College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 8, Sequences and Series - Section 8.3 - Geometric Sequences - 8.3 Exercises - Page 615: 64

Answer

$S_6=1330$

Work Step by Step

We need to find: $\sum_{k=1}^{6}64(\frac{3}{2})^{k-1}$ We see that this is a geometric sequence with $a=64$ and $r=\frac{3}{2}$. We know the partial sum of a geometric sequence is: $S_n=a_1\frac{1-r^n}{1-r}$ $S_6=64\frac{1-(\frac{3}{2})^{6}}{1-(\frac{3}{2})}=1330$
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