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: 33

Answer

$=\dfrac {2500}{19}$

Work Step by Step

$\sum ^{\infty }_{k=0}\left( \dfrac {1}{4}\right) ^{k}\times 5^{3-k}=\sum ^{\infty }_{k=0}\left( \dfrac {1}{20}\right) ^{k}\times 5^{3}=\dfrac {a_{1}}{1-r};a_{1}=5^{3};r=\dfrac {1}{20}\Rightarrow S_{\infty }=\dfrac {a_{1}}{1-r}=\dfrac {5^{3}}{1-\dfrac {1}{20}}=\dfrac {2500}{19}$
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