⋅ ⋅ = different types of
invitations
22. 4 2 3 5 120main starch vegetable appetizercourse
⋅ ⋅ ⋅ =
different dinner combinations 23. Each slot can have 0,1, ,9… (10 choices) Can allow repetition of digits and order is important. So, there are 410 10 10 10 10⋅ ⋅ ⋅ =
10, 000= different pin numbers
24. Each slot can have , ,a z… (26 choices). Can allow repetition of letters and order is important. So, there are
426 456,976= different passwords.
25. 15! 15 14 13 12 11!11!
⋅ ⋅ ⋅ ⋅=
11!32,760= different leadership teams
Chapter 12
1764
26. president vice secretary treasurer
president
3 2 4 1 24⋅ ⋅ ⋅ =
possible election outcomes
27. Each of 20 questions has 4 answer choices. So, there are 204 121.1 10≈ × possible ways to answer the questions on the exam.
28. Each of 25 questions has 5 answer choices. So, there are 25 85 ( 2,980,232,239 10 )= × different ways to complete the exam.
29. The number of 5-digit zip codes where any of 10 digits can be used in each slot is 510 =100,000. The number of 5-digit zip codes where 0 cannot be used in 1st and last
slot: 1st—9 digits
2nd—10 digits 3rd —10 digits 4th —10 digits 5th —9 digits
2 39 10× = 81,000 such zip codes.
31. 3230! 2.65 10≈ ×
30. 1st slot—24 (eliminate O and I) 2nd slot—24 3rd slot—24 4th slot—8 (eliminate 0 and 1) 5th slot—8 6th slot—8 3 324 8× = 7,077,888 license plate numbers
32. 4! 24= different seating arrangements
34. 50 49 48 117, 600⋅ ⋅ = different safe combinations
33. 40 39 38 59, 280⋅ ⋅ =
1st # 2nd # 3rd # (1 less digit (2 less digits to choose from) to choose from) (Note: Here, order is important.) So, there are 59,280 different locker combinations.
35.
( )1000 31000! 1000 999 998
1000 3 !P = = ⋅ ⋅ =
−
997, 002, 000 possible winning scenarios
36. 100 3 100 99 98 970, 200P = ⋅ ⋅ = different possible placings of 1st, 2nd, and 3rd. (Note: Here, order is important.)
37.
( )53 53! 53 52 51 50 49 48 47!6 6! 53 6 !
⎛ ⎞ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅= =⎜ ⎟ −⎝ ⎠ 6! 47!
22,957, 480= different 6-number combinations
Section 12.6
1765
38.
( )53 53! 53 52 51 50 49 48!5 5! 53 5 !
⎛ ⎞ ⋅ ⋅ ⋅ ⋅ ⋅= =⎜ ⎟ −⎝ ⎠ 5! 48!
2,869, 685= different 5-number combinations (Note: Here, order is not important.)
39.
( )52 52!5 5! 52 5 !
52 51 50 49 48 47!
⎛ ⎞=⎜ ⎟ −⎝ ⎠
⋅ ⋅ ⋅ ⋅ ⋅=
( )5 1 47!…
2,598,960= different 5-card hands (Note: Here, order is not important.)
40.
( )52 52!7 7! 52 7 !
52 51 50 49 48 47 46 45!
⎛ ⎞=⎜ ⎟ −⎝ ⎠
⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅=
( )7 1 45!…
133,784,560= different 7-card hands
41. ( )
52 52! 52 51 50!2 52 2 !2! 50!
⎛ ⎞ ⋅ ⋅= =⎜ ⎟ −⎝ ⎠ 2
1326= different blackjack hands
42. 4 aces—must be one of the cards 4 kings 4 queens 16 possible choices for 2nd card 4 jacks 4 tens So,
1 2
4 16 64st card nd card
⋅ = different hands yielding
21.
43.
# possiblescenarios forSweet Sixteen
1464 64! 4.9 1016 16!48!⎛ ⎞
= ≈ ×⎜ ⎟⎝ ⎠
44.
# possible scenariosfor Final Four
64 64! 64 63 62 61 60!4 60!4!
⎛ ⎞ ⋅ ⋅ ⋅ ⋅= =⎜ ⎟
⎝ ⎠ 60! ( )4 1
635,376=
…
45. 1 2
16 16 256st slot nd slot
AFC NFC
⋅ =
possible scenarios for Super Bowl
46. 6 6 36AFC NFC
⋅ = possible scenarios at this
stage
47.1 2 3 6
5 5 5 5player player player player
… (Nobody
votes themselves off) 65 15,625= = voting combinations for Survivor when down to 6 contestants
Chapter 12
1766
49. 26 6 6! 6
3 3 3!3!⎛ ⎞ ⎛ ⎞ ⎛ ⎞⋅ = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠⎝ ⎠ ⎝ ⎠
5 43!⋅ ⋅
2
220 400
⎛ ⎞⎜ ⎟⎝ ⎠
= =
50. 6! 6! 720 720 518,400⋅ = ⋅ =
48. 12!different orderings of 12 contestants If the contestants must alternate male/female, then the number of ways to do this are:
1 2 3 4 11 126 choices for 6 choices for 5 choices for
first male first female second male. And so on...
6 6 5 5 1 1…
So, the number of possibilities is: ( )22 2 2 26 5 1 6! 720 518,400⋅ ⋅ = = =… 51. The combination formula
n rC should be used instead.
52. Order is important here (since president, vice president, secretary, and treasurer are all different positions with different possibilities). So, should use permutation formula instead.
53. True
54. False. Suppose you have 6 objects, 4 of type A and 2 of type B. Then, the number of
distinguishable permutations is 6! 154!2!
= .
However, the number of combinations taken 3
at a time is 6! 203!3!
= .
55. False. Since there will be repetition when permuting the letters in ABBA that are not distinguishable while no such repetition occurs in permuting ABCD.
56. True 57. Observe that
1
!n r
n r
C nC +
=( )
( ) ( )1 ! 1 !! ! !
n r rn r r n
− − +−
( )1 !n r− −=
( )1 !r r+
( ) ( )1 !n r n r− − − !r1r
n r+
=−
58. Observe that
( )
( )( )1
!! !
!1 !
n r
n r
nn rP n
nPn r
−
−= =
− −( )
( )1 !! !
n rn r n
− +⋅
−( )1 ( )!n r n r− + −
=( )!n r−
( )1n r= − +
59. !( , ) !( )! !
nC n r rn r r
⋅ =− ⋅
!r⋅ !( )!
nn r
=−
60. ! !( , ) ( , )( )! ! ( ( ))! ( )!
n nC n r C n n rn r r n n r n r
= = = −− ⋅ − − ⋅ −
.
61. Use a calculator to do so, if you have one. 62. Use a calculator to do so, if you have one.
62. a. 5,040 b. 5,040 c. Yes. d. This is true because !n r n rP r C= .
63. a. 95,040 b. 95,040 c. Yes. d. This is true because !n r n rP r C= .
Section 12.7
1767
Section 12.7-------------------------------------------------------------------------------------------- 1. Sum the two dice. So, the possible rolls constitute the sample space: { }2,3 12S = …
2. { }, , , , , , ,S HHH HHT HTH HTT THH THT TTH TTT= H H T H T H T H H T T T H T 3. { , , , , , , , , , ,S BBBB BBBG BBGB BBGG BGBB BGBG BGGB BGGG GBBB GBBG=
}, , , , ,GBGB GBGG GGBB GGBG GGGB GGGG B B G B G B G B B G B G G B G B B G B G B G G B G B G G B G 4. { }1, 2, , 6, 1, 2, , 6S H H H T T T= … …
Chapter 12
1768
5. { }, , , , , , ,S RR RB RW BB BR BW WR WB= (Can’t have WW since there is only one white ball in the container.) R R B W B B R W R W B 6. { }, , , , , , , ,FF FS FJ SF SS SJ JF JS JJ F F S J F S S J F J S J Note: For problems 7-10, the sample space is given in Problem 2. (Note that there are 8 distinct elements all with the same probability of occurring.)
