Answer
$\displaystyle \sum_{j=1}^{6}\frac{i!}{2^i}$
Work Step by Step
$\frac{1}{2}+\frac{2}{4}+\frac{6}{8}+\frac{24}{16}+\frac{120}{32}+\frac{720}{64}=\frac{1!}{2^1}+\frac{2!}{2^2}+\frac{3!}{2^3}+\frac{4!}{2^4}+\frac{5!}{2^5}+\frac{6!}{2^6}=\displaystyle \sum_{j=1}^{6}\frac{i!}{2^i}$
