Answer
$\displaystyle \sum_{i=1}^{5}\frac{2^i-1}{2^{i+1}}$
Work Step by Step
$\frac{1}{4}+\frac{3}{8}+\frac{7}{16}+\frac{15}{32}+\frac{31}{64}=\frac{2^1-1}{2^{1+1}}+\frac{2^2-1}{2^{2+1}}+\frac{2^3-1}{2^{3+1}}+\frac{2^4-1}{2^{4+1}}+\frac{2^5-1}{2^{5+1}}=\displaystyle \sum_{i=1}^{5}\frac{2^i-1}{2^{i+1}}$