7. ( ) 1(all heads)8
P P HHH= = 8.
( )
(exactly one head) 3or or8
P
P TTH THT HTT
=
=
9.
( )
(at least one head) 718
P
P TTT
=
− =
10.
( )
(at least two heads)1or or or2
P
P HHT HTH THH HHH
=
=
Note: For Problems 11-16, there are 36 different possible outcomes of throwing 2 dice, assuming that the two dice are distinguishable from each other. The sample space is given in Example 4 on page 662 of the text.
Section 12.7
1769
11. ( )( ) ( )1,2 or 2,1
2 1336 18
P sum= = =
13. ( ) ( )sum is even 1 sum is oddP P= − 1 112 2
= − =
12. Note that the sum of the two dice is odd for the following rolls:
( ) ( ) ( )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
So, P (sum is odd) 18 136 2
= = .
14. Note that the sum of the two dice is prime if the sum is 3, 5, 7, or 11. Consulting the sample space in text (page 662), we see this can occur 18 distinct ways.
So, ( ) 15sum is prime36
P = .
15. Note that the sum of two dice can be greater than 7 only when the sum is 8, 9, 10, 11, or 12. Consulting sample space on page 662 of the text, we see that this can occur in 15 distinct ways. So, ( ) 15 5
36 127P sum > = =
16. Note that the sum of two dice can be less than 7 only when the sum is 2, 3, 4, 5, or 6. Consulting sample space, we see that this can occur in 15 distinct ways. So, ( ) 15 5
36 127P sum < = = Note: For Problems 17-20, the sample space is set of 52 distinct cardsS = . 17. There are 12 face cards (4 kings, 4 queens, 4 jacks). So, P(drawing a nonface card) 40 10
52 13= = .
18. There are 26 black cards. So, P(drawing a black card) 1
2= .
19. There are 4 twos, 4 fours, 4 sixes, 4 eights. So, P(drawing a 2, 4, 6, or 8) 16 4
52 13= = .
20. There are 4 threes, 4 fives, 4 sevens, 4 nines, 4 aces. So, P (drawing a 3, 5, 7, 9, or ace) 20 5
52 13= = .
21. ( ) ( )1 11 3not 1 14 4
P E P E= − = − = 22. ( ) ( )2 21 1not 1 12 2
P E P E= − = − =
23. If 1 2,E E are mutually exclusive, then
( ) ( ) ( )1 2 1 234
P E E P E P E∪ = + =
24. Since 1 2,E E are not mutually exclusive, we see that ( ) ( ) ( ) ( )1 2 1 2 1 2P E E P E P E P E E∪ = + − ∩
1 1 1 54 2 8 8
= + − =
Chapter 12
1770
25. If 1 2,E E are mutually exclusive, then
( )1 2 and 0P E E = . 26. Since 1 2,E E are independent, ( ) ( ) ( )1 2 1 2 and P E E P E P E= ⋅
1 1 14 2 8
= ⋅ =
27. a. 52 52! 52 51 50 49 48!4 4!48!
⎛ ⎞ ⋅ ⋅ ⋅ ⋅= =⎜ ⎟
⎝ ⎠ 4! 48!270, 725=
b. Number of ways to get 4
spades:13
7154
⎛ ⎞=⎜ ⎟
⎝ ⎠
So, P(getting 4 spades) 715 0.0026270,725
= ≅
c. Since there are 13 ways of getting 4 of a kind, we see
that ( ) 134 of a kind 0.00005270, 725
P = ≈
So, there is about a 0.005% chance of getting four of a kind.
28. a. 52 52! 52 51 50!2 50!2!
⎛ ⎞ ⋅ ⋅= =⎜ ⎟
⎝ ⎠ 50! ( )2 1⋅
1326= different 2-card hands b. Number of ways of getting
4 possibilities 12 possibilities
ace face card+
is card 1 card 2
4 12 48⋅ =
So, ( ) 48getting 21 0.0371326
P = ≈
So, there is about a 3.7% chance of getting 21.
29. Since drawing a 7 and drawing an 8 are mutually exclusive events, P (drawing 7 or 8) = P(draw 7) + P(draw 8)
4 4 8 2 0.15452 52 52 13
= + = = ≈
So, there is about a 15.4% chance of drawing a 7 or an 8.
30. Since drawing a red 7 or a black 8 are mutually exclusive events, P (drawing red 7 or black 8) = P (draw red 7) + P(draw black 8) = 2 2 4 1 0.077
52 52 52 13+ = = ≈
So, there is about a 7.7% chance of drawing a red 7 or a black 8.
31. Since these two events are independent, we see that
P (7 on 1st draw AND 8 on 2nd draw) 4 4 16 4 0.00652 51 2652 663
= ⋅ = = ≈ .
So, there is about a 0.6% chance of getting a 7 on the first draw and an 8 on the second draw. 32. Since these two events are independent, we see that
P (red 7 on 1st draw AND black 8 on 2nd draw) 2 2 4 1 0.00252 51 2652 663
= ⋅ = = ≈ .
So, there is about a 0.2% chance of getting a red 7 on the first draw and a black 8 on the second draw.
Section 12.7
1771
33. Sample space S has 52 elements, all elements are presumed equally likely.
So, P (5 daughters) 1 0.0312532
= = .
Thus, there is a 3.1% chance of getting 5 daughters.
34. Sample space S has 42 elements, all elements are presumed equally likely.
So, P (4 sons) 1 0.062516
= = .
Thus, there is a 6.25% chance of getting 4 sons.
35. Sample space has 52 elements, all elements presumed equally likely. So, ( ) ( )1 boy 1 0 boysP P≥ = −
( )1 5 daughtersP= −
1 311 0.9687532 32
= − = =
So, there is a 96.9% chance of having at least one boy.
36. Sample space has 62 elements, all elements presumed equally likely. So, ( ) ( )1 girl 1 0 girlsP P≥ = −
1 631 0.98437564 64
= − = =
So, there is about 98.4% chance of having at least one girl.
37. Assume independence of spins. Then, P(red on 1st spin AND 2nd AND 3rd AND 4th) = P(red on 1st) ⋅P(red on 2nd) …P(red on 4th) =
418 0.050338
⎛ ⎞ ≈⎜ ⎟⎝ ⎠
. So, about a 5.03% chance.
38. Again, assume independence of the spins. Then, P(green on 1st spin AND green on 2nd spin)
22 0.002838
⎛ ⎞= ≈⎜ ⎟⎝ ⎠
. So, about a 0.28%
chance. 39. P(not defective) = 9/10. Since you are drawing the 8 DVD players from different batches, the 8 choices are independent. So, the probability that none of the 8 are defective is ( )89
10 0.43≈ . So, about 43% chance.
40. P(non defective) 20 800 10
0.09510010
⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠= ≈⎛ ⎞⎜ ⎟⎝ ⎠
So, about 9.5% chance. 41. The events are independent since they are chosen at random. So, P(1st number even AND 2nd number even) =
( ) ( )1st number even 2nd number evenP P⋅ =
1 1 12 2 4⋅ =
So, 25% chance.
