Answer
$\displaystyle \sum_{i=1}^{7}\frac{i^2}{(i+1)!}$
Work Step by Step
$\frac{1^2}{2}+\frac{2^2}{6}+\frac{3^2}{24}+\frac{4^2}{120}+...+\frac{7^2}{40,320}=\frac{1^2}{(1+1)!}+\frac{2^2}{(2+1)!}+\frac{1^2}{(3+1)!}+\frac{1^2}{(4+1)!}+...+\frac{7^2}{(7+1)!}=\displaystyle \sum_{i=1}^{7}\frac{i^2}{(i+1)!}$
