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?

Starting with $\sum_{i=1}^ni=\frac{n(n+1)}{2}$, the sum simplifies to 
\begin{align*}
z=\frac{1}{2}\sum_{i=1}^\infty \frac{i(i+1)}{i!}&=\frac{1}{2}\sum_{i=1}^\infty \frac{i+1}{(i-1)!}\\
&=\frac{1}{2}\sum_{i=0}^\infty\frac{i+2}{i!}\\
&=\frac{1}{2}\sum_{i=0}^\infty\left(\frac{i}{i!}+\frac{2}{i!}\right)\\
&=\frac{1}{2}\left(\sum_{i=1}^\infty\frac{1}{(i-1)!}+\sum_{i=0}^\infty\frac{2}{i!}\right)\\
&=\frac{1}{2}\left(\sum_{i=0}^\infty\frac{1}{i!}+2\sum_{i=0}^\infty\frac{1}{i!}\right)\\
&=\frac{1}{2}(e+2e)=\frac{3e}{2}
\end{align*}
