Find $\textstyle{\frac{1\cdot 2}{3!} +\frac{2\cdot2^2}{4!}+\frac{3\cdot2^3}{5!}+\frac{4\cdot2^4}{6!}+\cdots}$ up to n terms? $$
\frac{(1)2}{3!}  + \frac{(2)2^2}{4!} + \frac{(3)2^3}{5!}  + \frac{(4)2^4}{6!} + \cdots =\sum\limits_{k=1}^{n}\frac{k\cdot 2^k}{(k+2)!}
$$
My attempt:
$$
\begin{align}
  e^x&=\sum_{n=1}^{\infty}\frac{x^n}{n!} \\
  (e^x)'&=\sum_{n=1}^{\infty}\frac{nx^{n-1}}{n!} \\
  x\cdot(e^x)'&=\sum_{n=1}^{\infty}\frac{nx^{n}}{n!} \\
  x\cdot(e^x)'&=\sum_{n=1}^{\infty}\frac{n(n+1)(n+2)x^{n}}{(n+2)!}
\end{align}
$$
After this attempt I realized exponential series is for infinite terms whereas the question concerns finite terms so approach may not work.
Can you please give any hints on the right approach to be tried?
 A: Using my hint in the comments,
\begin{align*}
\sum\limits_{k = 1}^n {\frac{{k2^k }}{{(k + 2)!}}} & = \sum\limits_{k = 1}^n {\frac{{(k + 2)2^k  - 2^{k + 1} }}{{(k + 2)!}}}  = \sum\limits_{k = 1}^n {\frac{{2^k }}{{(k + 1)!}}}  - \sum\limits_{k = 1}^n {\frac{{2^{k + 1} }}{{(k + 2)!}}} \\ & = \sum\limits_{k = 1}^n {\frac{{2^k }}{{(k + 1)!}}}  - \sum\limits_{k = 2}^{n + 1} {\frac{{2^k }}{{(k + 1)!}}}  = 1 - \frac{{2^{n + 1} }}{{(n + 2)!}}.
\end{align*}
A: Making the problem more general, write
$$\sum_{m=1}^n \frac{m\, x^m}{(m+2)!}=\frac 1 x\sum_{m=1}^n \frac{ x^{m+1}}{(m+1)!}-\frac 2 {x^2}\sum_{m=1}^n \frac{ x^{m+2}}{(m+2)!}$$
$$\sum_{m=1}^n \frac{ x^{m+1}}{(m+1)!}=e^x\frac{ \Gamma (n+2,x)}{(n+1)!}-x-1$$
$$\sum_{m=1}^n \frac{ x^{m+2}}{(m+2)!}=e^x\frac{ \Gamma (n+3,x)}{\Gamma (n+3)}-\frac{x^2}{2}-x-1$$ Recombining everything
$$\sum_{m=1}^n \frac{m\, x^m}{(m+2)!}=\frac{x+2}{x^2}-\frac{x^{n+1}}{(n+2)!}+e^x\frac {x-2}{x^2}\frac{  \Gamma (n+3,x)}{  (n+2)!}$$
You are lucky with $x=2$
$$\sum_{m=1}^n \frac{m\, 2^m}{(m+2)!}=1-\frac{2^{n+1}}{ (n+2)!}$$
A: This is not an answer: Just mistakenly thought that we need to find the sum to infinity.
Given, $\displaystyle \sum_{m=1}^\infty \frac{m\cdot 2^m}{(m+2)!}$. We put $m=n-2 $ and thus the series becomes $\displaystyle\sum_{n=3}^\infty \frac{(n-2)\cdot 2^{n-2}}{n!}=\frac{1}{4}\sum_{n=3}^\infty \frac{(n-2)\cdot 2^n}{n!}=\frac{1}{4}\sum_{n=3}^\infty \frac{n\cdot 2^n}{n!}-\frac{1}{4}\sum_{n=3}^\infty \frac{2\cdot 2^n}{n!}=A+B(say)$. Where $A,B$ denotes the first and second sum respectievly.
Now, $$A=\frac{1}{4}\sum_{n=3}^\infty\frac{n\cdot 2^n}{n!}=\frac{1}{2}\sum_{n=3}^\infty\frac{2^{n-1}}{(n-1)!}=\frac{1}{2}(e^2-1-2).$$ Recall that $\displaystyle e^x=\sum_{n=1}^\infty\frac{x^n}{n!}$ now put $x=2$.
Similarly, $$\displaystyle B=\frac{1}{2}\sum_{n=3}^\infty \frac{2^n}{n!}=\frac{1}{2}(e-1-2-\frac{2^2}{2!})=\frac{1}{2}(e^2-5).$$
So $\displaystyle A+B=\frac{1}{2}(e^2-3-e^2+5)=1.\square$
Hope this works.
