Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 11 - Infinite Series - 11.2 Summing and Infinite Series - Exercises - Page 547: 26



Work Step by Step

Given $$\sum_{n=2}^{n=\infty} \frac{7 \cdot(-3)^{n}}{5^{n}}=\sum_{n=2}^{n=\infty} 7\left(\frac{-3}{5}\right)^{n}$$ Since the series is a geometric series with $|r|=\left|\frac{3}{5}\right|<1$, then the series converges and has the sum \begin{align*} S&= \frac{a_1}{1-r}\\ &=\frac{ 7\left(\frac{-3}{5}\right)^2}{1-\frac{-3}{5}}\\ &= \frac{63}{40} \end{align*}
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