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: 43

Answer

$2032$

Work Step by Step

$\sum ^{10}_{k=4}2^{k}=\dfrac {a_{1}\times \left( r^{n}-1\right) }{r-1};a_{1}=2^{4}=16;r=2;n=10-4+1=7\Rightarrow S_{n}=\dfrac {16\times \left( 2^{7}-1\right) }{2-1}=16\times 127=2032$
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