Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 10: Infinite Sequences and Series - Section 10.1 - Sequences - Exercises 10.1 - Page 569: 17

Answer

$a_{n}= \frac{2^{n-1}}{3(n+2)},\quad n=1,2,3,...$

Work Step by Step

$n=1,\displaystyle \qquad a_{1}=\frac{2^0}{3\cdot 3}\qquad $ $n=2,:\displaystyle \qquad a_{2}=\frac{2^{1}}{3\cdot 4} $ $n=3,:\displaystyle \qquad a_{3}=\frac{2^{2}}{3\cdot 5} $ $n=4,:\displaystyle \qquad a_{4}=\frac{2^{3}}{3\cdot 6} \qquad $... a pattern emerges, $a_{n}=\displaystyle \frac{2^{n-1}}{3(n+2)}$ Testing $a_{5},$ $ a_{5}=\displaystyle \frac{2^{5-1}}{3(5+2)}=\frac{2^{4}}{21},\qquad$... yes, the pattern works. Testing on the previous terms: $ a_{1}=\displaystyle \frac{2^{1-1}}{3(1+2)}=\frac{1}{9},\qquad$... yes. $ a_{2}=\displaystyle \frac{2^{2-1}}{3(2+2)}=\frac{2}{12},\qquad$... yes. $ a_{3}=\displaystyle \frac{2^{3-1}}{3(3+2)}=\frac{2^{2}}{15},\qquad$... yes. The proposed formula works for all the listed terms. $a_{n}=\displaystyle \frac{2^{n-1}}{3(n+2)},\quad n=1,2,3,...$
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