42. The events are independent since chosen at random. P(1st number odd AND 2nd number odd)= ( ) ( )1st number odd 2nd number oddP P⋅
8 8 6415 15 225
⋅ =
Chapter 12
1772
43. Number of 2 card hands is given by: 52 52! 52 51 50!2 2!52!
⎛ ⎞ ⋅ ⋅= =⎜ ⎟
⎝ ⎠ 2 50!⋅1326=
Number of possible blackjack hands is
ace face or ten
4 16 64⋅ =
So, P(blackjack) 64 0.04831326
= ≈
So, about a 4.8% chance.
44. Assume independence of hands dealt and assume all cards are gathered and reshuffled. Then, probability of getting blackjack on 1st hand AND blackjack on 2nd hand is given by
( )( )0.0362 0.0362 0.0013≈ since the two events are independent. So, about a 0.13% chance.
45. Assume independence of games from week to week. Then, the probability of
going16 0− is 161 0.00001526
2⎛ ⎞ ≈⎜ ⎟⎝ ⎠
. So,
about a 0.001526% chance.
46. Assume independence of games from week to week. Then, the probability of winning at least 1 game =
( )1 lose allP−
1611
2⎛ ⎞= − ⎜ ⎟⎝ ⎠
1 0.0000153≈ − 0.9999847= So, it is practically a certainty.
47. a. An outcome is of the form: (gene from mother, gene from father)
So, the sample space is given by S = { (Brown, Blue), (Brown, Brown), (Blue, Brown), (Blue, Blue) } b. ¼ since outcomes are equally likely c. ¾
48. a. An outcome is of the form: (gene from mother, gene from father)
So, the sample space is given by S = { (Brown, Brown), (Blue, Brown) } b. 0 since (Blue, Blue) is not a possible outcome. c. 1 since Brown is dominant.
49. 52 52! 52 51 50 49 48 2,598,9605 47! 5! 5 4 3 2 1
⎛ ⎞ ⋅ ⋅ ⋅ ⋅= = =⎜ ⎟ ⋅ ⋅ ⋅ ⋅ ⋅⎝ ⎠
50. a.
Ways to choose Ways to choose 3 aces 3 non-aces
4 48 4! 48!2 3 2! 2! 3! 45!
103,776
⎛ ⎞ ⎛ ⎞⋅ = ⋅⎜ ⎟ ⎜ ⎟ ⋅ ⋅⎝ ⎠ ⎝ ⎠
=
b. Using #49 for total number of 5-card hands, the probability of a hand in (a) is
103,776 0.042,598,960
≈ . About 4% chance.
Section 12.7
1773
51. Using #49 for the total number of 5-card hands, we obtain
13!8! 5!
135 1287
2,598,960 2,598,960 2,598,960⋅
⎛ ⎞⎜ ⎟⎝ ⎠ = =
52. Using #49 for the total number of 5-card hands, we obtain
4! 4!2!2! 3!1!
4 42 3
0.000012,598,960 2,598,960
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⋅⎝ ⎠ ⎝ ⎠ = =
53. The events aren’t mutually exclusive. So,
probability =
2of spades
4 13 1 16 452 52 52 52 13
+ − = =
54. The tacit assumption made here is that the order in which they were born is important. This is not the case. It
should be 38
.
55. True. Since all events are mutually exclusive,
( ) ( ) ( ) ( )1 2 3 1 2 31 P E E E P E P E P E= ∪ ∪ = + + .
56. False
57. False 58. False. The probabilities are, in fact, equal since both scenarios can only
occur with probability 132
.
59. The probability that Person 1 has birthday on Day A is simply 1
365 . Now, to compute the probability that Persons 1 and 2 have the same birthday, we note, first and foremost, that these events are independent since the people are being chosen at random. Since there are 365 days in a year, we can see that the probability of two people having the same birthday is ( )1 1 1
365 365 365365 0.0027⋅ = = .
60. The probability of more than two people having the same birthday is equal to
1 1365 364
1 (no 2 people having same bday)1 0.999924733
P− =− ⋅ ≅
So, from a probabilistic point of view, in a group of 30 people, it is highly likely that at least two of them will have the same birthday.
61. Sample space { }31,32,33,34,35,36, 41, 4243, 44, 45, 46S =
So, ( )Sum is 2,5,or 6P ( ) ( ) ( )
2 212 12
13
Sumis 2 Sumis 5 Sumis 60
0.333
P P P= + +
= + +
= ≈
63-64. Do so if you have access to a computer.
62. Let (4)x P= . Then, (3) 2P x= . As such, since the only
events possible are getting 3 or getting 4, we know that
13(4) (3) 1 2 1 3 1P P x x x x+ = ⇒ + = ⇒ = ⇒ =
So, P(rolling 3) 23 0.667= ≈ .
Chapter 12
1774
65. Use n = 10 and k = 2 in the formula provided to obtain the approximate probability of 0.2907.
66. Use n = 8 and each of k = 2, k = 1, and k = 0 in the formula provided (and add the results) to obtain the approximate probability of 0.8652.
Chapter 12 Review Solutions ---------------------------------------------------------------------- 1. 1,8, 27, 64 2. 1! 2! 3! 4!, , ,
1 2 3 4,
which is equivalent to 1,1,2,6 (Note that
!n
nnan
= =( )1 !n
n− ( )1 !n= − .)
3. 5,8,11,14 4. ( ) ( ) ( ) ( )1 2 3 43 4 5 61 , 1 , 1 , 1x x x x− − − −
which is equivalent to 3 4 5 6, , ,x x x x− − .
5. 5
52 32 0.133 243
a ⎛ ⎞= = =⎜ ⎟⎝ ⎠
6. 864
6561a =
7. ( ) ( )( )
( )( )
15
15
1 15 1 ! 14 !15 15 1 ! 15 16 !
a− −
= = −+
14!= −
15 16 15 14!⋅ ⋅ ⋅1
3600= −
8. 101 111
10 10a = + =
9. ( ) 11 3 , 1nna n n+= − ≥
10. , odd
1 , evenn
n na
nn
⎧⎪= ⎨⎪⎩
11. ( )1 , 1nna n= − ≥ 12. 110 , 1n
na n−= ≥
13. 8! 8 7 6!6!
⋅ ⋅=
6!56= 14. 20! 20!
23!=
23 22 21 20!⋅ ⋅ ⋅1
10, 626=
15. ( )( )
( )1 !1 !1 !
n nn nn
−−=
+ ( ) ( )1 1 !n n n+ −1
1n=
+ 16. ( ) ( )2 !2 !
!
nnn
−−=
( ) ( )1 2 !n n n− − ( )1
1n n=
−
17. 5,3,1, 1− 18. 2 2 2
4 16 256
1, 2 1, 4 1,16 1 1, 4,16, 256= =
⋅ ⋅ ⋅ =
Chapter 12 Review
1775
19. ( ) ( )2 2
324
1, 2, 2 1 ,4 2=
=
⋅ ⋅
which is equivalent to 1, 2,4,32 .
20. ( )2 2
14
32
1 21,2, ,2 1
4==
⎛ ⎞⎜ ⎟⎝ ⎠
which is equivalent to 11,2, ,324
.
21. ( )5
13 3 5 15
n== =∑
22. 4
21
1 1 1 1 144 36 16 914 9 16 144n n=
+ + += + + + =∑ 205
144=
23. ( ) ( )6
1
63 1 4 19 692n
n=
+ = + =∑
24. 15
0
2 4 8 16 32 642! 1 2! 3! 4! 5!
k
k k
+
=
= + + + + +∑
8 4 8 2182 4 43 3 15 15
= + + + + + =
25. ( )7
11
12
n
nn
−=
−∑
26. 10
1
2n
n=∑ 27.
