Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 11 - Further Topics in Algebra - 11.3 Geometric Sequences and Series - 11.3 Exercises - Page 1034: 44

Answer

$29511$

Work Step by Step

$\sum ^{9}_{n=3}3^{k}=\dfrac {a_{1}\left( r^{n}-1\right) }{r-1};a_{1}=3^{3}=27;r=3;n=9-3+1=7\Rightarrow s_{n}=\dfrac {27\times \left( 3^{7}-1\right) }{3-1}=29511$
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