Finite Mathematics for Business, Economics, Life Sciences and Social Sciences, Barnett - Solutions, summaries, and outlines. 2022 updated

EXERCISE B-1 B-1


APPENDIX B SPECIAL TOPICS



EXERCISE B-1

2. an = 4n – 3; a1 = 4·1 – 3 = 1 2n  1 2 1  1 3
4. an = ; a1 = 
a2 = 4·2 – 3 = 5 2n 2 1 2
22 1 5
a3 = 4·3 – 3 = 9 a2 = 
22 4
a4 = 4·4 – 3 = 13
231 7
a3 = 
23 6
24 1 9
a4 = 
24 8

 1  n-1  1  1-1  1  0
6. an =    ; a1 =    =   = 1
 4  4  4

 1  2 - 1  1 1 1
a2 =    =   = –
 4  4 4

 1  3-1  1  2 1
a3 =    =   =
 4  4 16

 1  4-1  1  3 1
a4 =    =   = –
 4  4 64

8. an = 4n – 3; a15 = 4·15 – 3 = 57

2n  1 2  200  1 401
10. an = ; a200 = 
2n 2  200 400
5
12. k
k 1
2
= (1)2 + (2)2 + (3)2 + (4)2 + (5)2 = 1 + 4 + 9 + 16 + 25 = 55

4
14.  (2)
k 0
k
= (–2)0 + (–2)1 + (–2)2 + (–2)3 + (–2)4

= 1 – 2 + 4 – 8 + 16 = 11
4
1 1 1 1 1
16. 2
k 1
k
=   
21 22 23 24
1 1 1 1 8  4  2  1 15
=     
2 4 8 16 16 16
18. a1 = 7, a2 = 9, a3 = 9, a4 = 2, a5 = 4. Here n = 5 and the arithmetic mean is given by:
5
1 1 31
—
a =
5
a
i 1
i =
5
(7 + 9 + 9 + 2 + 4) =
5
= 6.2



Copyright © 2015 Pearson Education, Inc.

,B-2 APPENDIX B: SPECIAL TOPICS



20. a1 = 100, a2 = 62, a3 = 95, a4 = 91, a5 = 82, a6 = 87, a7 = 70, a8 = 75, a9 = 87, and a10 = 82. Here n =
10 and the arithmetic mean is given by:
10
1 1 831
—
a =
10
a
i 1
i =
10
(100 + 62 + 95 + 91 + 82 + 87 + 70 + 75 + 87 + 82) =
10
= 83.1

22. an = (–1)n(n – 1)2; a1 = (–1)1(1 – 1)2 = 0

a2 = (–1)2(2 – 1)2 = 1

a3 = (–1)3(3 – 1)2 = –4

a4 = (–1)4(4 – 1)2 = 9

a5 = (–1)5(5 – 1)2 = –16


1  ( 1)n 1  ( 1)1
24. an = ; a1 = =2
n 1
1  ( 1)2
a2 = =0
2
1  ( 1)3 2
a3 = =
3 3
1  ( 1)4
a4 = =0
4
1  ( 1)5 2
a5 = =
5 5

 1  n+1  1  1+1 1
26. an =    ; a1 =    =
 2  2 4
 1 2+1 1
a2 =    =–
 2 8
 1 3+1 1
a3 =    =
 2 16
 1  4+1 1
a4 =    =–
 2 32
 1 5+1 1
a5 =    =
 2 64
28. Given 4, 5, 6, 7, … The sequence is the set of successive integers beginning with 4. Thus, an = n + 3, n
= 1, 2, … .
30. Given –3, –6, –9, –12, … The sequence is the set of negative integers of the form –3n. Thus, an = –3n, n
= 1, 2, … .
1 2 3 4
32. Given , , , , The sequence is the set of all fractions of positive integers whose denominator is 1
2 3 4 5
plus the numerator. Thus,
n
an = , n = 1, 2, … .
n 1
Copyright © 2015 Pearson Education, Inc.

, EXERCISE B-1 B-3


34. Given –2, 4, –8, 16, … The sequence consists of positive integer powers of (–2). Thus,

an = (–2)n, n = 1, 2, … .

36. Given 3, –6, 9, –12, … The sequence consists of integer multiples of 3 with alternating sign. Thus,

an = (–1)n+13n, n = 1, 2, … .

