Precalculus (6th Edition)

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

Chapter 11 - Further Topics in Algebra - 11.1 Sequences and Series - 11.1 Exercises: 87

Answer

$=\sum ^{8}_{i=1}\left( -\dfrac {1}{2}\right) ^{i-1}$

Work Step by Step

$1-\dfrac {1}{2}+\dfrac {1}{4}-\dfrac {1}{8}+...+\dfrac {-1}{128}=\left( -\dfrac {1}{2}\right) ^{1-1}+\left( -\dfrac {1}{2}\right) ^{2-1}+\left( -\dfrac {1}{2}\right) ^{3-1}+\left( -\dfrac {1}{2}\right) ^{4-1}+\ldots \left( -\dfrac {1}{2}\right) ^{8-1}=\sum ^{8}_{i=1}\left( -\dfrac {1}{2}\right) ^{i-1}$
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