0 !
n
n
xn
∞
=∑ ( )Note: 2! 2, 3! 6, 4! 24= = =
28. ( )1
01
!
nn
n
xn
+∞
=
−∑ 29.
60
600.0430,000 1 $36,639.9012
A ⎛ ⎞= + =⎜ ⎟⎝ ⎠
So, the amount in the account after 5 years is $36,639.90. 31. Yes, it is arithmetic with 2d = − .
32. No
30. Since ( )180, 000 30, 000 1 , 1na n n= + − ≥ we have
[ ]4
1
4 180,000 270,0002900,000
nn
a=
= +
=
∑
So, the total salary for 4 years is $900,000. Also, the salary in the 4th year is given by
( )4 180,000 30,000 3 270,000a = + = .
33. Yes, it is arithmetic with 12
d = .
Chapter 12
1776
34. Yes, it is arithmetic with 1d = − . 35. ( ) ( )1 ! 1 !
!n
n n na
n+ +
= =!n
1n= +
Yes, it is arithmetic with 1d = .
36. 5 5na n= − Yes, it is arithmetic with 5d = .
37. ( )1 1na a n d= + −
( ) ( )4 1 5 5 9n n= − + − = −
38. ( ) ( )5 1 6 6 1na n n= + − = −
39. ( ) 2 2 51 13 3 3na n n⎛ ⎞= + − − = − +⎜ ⎟
⎝ ⎠
40. ( ) ( )0.001 1 0.01 0.01 0.009na n n= + − = − 41. We are given that
( )5 5 1 13ia a d= + − = ,
( )17 17 1 37ia a d= + − = . In order to find na , we must first find d. To do so, subtract the above formulae to eliminate 1a :
1
1
4 1316 37
12 24
a da d
d
+ =− + =
− = −
Hence, 2d = . So, ( )( )1 1 2na a n= + − . To find 1a , observe that
( )5 1 2 4 8a a− = = So, 113 8 so thata− = 1 5a = . Thus, ( ) ( )5 1 2 2 3, 1na n n n= + − = + ≥
42. We are given that ( )7 14 7 1ia a d= − = + − ,
( )10 23 10 1ia a d= − = + − . In order to find na , we must first find d. To do so, subtract the above formulae to eliminate 1a :
1
1
6 149 23
3 9
a da d
d
+ = −− + = −
− =
Hence, 3d = − . So, ( ) ( )1 1 3na a n= + − − .
To find 1,a observe that
( )( )7 1 3 6 18a a− = − = − So, 114 18a− − = − , so that 1 4a = . Thus
( ) ( )4 1 3 3 7, 1na n n n= + − − = − + ≥
Chapter 12 Review
1777
43. We are given that ( )8 152 8 1a a d= = + − ,
( )21 1130 21 1a a d= = + − .
In order to find na , we must first find d. To do so, subtract the above formulae to eliminate 1a :
1
1
7 5220 130
a da d+ =
− + =
13 78d− = − Hence, 6d = . So, ( ) ( )1 1 6na a n= + − . To find 1, observea that
( )8 1 6 7 42a a− = = So, 152 42a− = , so that 1 10a = . Thus, ( )10 1 6 6 4, 1na n n n= + − = + ≥ .
44. We are given that ( )11 130 11 1a a d= − = + − ,
( )21 180 21 1a a d= − = + − . In order to find na , we must first find d. To do so, subtract the above formulae to eliminate 1a :
1
1
10 3020 80
a da d+ = −
− + = −
10 50d− = Hence, 5d = − . So,
( ) ( )1 1 5na a n= + − − . To find 1,observea that
( )11 1 5 10 50a a− = − = − So, 130 50a− − = − , so that 1 20a = . Thus,
( )( )20 1 5 5 25, 1na n n n= + − − = − + ≥ .
45. ( )20
1
203 3 60 6302k
k=
= + =∑ 46. ( ) ( )15
1
155 6 20 1952n
n=
+ = + =∑
47. The sequence is arithmetic with 12n = , 1 2a = , and 12 68a = . So, the given sum is
( )12 2 68 4202
+ = .
48. The sequence is arithmetic with
17n = , 114
a = , and 17314
a −= .
So, the given sum is 17 1 31 17 30 63.752 4 4 2 4
−⎛ ⎞ ⎛ ⎞− = = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
.
49. Bob Tania
( )45,000 1 2000na n= + − ( )38,000 4000 1na n= + −
15
15
1
15 45,000 73,0002n
n a
a= =
⎡ ⎤⎢ ⎥= +⎢ ⎥⎣ ⎦
∑ = 885,000 15
15
1
15 38,000 94,0002n
n a
a= =
⎡ ⎤⎢ ⎥= +⎢ ⎥⎣ ⎦
∑ = 990,000
So, Bob earns $885,000 in 15 years, while Tania earns $990,000 in 15 years. 50. The sequence 16,48,80, ,… is arithmetic with 1 16a = and 32d = . So, the nth term is given by ( )16 1 32 32 16na n n= + − = − . So, 5 144a = . Further,
[ ]5
1
5 16 144 4002n
na
=
= + =∑ . So, he would have fallen 400 feet in 5 seconds.
Chapter 12
1778
51. Yes, it is geometric with 2r = − . 52. No
53. Yes, it is geometric with 12
r = . 54. Yes, it is geometric with 10r = .
55. 3, 6,12, 24, 48 56. 10 10 10 1010, , , ,4 16 64 256
57. 100, 400, 1600, 6400, 25,600− − 58. 15 1560, 30, 15, ,
2 4− − −
59. 1 1
1 7 2 , 1n nna a r n− −= = ⋅ ≥
60. 1112 , 1
3
n
na n−
⎛ ⎞= ≥⎜ ⎟⎝ ⎠
61. ( ) 11 2 , 1n
na n−= − ≥ 62.
132 1 , 15 4
n
na n−−⎛ ⎞= ≥⎜ ⎟
⎝ ⎠
63. ( ) 12 2 , 1n
na n−= ≥
So, ( )2425 2 2 33,554, 432a = =
64. ( ) 11 2 , 12
nna n−= ≥
( )910
1 2 2562
a = =
65. 11100 , 1
5
n
na n−−⎛ ⎞= ≥⎜ ⎟
⎝ ⎠
So, in particular, 11
612
1100 2.048 105
a −−⎛ ⎞= = − ×⎜ ⎟⎝ ⎠
66. 111000 , 1
2
n
na n−−⎛ ⎞= ≥⎜ ⎟
⎝ ⎠
So, in particular, 10
111 10001000
2 1024a −⎛ ⎞= =⎜ ⎟
⎝ ⎠.
67. 99
1
1
1 1 1 3 19,6823 4920.502 2 1 3 4
n
n
−
=
⎡ ⎤−⎛ ⎞ = = =⎜ ⎟ ⎢ ⎥−⎝ ⎠ ⎣ ⎦∑
68.
11
111
1
111 21 1 12 1
2
1 40942 12148 2048
n
n
−
=
⎡ ⎤⎛ ⎞−⎢ ⎥⎜ ⎟⎛ ⎞ ⎝ ⎠⎢ ⎥=⎜ ⎟ ⎢ ⎥⎝ ⎠ −⎢ ⎥⎣ ⎦
⎛ ⎞= − =⎜ ⎟⎝ ⎠
∑
Chapter 12 Review
1779
69. ( )88
1
1
1 35 3 5 16, 4001 3
n
n
−
=
⎡ ⎤−= =⎢ ⎥−⎣ ⎦
∑ 70. First, note that the given sum is not in the standard form since the exponent on the common ratio is n rather than
1n − . So, you must first simplify the expression to put it into standard form, and then apply the known formula, as follows:
( ) ( )7 7
1
1 1
7
2 105 53 3
10 1 5 195,3103 1 5 3
n n
n n
−
= =
=
⎡ ⎤−= =⎢ ⎥−⎣ ⎦
∑ ∑
71. 1 3213
=−
72.
