Answer
$\displaystyle \frac{341}{512}$
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 $a=1,\displaystyle \quad r=-\frac{1}{2}.$
So, $a_{n}=ar^{n-1} =(-\displaystyle \frac{1}{2})^{n-1}$
Given the last term, we find n:
$512=(-\displaystyle \frac{1}{2})^{n-1}$
... recognizing 512 as a power of 2,
.... or factoring, 512=2(256)=$2^{2}(128)...=2^{9}$
$(-\displaystyle \frac{1}{2})^{n-1}=(-\frac{1}{2})^{9}$
$n-1=9$
$n=10$
So,
$S_{10}=(1)\displaystyle \frac{1-(-\frac{1}{2})^{10}}{1-(-\frac{1}{2})}$
$=\displaystyle \frac{1-\frac{1}{1024}}{\frac{3}{2}}=\frac{\frac{1023}{1024}}{\frac{3}{2}}=\frac{1023\cdot 2}{1024\cdot 3}$
$=\displaystyle \frac{341}{512}$.
