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: 7

Answer

$9841$

Work Step by Step

$\sum ^{8}_{k=0}3^{k}=\dfrac {a_{1}\left( r^{n}-1\right) }{r-1};a_{1}=3^{0}=1;r=3;n=9\Rightarrow S_{n}=\dfrac {1\times \left( 3^{9}-1\right) }{3-1}=9841$
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