Answer
$\sum_{k=0}^7 (-1)^k \frac{1}{2^k}$
Work Step by Step
$$1-\frac12+\frac14-\frac18+\dots-\frac{1}{128}=$$
$$=(-1)^0\frac{1}{2^0}+(-1)^1\frac{1}{2^1}+(-1)^2\frac{1}{2^2}+(-1)^3\frac{1}{2^3}+(-1)^4\frac{1}{2^4}+(-1)^5\frac{1}{2^5}+(-1)^6\frac{1}{2^6}+(-1)^7\frac{1}{2^7}=\sum_{k=0}^7 (-1)^k \frac{1}{2^k}$$
