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: 56

Answer

$S_{9}=10220$

Work Step by Step

We are asked to find the sum of: $5120+2560+1280+...+20$ We see that this is a geometric sequence with $a_1=5120$. We find $r$: $r=2560/5120=0.5$ We know that a geometric sequence has the form: $a_{n}=ar^{n-1}$ We use this to find $n$: $a_n=(5120)(0.5)^{n-1}$ $20=(5120)(0.5)^{n-1}$ $20=5120*(0.5)^n/0.5$ $10=5120*(0.5)^n$ $\frac{1}{512}=(\frac{1}{2})^n$ $512=2^n$ $n=\log_2 512=9$ We know the partial sum of a geometric sequence is: $S_n=a_1\frac{1-r^n}{1-r}$ $S_{9}=5120 \frac{1-(\frac{1}{2})^{9}}{1-\frac{1}{2}}=10220$
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