Find the summation $\frac{1}{1!}+\frac{1+2}{2!}+\frac{1+2+3}{3!}+ \cdots$ What is the value of the following sum?

$$\frac{1}{1!}+\frac{1+2}{2!}+\frac{1+2+3}{3!}+ \cdots$$

The possible answers are:

A. $e$
B. $\frac{e}{2}$
C. $\frac{3e}{2}$
D.  $1 + \frac{e}{2}$

I tried to expand the options using the series representation of $e$ and putting in $x=1$, but I couldn't get back the original series. Any ideas?

You want calculate $\sum_{k=1}^\infty\frac{k(k+1)}{2(k!)}=\frac{1}{2}\sum_{k=1}^\infty\frac{k+1}{(k-1)!}=\frac{1}{2}\sum_{k=0}^\infty\frac{k+2}{k!}$...(*)
But, as $e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots$, then $x^2e^x=x^2+x^3+\frac{x^4}{2!}+\frac{x^5}{3!}+\cdots$. Deriving both sides give us $x^2e^x+2xe^x=2x+3x^2+\frac{4x^3}{2!}+\frac{5x^4}{3!}+...$. Thus, evaluating at $x=1$: $e+2e=\sum_{k=0}^\infty\frac{k+2}{k!}$.
Substituying the last in (*) give us finally $\frac{3e}{2}$.
