Answer
$9841$
Work Step by Step
$\sum ^{8}_{k=0}3^{k}=\dfrac {a_{1}\left( r^{n}-1\right) }{r-1};a_{1}=3^{0}=1;r=3;n=9\Rightarrow S_{n}=\dfrac {1\times \left( 3^{9}-1\right) }{3-1}=9841$
Calculus: Early Transcendentals (2nd Edition)
by Briggs, Bill L.; Cochran, Lyle; Gillett, Bernard
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 - Page 623: 7
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