Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 11 - Further Topics in Algebra - 11.3 Geometric Sequences and Series - 11.3 Exercises - Page 1034: 33

Answer

$682$

Work Step by Step

$r=\dfrac {128}{32}=\dfrac {32}{8}=\dfrac {8}{2}=4;a_{1}=2\Rightarrow S_{n}=\dfrac {a_{1}\left( r^{n}-1\right) }{r-1}=\dfrac {2\times \left( 4^{5}-1\right) }{4-1}=682$
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