Sequences Series and Mathematical Induction

Sequences Series and Mathematical Induction
Sequences Series and Mathematical Induction
Arithmetic sequences have the form a1
,a1 d,a1 2d,a1 3d, a1 4d, …,
a1 (n – 1)d, …, where a1
is the first term and d = an 1 – an
is the common
difference between successive terms. The closed formula for the nth term is an =
a1 (n – 1)d, n = 1, 2, 3, …. Inspection of the arithmetic sequence 2, 5, 8, 11,
14, … reveals that a1 = 2 and d = 3. Thus, the closed formula for its nth term is
an = a1 (n – 1)d = 2 (n – 1)3 = 3n – 1.
344. Inspection of the arithmetic sequence 6, 4, 2, 0, … reveals that a1 = 6 and
d = –2. Thus, the closed formula for its nth term is a = a (n – 1)d = 6 (n – 1)
(–2) = –2n 8.
345. Inspection of the arithmetic sequence , … reveals that a1 =
and . Thus, the closed formula for its nth term is
Sequences Series and Mathematical Induction
346. The recursive formula for an arithmetic sequence is a1 = a1 and an 1 = an
d, where a1
is the first term and n = 1, 2, 3, …. From question 343, a1 = 2 and d
= 3, so the recursive formula for 2, 5, 8, 11, 14, … is a1 = 2 and an 1 = an 3.
347. From question 344, a1 = 6 and d = –2, so the recursive formula for 6, 4, 2,
0, … is a = 6 and a =a (–2).
348. Geometric sequences have the form a1
,a1r,a1r
2
,a1r
3
,…, a1r
n–1
, …, where a1
is the first term and is the common ratio between successive terms. The
closed formula for the nth term is an = a1r
n–1
, n = 1, 2, 3, …. Because the
sequence 2, 6, 18, 54, … is geometric with a1 = 2 and a2 = 6, the common ratio r
= = 3. Thus, the closed formula of the sequence is an = a1r
n–1 = 2(3)
n–1
.
349. Because the sequence , … is geometric with a1 = and a2 = ,
the common ratio . Thus, the closed formula of the sequence is an =
a1r
n–1 = .
350. Because the sequence 3,–6,12,–24, … is geometric with a = 3 and a = –6,
the common ratio = –2. Thus, the closed formula of the sequence is an =
a1r
n–1 = 3(–2)
n–1
.
351. The recursive formula for a geometric sequence is a1 = a1 and an 1 = ran
,
Sequences Series and Mathematical Induction
n = 1, 2, 3, …. From question 348, a1 = 2 and r = 3, so the recursive formula for
2, 6, 18, 54, … is a1 = 2 and an 1 = 3an
.
352. From question 349, a1 = and r = , so the recursive formula for
, … is a1 = and an 1 = an
.
353. (A) a1 = 3(1) 2, a2 = 3(2) 2, a3 = 3(3) 2, a4 = 3(4) 2 : 5, 8, 11, 14
(B) a17 = 3(17) 2 = 53
354.
355. (A) a1 = –2(1)– 1, a2 = –2(2)– 1, a = –2(3)– 1, a4 = –2(4)– 1 : –3,–5,–7,–
9
(B) a100 = –2(100) – 1 = –201
356.
357.
358. Use the common difference d = an 1 – a or common ratio formulas
to check.
0.5 – 1 = –0.5; 0.25 – 0.5 = –0.25
These differences are not the same, so the sequence is not arithmetic.
The ratios are the same so the sequence is geometric.
359. Use the common difference d = an 1 – a or common ratio formulas
to check.
These differences are not the same, so the sequence is not arithmetic.
The ratios are not the same so the sequence is not geometric.
Thus, the sequence , is neither arithmetic nor geometric.
360. The sequence 2, 2, 2, 2, … is arithmetic with d = 0 and geometric with r =
1.
361. These are terms of an arithmetic sequence whose closed formula is an = 2
(n – 1)5 = 5n – 3. The number 147 must satisfy this formula. Therefore, 147 =
5n – 3. Solving this equation for n yields = 30. Thus, there are 30
terms in the list.
362. These are terms of a geometric sequence whose closed formula is an =
2(2)
n–1 = 2
n
. The number 4,096 must satisfy this formula. Therefore, 4,096 = 2
n
.
Solving using logarithms,
Thus, there are 12 terms included in the list.
(Note: Trial-and-error guessing also leads to the solution n = 12.)
363.
364. Make the following symbol substitution: k = j – 1. Then when k = 1, j = 2,
and when k = n, j = n 1. So you have
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