21 115 25
1 6 3015 5
−⎛ ⎞⎜ ⎟⎝ ⎠ = =+
73. ( ) 148,000 1.02 , 1n
na n−= ≥ So, in particular,
( )13 112 48000 1.02 60,875.61a −= = .
So, the salary after 12 years is $60,875.61.
74. ( ) 115000 0.80 , 1nna n−= ≥
So, in particular, ( )3 1
3 15,000 0.80 9600a −= = . So, the boat will be worth $9600 after 3 years.
75. Claim: 3 3 , for all 1nn n≤ ≥ . Proof. Step 1: Show the statement is true for 1n = .
13(1) 3≤ is true since both terms equal 3. Step 2: Assume the statement is true for n k= : 3 3kk ≤ Show the statement is true for 1n k= + : 13( 1) 3kk ++ ≤
1
1
3( 1) 3 33 3 (by assumption)3 3 (since 3 3 , for 1)2(3 ) 3(3 ) 3
k
k k k
k k k
k k
k+
+ = +
≤ +
≤ + ≤ ≥
= < =
This completes the proof. ■
Chapter 12
1780
76. Claim: 14 4 , for all 1n n n+< ≥ . Proof. Step 1: Show the statement is true for 1n = .
1 1 1 24 4 4 16+< = = is clearly true. Step 2: Assume the statement is true for n k= : 14 4k k+< Show the statement is true for 1n k= + : 1 24 4k k+ +<
1
1
2
4 4 44 4 (by assumption)4
k k
k
k
+
+
+
= ⋅
< ⋅
=
This completes the proof. ■ 77. Claim: 22 7 ... (5 3) (5 1), for all 1nn n n+ + + − = − ≥ . Proof. Step 1: Show the statement is true for 1n = .
1 12 22 (5(1) 1) (4)= − = is clearly true.
Step 2: Assume the statement is true for n k= : 22 7 ... (5 3) (5 1)kk k+ + + − = − Show the statement is true for 1n k= + :
122 7 ... (5( 1) 3) (5( 1) 1)kk k++ + + + − = + −
Observe that ( )
( )( )2
2(5 1) 2(5 2)
2
1 5 45 9 4 12 2 2
2 7 ... (5 3) (5( 1) 3) 2 7 ... (5 3) (5( 1) 3)(5 1) (5( 1) 3) (by assumption)
(5( 1) 1)
k
k k k
k kk k k
k k k kk k
k
− + +
+ ++ + +
+ + + − + + − = + + + − + + −
= − + + −
=
= = = + −
This completes the proof. ■
Chapter 12 Review
1781
78. Claim: 22 ( 1) , for all 3n n n> + ≥ . Proof. Step 1: Show the statement is true for 3n = .
2 218 2(3) (3 1) 16= > + = is clearly true. Step 2: Assume the statement is true for n k= : 2 22 ( 1)k k> + Show the statement is true for 1n k= + : 2 22( 1) ( 1 1)k k+ > + +
2 2
2
2
2
1 since 3
2
2( 1) 2 4 1( 1) 4 1
2 1 4 14 4 (2 2)
( 2)k
k k kk k
k k kk k k
k> ≥
+ = + +
> + + +
= + + + +
= + + + −
> +
This completes the proof. ■ 79.
( )11 11! 11! 11 10 9 8!8 11 8 !8! 3!8!
⎛ ⎞ ⋅ ⋅ ⋅= = =⎜ ⎟ −⎝ ⎠ 8! ( )
1653 2 1
=⋅ ⋅
80.
( )10 10! 10!0 10 0 !0!
⎛ ⎞= =⎜ ⎟ −⎝ ⎠ 10!
1 10!0!
= =
81.
( )22 22! 22!22 22 22 !22!⎛ ⎞
= =⎜ ⎟ −⎝ ⎠ 22!1 10!0!
= =
82.
( )47 47! 47!45 47 45 !45! 2!45!
47 46 45!
⎛ ⎞= =⎜ ⎟ −⎝ ⎠
⋅ ⋅=
45! ( )1081
2 1=
⋅
83. 4
4 4
0
4 0 3 1 2 2 1 3 0 4
4 0 3 1 2 2 1 3 0 44! 4! 4! 4! 4!4!0! 3!1! 2!2! 1!3! 0!4!
1 4 6 4 1
4 3
4( 5) ( 5)
4 4 4 4 4( 5) ( 5) ( 5) ( 5) ( 5)
0 1 2 3 4
( 5) ( 5) ( 5) ( 5) ( 5)
20 15
k k
k
x xk
x x x x x
x x x x x
x x
−
=
= = = = =
⎛ ⎞− = −⎜ ⎟
⎝ ⎠⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
= − + − + − + − + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
= − + − + − + − + −
= − +
∑
20 500 625x x− +
Chapter 12
1782
84. 5
5 5
0
5 0 4 1 3 2 2 3 1 4 0 5
5 0 4 1 3 2 2 3 1 4 0 55! 5! 5! 5! 5! 5!5!0! 4!1! 3!2! 2!3! 1!4! 0!5!
1 5 10 10 5 1
5 4 3 2
5( )
5 5 5 5 5 50 1 2 3 4 5
5 10
k k
kx y x y
k
x y x y x y x y x y x y
x y x y x y x y x y x y
x x y x y
−
=
= = = = = =
⎛ ⎞+ = ⎜ ⎟
⎝ ⎠⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
= + + + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
= + + + + +
= + + +
∑
2 3 4 510 5x y xy y+ +
85. 3
3 3
0
3 0 2 1 1 2 0 3
3 0 2 1 1 2 0 33! 3! 3! 3!3!0! 2!1! 1!2! 0!3!
1 3 3 1
3 2
3(2 5) (2 ) ( 5)
3 3 3 3(2 ) ( 5) (2 ) ( 5) (2 ) ( 5) (2 ) ( 5)
0 1 2 3
(2 ) ( 5) (2 ) ( 5) (2 ) ( 5) (2 ) ( 5)
8 60 150
k k
k
x xk
x x x x
x x x x
x x x
−
=
= = = =
⎛ ⎞− = −⎜ ⎟
⎝ ⎠⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
= − + − + − + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
= − + − + − + −
= − + −
∑
125
86.
( )
( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( )
4 42 3 4 2 3
0
4 3 2 1 02 3 0 2 3 1 2 3 2 2 3 3 2 3 4
4 3 2 1 02 3 0 2 3 1 2 3 2 2 3 3 2 34! 4! 4! 4! 4!4!0! 3!1! 2!2! 1!3! 0!4!
1 4 6 4 1
4( ) ( )
4 4 4 4 4( ) ( ) ( ) ( ) ( )
0 1 2 3 4
( ) ( ) ( ) ( ) (
k k
k
x y x yk
x y x y x y x y x y
x y x y x y x y x y
−
=
= = = = =
⎛ ⎞+ = ⎜ ⎟
⎝ ⎠⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
= + + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
= + + + +
∑
4
8 6 3 4 6 2 9 12
)
4 6 4x x y x y x y y= + + + +
Chapter 12 Review
1783
87. 5
5 5
0
5 0 4 1 3 2 2 3
1 4 0 5
5 0 4 1 3 25! 5! 5! 5!5!0! 4!1! 3!2! 2!3!
