Answer
$10,220$
Work Step by Step
See p. 861.
For the geometric sequence $a_{n}=ar^{n-1}$
the nth partial sum$ S_{n}=\displaystyle \sum_{k=1}^{n}ar^{k-1}$ (where $r\neq 1$)
is given by$ \displaystyle \quad S_{n}=a\cdot\frac{1-r^{n}}{1-r}$
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We see that $a=5120,\displaystyle \quad r=\frac{1}{2}$
So, $a_{n}=ar^{n-1} =5120(\displaystyle \frac{1}{2})^{n-1}$
Given the last term, we find n:
$20=5120(\displaystyle \frac{1}{2})^{n-1}\qquad/\div(5120)$
$\displaystyle \frac{1}{256}=(2)^{n-1}$
... recognizing $256$ as a power of 2,
.... or factoring, $256$=2($128$)=$2^{2}(64)...=2^{8}$
$(\displaystyle \frac{1}{2})^{8}=(\frac{1}{2})^{n-1}$
$n-1=8$
$n=9$
So
$S_{9}=5120\displaystyle \cdot\frac{1-(0.5)^{9}}{1-(0.5)}=10,220$
