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: 14



Work Step by Step

As we are given that $\sum_{i=1}^{6}(\dfrac{3}{2})^i$ Now, terms will be {$\dfrac{3}{2},\dfrac{9}{4},\dfrac{27}{8}...$} This shows an geometric sequence. whose sum will be $S_n=\dfrac{a_1(1-r^n)}{1-r}$ Thus, $S_6=\dfrac{(\dfrac{3}{2})(1-(\dfrac{3}{2})^6)}{1-\dfrac{3}{2}}=\dfrac{\dfrac{3}{2}(\dfrac{-665}{64})}{-1/2}$ $S_6=\dfrac{1995}{64}$
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