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: 14

Answer

$\Sigma_{k=1}^{10}(\frac{4}{7})^k =1.3284$

Work Step by Step

$\Sigma_{k=1}^{10}(\frac{4}{7})^k =\frac{a(1-(r)^n)}{1-r}$ $a=(\frac{4}{7})^1=\frac{4}{7}$ $r=\frac{4}{7}$ $n= 10$ $\Sigma_{k=1}^{10}(\frac{4}{7})^k =\frac{(\frac{4}{7})(1-(\frac{4}{7})^{10})}{1-(\frac{4}{7})}=1.3284$
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