1 5 10 10
5( 1) ( ) (1) , 0
5 5 5 5( ) (1) ( ) (1) ( ) (1) ( ) (1)
0 1 2 3
5 5( ) (1) ( ) (1)
4 5
( ) (1) ( ) (1) ( ) (1) (
k k
k
x x xk
x x x x
x x
x x x x
−
=
= = = =
⎛ ⎞+ = ≥⎜ ⎟
⎝ ⎠⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
= + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
= + + +
∑
5 3 12 2 2
2 3
1 4 0 55! 5!1!4! 0!5!
5 1
2
) (1)
( ) (1) ( ) (1)
5 10 10 5 1
x x
x x x x x= =
+ +
= + + + + +
88.
( ) ( )
( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( )
2 1 2 13 3 3 3
2 1 2 1 2 13 3 3 3 3 3
2 1 2 1 2 1 2 13 3 3 3 3 3 3 3
2 1 2 1 23 3 3 3 3
6 66
0
6 0 5 1 4 2
3 3 2 4 1 5 0 6
6 0 5 16! 6! 6!
6!0! 5!1! 4!2!
1 6 15
6( )
6 6 60 1 2
6 6 6 63 4 5 6
k k
kx y x y
k
x y x y x y
x y x y x y x y
x y x y x
−
=
= = =
⎛ ⎞+ = ⎜ ⎟
⎝ ⎠⎛ ⎞ ⎛ ⎞ ⎛ ⎞
= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
+ + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
= + +
∑
( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( )
1 2 13 3 3
2 1 2 1 2 13 3 3 3 3 3
10 8 51 2 4 4 23 3 3 3 3 3 3 3
4 2 3 36!
3!3!
20
2 4 1 5 0 66! 6! 6!
2!4! 1!5! 0!6!15 6 1
4 2 26 15 20 15 6
y x y
x y x y x y
x x y x y x y x y x y y
=
= = =
+
+ + +
= + + + + + +
89. 5 5 0 4 3 2 2 3 4 5 0
5 4 3 2 2 3 4 5
( ) 5 ( ) 10 ( ) 10 ( ) 5 ( ) ( )5 10 10 5
r s r s r s r s r s r s s rr r s r s r s rs s
− = + − + − + − + − + −
= − + − + −
90. 4 4 0 3 1 2 2 1 3 0 4
4 4 3 3 2 2 2 2 3 3 4 4
( ) ( ) ( ) 4( ) ( ) 6( ) ( ) 4( ) ( ) ( ) ( )4 6 4
ax by ax by ax by ax by ax by ax bya x a bx y a b x y ab xy b y
+ = + + + +
= + + + +
91. This term is 8 2 2 68 8! 8 7 6!( 2) (4)2 2!6!
x x−⎛ ⎞ ⋅ ⋅− = =⎜ ⎟
⎝ ⎠ (2 1) 6!⋅6 6(4) 112x x=
So, the desired coefficient is 112.
Chapter 12
1784
92. This term is 3 7 3 47 7! 7 6 5 4!3 (27)3 3!4!
y y−⎛ ⎞ ⋅ ⋅ ⋅= =⎜ ⎟
⎝ ⎠ (3 2 1) 4!⋅ ⋅4 4(27) 945y y=
So, the desired coefficient is 945. 93. This term is
6 4 4 2 46 6! 6 5 4!(2 ) (5 ) (4 625)4 4!2!
x y x y−⎛ ⎞ ⋅ ⋅= ⋅ =⎜ ⎟
⎝ ⎠ (2 1) 4!⋅2 4 2 4(2500) 37,500x y x y=
So, the desired coefficient is 37,500.
94. This term is 2 8 4 4 8 48 8! 8 7 6 5 4!( ) ( )4 4!4!
r s r s−⎛ ⎞ ⋅ ⋅ ⋅ ⋅− = =⎜ ⎟
⎝ ⎠ (4 3 2 1) 4!⋅ ⋅ ⋅8 4 8 470r s r s=
So, the desired coefficient is 70.
95. Since 53 53! 53 52 51 50 49 48 47!6 6!47!
⎛ ⎞ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅= =⎜ ⎟
⎝ ⎠ 6! 47!22,957,480= , there are
22,957,480 different 6-number combinations.
96. Since 108 108! 108 107 ... 96 95!13 95!13!
⎛ ⎞ ⋅ ⋅ ⋅ ⋅= =⎜ ⎟
⎝ ⎠ 13! 95!20,592,957,740,000,000= , there are
20,592,957,740,000,000 different 13-card Canasta hands.
97. ( )7 4
7! 7! 7 6 5 4 3!7 4 ! 3!
P ⋅ ⋅ ⋅ ⋅= = =
− 3!840=
98. ( )9 9
9! 9! 9! 362,8809 9 ! 0!
P = = = =−
99. ( )12 5
12! 12! 12 11 10 9 8 7!12 5 ! 7!
P ⋅ ⋅ ⋅ ⋅ ⋅= = =
− 7!95,040=
100. ( )10 1
10! 10! 10 9!10 1 ! 9!
P ⋅= = =
− 9!10=
101. ( )12 7
12! 12! 12 11 10 9 8 7!12 7 !5! 7!5!
C ⋅ ⋅ ⋅ ⋅ ⋅= = =
− 7! ( )792
5 4 3 2 1=
⋅ ⋅ ⋅ ⋅
102. ( )40 5
40! 10! 40 39 38 37 36 35!40 5 !5! 35!5!
C ⋅ ⋅ ⋅ ⋅ ⋅= = =
− 35! ( )658,008
5 4 3 2 1=
⋅ ⋅ ⋅ ⋅
103. ( )9 9
9! 9! 19 9 !9! 0!9!
C = = =−
104. 53 653! 53 52 51 50 49 48 47!
6!47!C ⋅ ⋅ ⋅ ⋅ ⋅ ⋅
= =6! 47!
22,957, 480=
Chapter 12 Review
1785
105. There areModels Interior Exterior
3 2 5 30⋅ ⋅ = different cars to choose from.
106. Since each slot can have one of 26 letters, and order is important, there are 626 = 30,895,776 different passwords. 107. Since all 4 jobs are different with different responsibilities, we know order is important in how we fill the slots. As such, there are
( )10 410! 10! 10 9 8 7 6!
10 4 ! 6!P ⋅ ⋅ ⋅ ⋅= = =
− 6!5040= different leadership teams.
108. Since each of the 6 slots can be filled with one of 24 letters and 8 digits, and order is important, there are 632 = 1,073,741,824 different license plate numbers. 109. There are
Seat 1 Seat 2 Seat 3 Seat 4 Seat 5
5 4 3 2 1 120⋅ ⋅ ⋅ ⋅ = different seating arrangements.
Since there are 8 home games in a season, it would take 1208 15= years to go through all
possible arrangements.
110. ( )60 3
60! 60! 60 59 58 57!60 3 ! 57!
P ⋅ ⋅ ⋅= = =
− 57!205,320=
111. Since the prizes are different, we know order is important. So, there are
( )100 4100! 100! 100 99 98 97 96!
100 4 ! 96!P ⋅ ⋅ ⋅ ⋅= = =
− 96!94,109,400= different ways to
distribute the prizes.
112. Since 117 117! 117 116 115!
2 2!115!⎛ ⎞ ⋅ ⋅
= =⎜ ⎟⎝ ⎠ 2!115!
6786= , there are 6786 different matchups.
