Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 8 - Sequences and Infinite Series - 8.3 Infinite Series - 8.3 Exercises: 9


$\dfrac {1\times \left( \dfrac {4}{25}\right) ^{21}-1}{\dfrac {4}{25}-1}=$

Work Step by Step

$\sum ^{20}_{k=0}\left( \dfrac {2}{5}\right) ^{2k}=a_{1}\dfrac {r^{n}-1}{r-1};a_{1}=1;r=\dfrac {4}{25};n=21\Rightarrow S_{n}=\dfrac {1\times \left( \dfrac {4}{25}\right) ^{21}-1}{\dfrac {4}{25}-1}=$ Using calculator as indicated in question we get
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