Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 10 - Mid-Chapter Check Point - Page 1077: 15



Work Step by Step

As we are given that $\sum_{i=1}^{\infty}(\dfrac{-2}{5})^{i-1}$ we have $r=\dfrac{-2}{5}$ and $a_1=1$ This shows an infinite geometric sequence whose sum will be $S_{\infty}=\dfrac{a_1}{1-r}=\dfrac{1}{1-(\dfrac{-2}{5})}$ and $=\dfrac{1}{7/5}$ Thus, $S_{\infty}=\dfrac{5}{7}$
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