4 16 64 256 4
38. Given , , , , The sequence consists of the positive integer powers of   . Thus,
3 9 27 81  3
4n
an =   , n = 1, 2, … .
 3

40. Given 1, 2x, 3x2, 4x3, … The sequence consists of non-negative integer powers of x multiplied by a
number which is one more than the power. Thus, an = nxn-1, n = 1, 2, … .

x2 x3 x 4
42. Given x, , , , The sequence consists of positive integer powers of x divided by the power.
2 3 4
Thus,
xn
an = , n = 1, 2, … .
n
4
( 2)k 1 ( 2)11 ( 2) 21 ( 2)31 ( 2) 41 4 8 16 32 4 8 16 32
44. 
k 1 2k  1
=    = 
2 1  1 2  2  1 2  3  1 2  4  1 3 5
 
7 9
=  
3 5 7

9
7
( 1)k ( 1)3 ( 1) 4 ( 1)5 ( 1)6 ( 1)7 1 1 1 1 1
46. 
k 3
2
k k
= 2
 2  2  2
3 3 4 4 5 5 6 6 7 7
 2 =–    
6 12 20 30 42
3
1 k 1 k + 1 1 1 1 x3 x4
48. 
k 1 k
x x = x1+1 + x2+1 + x3+1 = x2 +
1 2 3 2

3
4
( 1)k x 2 k ( 1)0 x 2(0) ( 1)1 x 2(1) ( 1) 2 x 2(2) ( 1)3 x 2(3) ( 1)4 x 2(4)
50. 
k 0 2k  2
=
2(0)  2
+
2(1)  2
+
2(2)  2
+
2(3)  2
+
2(4)  2
1 x 2 x 4 x 6 x8
=    
2 4 6 8 10
4 3
52. (A) 1 2 + 22 + 32 + 42 = k
k 1
2
(B) 12 + 2 2 + 3 2 + 4 2 =  ( j  1)
j 0
2




1 1 1 1 5
( 1) k 1 1 1 1 1 4
( 1) j
54. (A) 1–    =
3 5 7 9
 2k  1
k 1
(B) 1–    =
3 5 7 9
 2 j 1
j 0


n n
1 1 1 1
56. 1+
2 2
 2   2 =
3 n
k
k 1
2
58. 1 – 4 + 9 – … + (–1)n+1n2 =  (1)
k 1
k 1 2
k



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, B-4 APPENDIX B: SPECIAL TOPICS


1 1 1 1 1 1 1 1 1  1 1  1 
60. True. Let I = + + +…+ , then I = +      n1  = + I  n 
2 4 8 2n 2 2 2 4 2  2 2  2 
1 1 1 1 1 1
So, I = + I – n1 or I= – n1 .
2 2 2 2 2 2
1
Thus, I = 1 – n < 1.
2
62. True. Observe that if n is even, then
1 1 1 ( 1) n 1  1   1 1   1 1 1 1
1– + – +…+ = 1   +    + … +    >1– = ;
2 3 4 n  2 3 4  n 1 n  2 2
if n is odd, then
1 1 1 ( 1) n 1  1   1 1   1 1  1 1 1
1– + – +…+ = 1   +    + … +   + >1– = .
2 3 4 n  2   3 4   n  2 n  1  n 2 2

1
64. a1 = 3 and an = 2an-1 – 2 66. a1 = 1 and an = – an-1 for n ≥ 2.
3
for n ≥ 2.
a1 = 1
a1 = 3 1 1
a2 = 2·3 – 2 = 4 a2 = – ·1 = –
3 3
a3 = 2·4 – 2 = 6 1  1 1
a3 = – ·    =
a4 = 2·6 – 2 = 10 3  3 9
a5 = 2·10 – 2 = 18 1 1 1
a4 = – ·   = –
3 9 27
1  1  1
a5 = – ·    =
3  27  81
A 1  A 
68. In a1 = , an =  an 1   , n ≥ 2, let A = 6. Then:
2 2  an 1 
6
a1 = =3
2
1  A 1  6 1 5
a2 =  a1   = 3  = (3  2) =
2  a1  2  3 2 2
1  A 1  5 6  1  5 12  49
a3 =  a2   =    =   =
2  a2  2  2 5 2  2  2 5  20
1  A 1  49 6  4,801
a4 =  a3   =    = ;
2  a3  2  20 49 20  1,960
4,801
a4 = ≈ 2.4494898, 6 ≈ 2.4494897
1,960




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