Precalculus (6th Edition) Blitzer

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

Chapter 10 - Section 10.3 - Geoetric Sequences and Series - Exercise Set - Page 1075: 71



Work Step by Step

From the given information, we can observe that it is a geometric series $1,2,4\cdots $ Here ${{a}_{1}}=1$ and common ratio $r=2$ and $n=15$. Using the formula for the sum of a geometric series, we can find the sum of finite terms ${{S}_{n}}=\frac{{{a}_{1}}\left( 1-{{r}^{n}} \right)}{\left( 1-r \right)}$. So; $\begin{align} & {{S}_{n}}=\frac{1\left( 1-{{2}^{15}} \right)}{\left( 1-2 \right)} \\ & =\frac{1\left( 1-32768 \right)}{-\left( 1 \right)} \\ & =32767 \end{align}$
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