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 - 12.1 Sequences - 12.1 Assess Your Understanding - Page 807: 55



Work Step by Step

$\sum_{k=0}^{n-1} \left(\frac{1}{3^{k+1}}\right)=\left(\frac{1}{3^{0+1}}\right)+\left(\frac{1}{3^{1+1}}\right)+...+\left(\frac{1}{3^{(n-1)+1}}\right)=\frac{1}{3}+\frac{1}{9}+...+\frac{1}{3^n}$
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