Answer
$\displaystyle \sum_{i=0}^{7}(-1)^i\frac{1}{2^i}$
Work Step by Step
$1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{1}{16}-\frac{1}{32}+\frac{1}{64}-\frac{1}{128}=(-1)^0\frac{1}{2^0}+(-1)^1\frac{1}{2^1}+(-1)^2\frac{1}{2^2}+(-1)^3\frac{1}{2^3}+(-1)^4\frac{1}{2^4}+(-1)^5\frac{1}{2^5}+(-1)^6\frac{1}{2^6}+(-1)^7\frac{1}{2^7}=\displaystyle \sum_{i=0}^{7}(-1)^i\frac{1}{2^i}$
