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?

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 $x=2$
$$\sum_{m=1}^n \frac{m\, 2^m}{(m+2)!}=1-\frac{2^{n+1}}{ (n+2)!}$$
