Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 12 - Sequences; Induction; the Binomial Theorem - 12.3 Geometric Sequences; Geometric Series - 12.3 Assess Your Understanding - Page 825: 62

Answer

$12$

Work Step by Step

We are given the geometric series: $\sum_{k=1}^{\infty} 8\left(\dfrac{1}{3}\right)^{k-1}$ Determine the elements of the geometric series: $a_1=8\left(\dfrac{1}{3}\right)^{1-1}=8$ $r=\dfrac{1}{3}$ We compute $|r|$: $|r|=\left|\dfrac{1}{3}\right|=\dfrac{1}{3}$ Because $|r|=\dfrac{1}{3}<1$, the series converges. Determine its sum: $S=\dfrac{a_1}{1-r}==\dfrac{8}{1-\dfrac{1}{3}}=\dfrac{8}{\dfrac{2}{3}}=12$
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