• # question_answer $1+\frac{1+2}{1\,!}+\frac{1+2+3}{2\,!}+\frac{1+2+3+4}{3\,!}+....\infty =$ A) 0 B) 1 C) $\frac{7e}{2}$ D) $2\,e$

$\frac{1}{0!}+\frac{1+2}{1!}+\frac{1+2+3}{2!}+....\infty$ ${{n}^{th}}$term ${{T}_{n}}=\frac{1+2+3+4+.....+n}{(n-1)!}=\frac{n(n+1)}{2(n-1)!}$ ${{T}_{n}}=\frac{1}{2}\left[ \frac{1}{(n-3)!}+\frac{4}{(n-2)!}+\frac{2}{(n-1)!} \right]$ Therefore sum${{S}_{\infty }}=\frac{7e}{2}$.