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: 32

Answer

$=\dfrac {1}{4}$

Work Step by Step

$\dfrac {1}{16}+\dfrac {3}{64}+\dfrac {9}{256}+\dfrac {27}{1024}=\dfrac {1}{16}+\dfrac {1}{16}\times \left( \dfrac {3}{4}\right) +\dfrac {1}{16}\times \left( \dfrac {3}{4}\right) ^{2}+\dfrac {1}{16}\times \left( \dfrac {3}{4}\right) ^{3}\ldots =\dfrac {a_{1}}{1-r};a_{1}=\dfrac {1}{16};r=\dfrac {3}{4}\Rightarrow S_{\infty }=\dfrac {\dfrac {1}{16}}{1-\dfrac {3}{4}}=\dfrac {1}{4}$
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