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: 28

Answer

$=\dfrac {64}{49}$

Work Step by Step

$\sum ^{\infty }_{k=3}\dfrac {3\times 4^{k}}{7^{k}}=\sum ^{\infty }_{n=3}3\times \left( \dfrac {4}{7}\right) ^{k}=\dfrac {a_{1}}{1-r};a_{1}=3\times \left( \dfrac {4}{7}\right) ^{3};r=\dfrac {4}{7}\Rightarrow S_{\infty }=\dfrac {a_{1}}{1-r}=\dfrac {3\times \left( \dfrac {4}{7}\right) ^{3}}{1-\dfrac {4}{7}}=\dfrac {64}{49}$
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