Question

$1+\frac{2^3}{2!}+\frac{3^3}{3!}+\frac{4^3}{4!}+...=ce$.
Find $c$

Answer 1

General term, $T_n=\frac{n^3}{n!}=\frac{n^2}{(n-1)!}=\frac{(n^2-1)+1}{(n-1)!}=\frac{(n-1)(n+1)+1}{(n-1)!}$

$=\frac{(n-2)+3}{(n-2)!}+\frac{1}{(n-1)!}=\frac{1}{(n-3)!}+\frac{3}{(n-2)!}+\frac{1}{(n-1)!}$

$\therefore$ Sum $=\sum T_n={\sum }_{n=3}^{\infty }\frac{1}{(n-3)!}+{\sum }_{n=2}^{\infty }\frac{3}{(n-2)!}+{\sum }_{n=1}^{\infty }\frac{1}{(n-1)!}=5e$