113. Since 52 52! 52 51 50 49 48 47 46!6 6!46!
⎛ ⎞ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅= =⎜ ⎟
⎝ ⎠ 6! 46!20,358,520= , there are
20,358,520 different 6-card hands. 114. Since
Ace Face or 10
8 32 256⋅ = , there are 256 different 2-card combinations (using two
decks) that lead to blackjack. 115. The sample space has 42 16= elements and there is only one way to get all heads. So, 1
16( ) 0.0625P HHHH = = . So, there is a 6.25% chance of getting all heads. 116. Since the sum of two dice is odd for the following rolls, we know that
P (sum is odd) 18 136 2
= = .
Odd sum rolls: ( ) ( ) ( )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
Chapter 12
1786
117. (The sample space for this problem can be found in Example 4 of Section 9.7 of the text.)
6 3036 36(not getting 7) 1 (getting 7) 1 0.833P P= − = − = =
So, there is about an 83.3% chance of not getting 7. 118. 13 1
52 4(diamond)P = = 119. 1 2
1 1 3 3(not ) 1 ( ) 1P E P E= − = − = ≈ 66.7% 120. Since 1E and 2E are mutually exclusive,
51 11 2 1 2 3 2 6( ) ( ) ( )P E E P E P E∪ = + = + =
121. Since 1E and 2E are not mutually exclusive,
1 2 1 2 1 271 1 1
3 2 4 12
( ) ( ) ( ) ( )58.3%
P E E P E P E P E E∪ = + − ∩
= + − = ≈
122. Since 1E and 2E are independent, 1 1 11 2 1 2 3 2 6( ) ( ) ( )P E E P E P E∩ = ⋅ = ⋅ = .
123. Since these two events are mutually exclusive, we see that P(drawing ace or drawing 2) = P(drawing ace) + P(drawing 2)
84 4 252 52 52 13 15.4%= + = = ≈ .
124. Since these two events are independent, we see that P(drawing ace on the first draw and drawing 2 on the second draw) = P(drawing ace on the first draw) iP(drawing 2 on the second draw) =
164 452 51 2652⋅ = .
125. The sample space in this problem has 52 32= elements. So, we see that
31132 32
(having at least one girl) 1 (having no girls)1 (having all boys)1 96.88%
P PP
= −= −= − = =
126. ( ) ( )111 1 12 2 4096
Win Lose
⋅ = 127. 53693600 128. The sum is infinite.
129. 34,87514 130. 6556
131. The graphs of the two curves are given by:
The series will sum to 1
1 2x+.
132. Yes, and the sum is approximately 7.4215 since
0
1 7.42151
n
n
ee e
ππ π
π
∞
=
⎛ ⎞ = = ≈⎜ ⎟ −⎝ ⎠ −∑
Chapter 12 Practice Test
1787
133. The sum is 99,900. Yes, it agrees with Exercise 77.
134. The graphs are given by
Yes, it confirms the proof in Exercise 78.
135. As each new term is added, the graphs become better approximations of the graph of 4(1 2 )y x= + for
0.1 0.1x− < < . The series does not get closer to this graph for 0.1 1x< < .
136. As each new term is added, the graphs become better approximations of the graph of 4(1 2 )y x= − for
0.1 0.1x− < < . The series does not get closer to this graph for 0.1 1x< < .
137. a. 11,440 b. 11,440 c. Yes d. This is true because !
n rPn rr C= .
138. a. 20,358,520 b. 20,358,520 c. Yes d. This is true because !
n rPn rr C= .
139. Using the formula with n = 10 and k = 9 (and then cubing the result since the events are independent) yields the approximate probability of 0.0722.
140. Using the formula with n = 8 and k = 9 (and then squaring the result since the events are independent) yields the approximate probability of 0.7795.
Chapter 12 Practice Test---------------------------------------------------------------------------- 1. 1nx − 2. geometric with r x=
3. 11
n
nxSx
−=
− 4.
0
n
n
x∞
=∑
5. We must have 1x < in order for the sum in Problem 4 to exist. In such case, the sum is 1
1 x− .
6. ( )131
3 11 3
11 2
n
n
∞
=
= =−∑ 7. ( ) ( ) ( ) ( )
101101041 1 1
4 4 411 4
13 3 1 1
1n
n=
−⋅ = ⋅ ⋅ = − ≈
−∑
8. ( ) ( )502 3 2(50) 1 25 104 2600+ + = = 9. ( ) ( )
100
1
1005 3 2 497 24,9502n
n=
− = + =∑
Chapter 12
1788
10. Claim: 2
( 1)
2 4 ... 2 , for all 1n n
n n n n= +
+ + + = + ≥ .
Proof. Step 1: Show the statement is true for 1n = .
22 1 1 1 1 2= + = + = , which is clearly true. Step 2: Assume the statement is true for n k= : 2 4 ... 2 ( 1)k k k+ + + = + Show the statement is true for 1n k= + : 2 4 ... 2( 1) ( 1)( 2)k k k+ + + + = + + Observe that
( )2 4 ... 2 2( 1) 2 4 ... 2 2( 1)( 1) 2( 1) (by assumption)
( 2)( 1)
k k k kk k kk k
+ + + + + = + + + + +
= + + += + +
This completes the proof. ■
11. 7! 7 6 5 4 3 2!2!
⋅ ⋅ ⋅ ⋅ ⋅=
2!2520=
12. Note that 5 5 4 3 25 5(2 ) (2 ) (2 ) (2 ) ...
1 2x y x x y x y⎛ ⎞ ⎛ ⎞+ = + + +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
Hence, 3 2 3 2 3 25 5!(2 ) 8 802 2!3!
x y x y x y⎛ ⎞= =⎜ ⎟
⎝ ⎠.
13. 15 15! 15 14 13 12!12 12!3!⎛ ⎞ ⋅ ⋅ ⋅
= =⎜ ⎟⎝ ⎠ 3!12!
455= 14. ( )
! ! 1! ! 0! !
k k kk k k k k⎛ ⎞
= = =⎜ ⎟ −⎝ ⎠
15. ( )14 3
14! 14! 14 13 12 11!14 3 ! 11!
P ⋅ ⋅ ⋅= = =
− 11!2184=
16. ( )200 3
200! 200! 200 199 198 197!200 3 !3! 197!3!
C ⋅ ⋅ ⋅= = =
− 197! ( )1,313, 400
3 2 1=
⋅ ⋅
17.
( ) ( )
( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
5 52 5 21 1
0
5 4 30 1 22 2 21 1 1
2 1 03 4 52 2 21 1 1
5 4 30 1 22 2 25! 5! 5!1 1 15!0! 4!1! 3!2!
1 5 10
2 132 25! 5!1 12!3! 1!4!
10 5
5( ) , 0
5 5 50 1 2
5 5 53 4 5
k kx x
k
x x x
x x x
x x x
x x
x x xk
x x x
x x x
x x x
x x
−
=
= = =
= =
⎛ ⎞+ = ≠⎜ ⎟
⎝ ⎠⎛ ⎞ ⎛ ⎞ ⎛ ⎞
= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
= + +
+ +
∑
( ) ( )
2 5
04 525! 10!5!
1
10 7 4 5 15 10 10
x
x x
x
x x x x=
+
= + + + + +
Chapter 12 Practice Test
1789
18. 4 4 0 3 2 2 3 0 4
4 3 2
4 3 2
4 4 4 4(3 2) (3 ) (2) (3 ) ( 2) (3 ) ( 2) (3 )( 2) (3 ) ( 2)
1 2 3 44! 4! 4!81 54 36 24 16
3!1! 2!2! 1!3!81 216 216 96 16
x x x x x x
x x x
x x x x
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞− = + − + − + − + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
= − + − +
= − + − +
19. Intuitively, there are more permutations than combinations since order is taken into account when determining the number of permutations, while order does not increase the number of combinations. More precisely, observe that
! 1 ! 1!( )! ! ( )! !n r n r
n nC Pr n r r n r r
⎛ ⎞= = =⎜ ⎟− −⎝ ⎠
.
