Answer
$=\sum ^{8}_{i=1}\left( -\dfrac {1}{2}\right) ^{i-1}$
Work Step by Step
$1-\dfrac {1}{2}+\dfrac {1}{4}-\dfrac {1}{8}+...+\dfrac {-1}{128}=\left( -\dfrac {1}{2}\right) ^{1-1}+\left( -\dfrac {1}{2}\right) ^{2-1}+\left( -\dfrac {1}{2}\right) ^{3-1}+\left( -\dfrac {1}{2}\right) ^{4-1}+\ldots \left( -\dfrac {1}{2}\right) ^{8-1}=\sum ^{8}_{i=1}\left( -\dfrac {1}{2}\right) ^{i-1}$
