B-1
APPENDIX B SPECIAL TOPICS
EXERCISE B-1
2. an = 4n – 3; a1 = 4·1 – 3 = 1
a2 = 4·2 – 3 = 5
4. an = 21
2
n
n
; a1 = 21 1 3
21 2
6. an = 1
4
n-1
; a1 = 1
4
1-1
= 1
4
0
= 1
2-1
1
3-1
2
4-1
3
10. an = 21
n
; a200 = 2 200 1 401
12.
5
2
k
14.
(2)
k
16.
4
1
=
1234
1111
18. a1 = 7, a2 = 9, a3 = 9, a4 = 2, a5 = 4. Here n = 5 and the arithmetic mean is given by:
1
i
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:
22. an = (–1)n(n – 1)2; a1 = (–1)1(1 – 1)2 = 0
24. an = 1(1)
n
n
; a1 =
1
1(1)
1
= 2
2
1(1)
= 0
26. an = 1
2
n+1
; a1 = 1
2
1+1
= 1
4
2+1
28. Given 4, 5, 6, 7, … The sequence is the set of successive integers beginning with 4. Thus, an = n + 3, n
30. Given –3, –6, –9, –12, … The sequence is the set of negative integers of the form –3n. Thus, an = –3n, n
32. Given
1234
,,,,
2345
The sequence is the set of all fractions of positive integers whose denominator is 1
EXERCISE B-1 B-3
34. Given –2, 4, –8, 16, … The sequence consists of positive integer powers of (–2). Thus,
36. Given 3, –6, 9, –12, … The sequence consists of integer multiples of 3 with alternating sign. Thus,
38. Given
41664256
,,, ,
392781
The sequence consists of the positive integer powers of 4
3
. Thus,
40. Given 1, 2x, 3x2, 4x3, … The sequence consists of non-negative integer powers of x multiplied by a
42. Given x,
234
,,,
234
xxx
The sequence consists of positive integer powers of x divided by the power.
x
44.
41
(2)
k
11 21 31 41
(2) (2) (2) (2)
= 4 8 16 32
46.
7
(1)
k
34567
(1) (1) (1) (1) (1)
= – 11111
48.
3
1
1k
x
=
1
34
x
x
50.
42
(1)
kk
x
02(0)
(1)
x
12(1)
(1)
x
22(2)
(1)
x
32(3)
(1)
x
42(4)
(1)
x
2 + 22 + 32 + 42 =
4
2
3
2
54. (A) 1 –
1111
=
51
(1)
k
=
4
(1)
j
56. 1 + 22 2
11 1
= 2
1
n
(1)
n
k
k
B-4 APPENDIX B: SPECIAL TOPICS
60. True. Let I = 1
11 1
= 1
I
62. True. Observe that if n is even, then
1 – 1
1
(1)
n
= 1
12
+ 11
+ … + 11
> 1 – 1
64. a1 = 3 and an = 2an-1 – 2
66. a1 = 1 and an = – 1
3an-1 for n ≥ 2.
a1 = 1
68. In a1 = 2
A, an = 1
1
1
2n
n
A
aa
, n ≥ 2, let A = 6. Then:
a1 = 6
EXERCISE B-2 B-5
70. b1 = 5
5
15
2
= 5
5(1.618034) ≈ 0.724
EXERCISE B-2
2. (A) 5, 20, 100, …
(B) –5, –5, –5, …
(C) 7, 6.5, 6, …
(D) 512, 256, 128, …
4.
200
3333 3
(200 times). This series is arithmetic with d = 3 and geometric with r = 1.
6. This is a geometric series with r = –3. Using sum formula we have S20 =
20
3(( 3) 1)
8. Neither arithmetic nor geometric, since an – an-1 is not the same for all n and
n
a
is not the same for all
10. a1 = –2; d = –3
B-6 APPENDIX B: SPECIAL TOPICS
12. a1 = 8; d = –10:
14. a1 = 203; a30 = 261:
18. a1 = 3; a7 = 2,187; r = 3:
ra a
20. a1 = 240; r = 1.06:
22. a1 = 100; a10 = 300:
24. a1 = 8,000; r = 0.4:
26. S50 =
50
(2 3)
k
. The sequence of terms is an arithmetic sequence. Therefore,
28. S8 =
8
2k
. The sequence of terms is a geometric sequence with common ratio r = 2 and a1 = 21 = 2.
30. Let a1 = 24, d = 2. Then, using the formula an = a1 + (n – 1)d, we can find n.
32. (A) 16, 4, 1, …
. Since r = 4
16 = 1
4 and |r| < 1, the sum exists:
EXERCISE B-2 B-7
(B) 1, –3, 9, …
34. g(t) = 18 – 3t:
36. g(x) = 2x
38. Use a1 = 2 and d = 2 in Sn = 2
n[2a1 + (n – 1)d]:
40. Yes. For example, let a1 = 1
2 and let 0 < r < 1, then S =
1
2
1r. For S ≥ 1000, we should solve the
inequality
42. Yes. Using the sum formula Sn = 2
n(a1 + an) with a1 = 1, an = 1.1 and Sn = 105, we obtain n = 100.
44. No. Using the sum formula for an infinite geometric series 1
1
a
Sr
with a1 = 10 and S = 5, we
46. Consider the time line:
$5400 $5100 $600 $300
50. $P
EXERCISE B-3
16. 74
C = 7!
18. 55
C = 5! 5! 5!
EXERCISE B-3 B-9
20. 18 3
C = 18!
= 18 17 16
22. (m + n)5 = 50
Cm5 + 51
Cm4n + 52
Cm3n2 + 53
Cm2n3 + 54
Cmn4 + 55
Cn5
24. (u – 2)5 = 50
Cu5 + 51
Cu4(–2) + 52
Cu3(–2)2 + 53
Cu2(–2)3 + 54
Cu(–2)4 + 55
C (–2)5
26. (x – 2y)5 = 50
Cx5 + 51
Cx4(–2y) + 52
Cx3(–2y)2 + 53
Cx2(–2y)3 + 54
Cx(–2y)4 + 55
C (–2y)5
28. The third term in the expansion of (x – 3)20 is:
30. The 13th term in the expansion of (p + q)15 is:
32. The third term in the expansion of (2x + y)12 is:
34. nr
C = !
n
n
n
= nnr
C
36. According to the Binomial Theorem:
38. nr
C = !
n
·!
n
= 1nr
·!
n