What is the sum of this series involving factorial in denominator? $$1 + \frac{1^2 + 2^2}{2!} + \frac{{1}^2 + {2}^2 + 3^2}{3!} + \cdots$$
I can't figure out how to do summations which involve a factorial term in the denominator. Please help.
This is a past year IITJEE question by the way.

HINT:
$$\sum_{r=1}^n\dfrac{1^2+2^2+\cdots+r^2}{r!}=\frac16\sum_{r=1}^n\dfrac{r(r+1)(2r+1)}{r!}=\frac16\sum_{r=1}^n\dfrac{r(r+1)(2r+1)}{r!}$$
Now for $r>0,$ $$\dfrac{r(r+1)(2r+1)}{r!}=\dfrac{(r+1)(2r+1)}{(r-1)!}$$
Let $(r+1)(2r+1)=2(r-1)(r-2)+a(r-1)$
Set $r=2$ to get $a$
Now, $$e^x=\sum_{r=0}^\infty\dfrac{x^u}{u!}$$
