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?
 A: $$\frac{k(k+1)}{2k!}=\frac{k(k-1)+2k}{2k!}=\frac1{2(k-2)!}+\frac1{(k-1)!}$$
Hence the sum is
$$\frac e2+e.$$

Note that the first summation in the RHS must be started at $k=2$ (or use $1/(-1)!:=0$).
A: Clearly the $r^{th}$ numerator is $1+2+3+...+r= \frac{r(r+1)}{2}$ . 
And the $r^{th}$ denominator is $r!$.
Thus $$\displaystyle U_r=\frac{\frac{r(r+1)}{2}}{r!}=\frac{r(r+1)}{2r!}$$
Since the degree of the numerator is $2$ , use partial fractions to find $A,B,C$ such that (If you use partial fractions up to $(r-3)!$ , its' coefficient will be zero when comparing coefficients.)
$\displaystyle U_r=\frac{r(r+1)}{2r!}=\frac{A}{(r-2)!}+\frac{B}{(r-1)!}+\frac{C}{r!}$
$\displaystyle (2r!)\times U_r=(2r!)\times \frac{r(r+1)}{2r!}=(2r!)\times \frac{A}{(r-2)!}+(2r!)\times \frac{B}{(r-1)!}+(2r!)\times \frac{C}{r!}$
So $\displaystyle r(r+1)=r!\times \frac{2A}{(r-2)!}+r!\times \frac{2B}{(r-1)!}+r!\times \frac{2C}{r!}$
.............................................................................
Now observe that 
$r!=1\times 2\times 3\times .... \times (r-2)\times(r-1)\times r $
$\Rightarrow r!=(r−2)! ×(r−1)r $ and
$ \Rightarrow  r!=(r−1)!×r $
...............................................................................
So $\displaystyle r(r+1)=(r−2)! ×(r−1)r \times \frac{2A}{(r-2)!}+(r−1)!×r\times \frac{2B}{(r-1)!}+r!\times \frac{2C}{r!}$
So  $\displaystyle r^2+r = 2A(r-1)r+2Br+ 2C $
Clearly $C=0$ , $B=1$ and $A=\frac{1}{2}$
So $\displaystyle U_r=\frac{r(r+1)}{2r!}=\frac{1}{2(r-2)!}+\frac{1}{(r-1)!}$
$\displaystyle \sum_{r=2}^{\infty}U_r= \frac{1}{2} \left( \frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+.....\right)+\left( \frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+.....\right)$
$\displaystyle \sum_{r=2}^{\infty}U_r= \frac{1}{2} \left( e\right)+\left( e-1\right)$
$\displaystyle \sum_{r=1}^{\infty}U_r= U_1+\frac{1}{2} \left( e\right)+\left( e-1\right)=1+\frac{e}{2}+e-1 =\frac{3e}{2}$
A: 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*}
A: Your sum is $$\sum_{n = 1}^{\infty} \sum_{k = 1}^{n} \frac {k} {n!} = \sum_{n = 1}^{\infty} \frac {n + 1} {2 (n - 1)!} = \sum_{n = 0}^{\infty} \frac {n + 2} {2 n!} = \frac {3e} {2}.$$
A: 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 $\frac{3e}{2}$.
A: Hint...start by writing out the series for $xe^x$ and differentiate twice