Since ! 1,r ≥ it follows that 1! 1r ≤ , so that n r n rP C≥ .
20. Since order is important here, the number of ways of choosing 3 horses out of 15 is
( )15 315! 15! 15 14 13 12!
15 3 ! 12!P ⋅ ⋅ ⋅= = =
− 12!2730= .
So, the probability of NOT winning is 273015!1 1− ≈ .
21. 1838(red) 0.47P = ≅ . So, about 47% chance.
22. Since all spins are mutually independent, we see that P(red on 1st AND red on 2nd AND red on 3rd AND red on 4th AND red on 5th) =
P(red on 1st) iP(red on 2nd ) iP(red on 3rd) iP(red on 4th) iP(red on 5th) = ( )518
38 0.0238≅ So, about 2.38% chance. 23. Since spins are independent, the previous history has no effect on the probability of a particular spin. So, this probability is again 18
38 0.47≅ . 24. The desired probability is as follows:
3 31 4 2 110 9 8 7 6 420
Black on Blue on Red on Red on Green on first draw second draw third draw fourth draw fifth draw
⋅ ⋅ ⋅ ⋅ =
25. Since these two events are not mutually exclusive (since there is an ace of diamonds), the probability is:
13 164 1 452 52 52 52 13
(ace or diamond) (ace) (diamond) (ace AND diamond)0.308
P P P P= + −= + − = = ≈
26. a. 8! = 40,320 b. 18! 0.00002≈
27. Expanding this using a calculator reveals that the constant term is 184,756.
28. Using a calculator to compute this sum yields 73,375
12 .
Chapter 12
1790
Chapter 12 Cumulative Review------------------------------------------------------------------- 1.
( )
( )
10 1 5 4 10 1 5 43 2 2 3 3 2 3 2
10 1 5 43 2
6 43 22 3 2
23 2
x x x xx x x x
x xx
xx
xx
+ + + ++ = −
− − − −+ − +
=−
−=
−−
= =−
2. Let w = width of rectangle. Then, 2 5l w= − . Given that the perimeter is 38
inches, we have: 2 2(2 5) 38
2 4 10 386 48
8
w ww w
ww
+ − =+ − =
==
Thus, 11l = and so, the dimensions are 8 inches by 11 inches.
3. 25 ( 5) 4(2)(11) 5 63
2(2) 4
5 3 74
x
i
± − − ± −= =
±=
4. 5 3
5 3 or 5 38 or 2
xx x
x x
− >
− > − < −> <
So, the solution set is ( ) ( ), 2 8,−∞ ∪ ∞ . 5. Since the slope is undefined, the line is vertical and hence, is described by an equation of the form x a= . Since the point (-8,0) is on the line, the equation is 8x = − .
6. x-intercept: 3 5(0) 15 5. So, (5,0).x x− = ⇒ =
y-intercept: 3(0) 5( ) 15 3. So, (0, 3).y y− = ⇒ = − −
Slope 35m =
7.
( )
( )
2 2
2 2 2
( ) ( )
( ) 3( ) 2 3 2
2 3 3 2 3 2
2 32 3
f x h f xh
x h x h x xh
x hx h x h x xh
h x hx h
h
+ −
+ − + + − − +=
+ + − − + − + −=
+ −= = + −
8.
12 1
1( ( )) 4 4 (2 1) 2 3x
f g x x x+
= − = − + = − +
provided that 12x ≠ − .
Domain: ( ) ( )1 12 2, ,−∞ − ∪ − ∞
9.
( )( )
2
2
2
2
( ) 0.04 1.2 3
0.04 30 3
0.04 30 225 3 9
0.04( 15) 6
f x x x
x x
x x
x
= − + −
= − − −
= − − + − +
= − − +
So, the vertex is (15,6).
10. ( )( )2 2( ) 9 1
( 3 )( 3 )( 1)( 1)
P x x x
x i x i x x
= + −
= + − − +
Chapter 12 Cumulative Review
1791
11. Vertical asymptote: 3x = Horizontal asymptote: 5y = −
12. Use ( )1 ntrnA P= + .
Here, 65,000, 0.047, 52, 17A r n t= = = = .
So, we have ( )
( )52(17)0.04752
52(17)0.04752
65,000
1
65,000 1
29, 246.21
P
P+
= +
≈ =
So, about $29,246.21. 13. ln 6
2 ln 2log 6 2.585= ≈ 14.
2
2
ln(5 6) 25 6
6 2.6785
xx e
ex
− =
− =
+= ≈
15. Solve the system:
85
8 5 1510 ( )
x yy x− =⎧
⎨ = +⎩
(1)2
Substitute (2) into (1): ( )8
58 5 10 15 0 65x x− + = ⇒ = Since this results in a false statement, the system has no solution.
16. Using matrix methods, we have
1 2 2
11 12
2
1 1 12 2 2
2 1 1 1 2 1 1 11 1 4 3 0 1 7 5
150 1 7
R R R
R R
− →
→
⎡ − ⎤ ⎡ − ⎤⎯⎯⎯⎯⎯→⎢ ⎥ ⎢ ⎥− − −⎣ ⎦ ⎣ ⎦
⎡ ⎤−⎯⎯⎯→⎢ ⎥−−⎣ ⎦
Let y = a. Then, 7 5x a= − (using 2nd row). Hence, substituting these into the equation corresponding to the first row, we see that:
1 1 12 2 2(7 5) 11 13a a z z a− − + = ⇒ = −
17. The region is as follows:
Note on the graph: C1: y = 5 – x
Vertex 4 5z x y= + (1,4) 24 MAX (1,2) 14 (3,2) 22
So, the maximum is z(1,4) = 24.
Chapter 12
1792
18.
2 1 23 1 3
2 3 3
32
1 5 2 3 1 5 2 33 1 2 3 0 14 8 122 4 4 10 0 14 8 4
1 5 2 30 14 8 120 0 0 16
R R RR R R
R R R
− →− →
− →
⎡ − ⎤ ⎡ − ⎤⎢ ⎥ ⎢ ⎥− ⎯⎯⎯⎯→ − −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦
⎡ − ⎤⎢ ⎥⎯⎯⎯⎯→ − −⎢ ⎥⎢ ⎥−⎣ ⎦
The last row suggests the system has no solution.
19. 9 0 11 2 1 99 18 91 2 9 1 4 29 4 7
− −⎡ ⎤ ⎡ ⎤ ⎡ ⎤=⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦ ⎣ ⎦
20. 2 5 1
4 0 1 0 1 41 4 0 2 5 1
1 3 2 3 2 12 1 3
2(12) 5(3) 1(9) 0
−= − −
− −−
= − − =
21. Since (3,5) is the vertex and the directrix is x = 7, the parabola opens to the left and has general equation
- 24 ( 3) ( 5)p x y− = − To find p, note the distance from vertex to directrix is 4 units, so p = 4. Hence, the equation is
2
2
16( 3) ( 5)1 ( 5) 3
16
x y
x y
− − = −
= − − +
23. 0 1 2 3
2 13 3
2 2 2 2 1 1 31! 2! 3! 4!
+ + + = + + + =
24. The sequence can be described by
15 3 , 1nna n−= ⋅ ≥ .
So, it is geometric.
22.
25. 102 1024=
26. Using a calculator to compute this sum yields 20,090.
27. Using a calculator to expand this quantity then reveals that the constant term is -3,432.
