<s>[INST] Find the sum of the power series $\sum\limits_{n=1}^\infty \frac{(n+2)!}{(2!)(n!)}x^n$ 
Find the sum of the power series $\sum\limits_{n=1}^\infty \frac{(n+2)!}{(2!)(n!)}x^n$

frist I use $$\sum_{n=1}^\infty x^n = \frac{x}{1-x}$$
and multiple two side by $x^2$
can get $$\sum_{n=1}^\infty x^{n+2} = \frac{x^3}{1-x}$$
then diff each side two times
we can obtain $$\sum_{n=1}^\infty (n+1)(n+2)x^n= \frac{2x^3-6x^2+6x}{(1-x)^3}$$
but not the solution  $=\frac{1}{(1-x)^3}$
 [/INST] Starting from zero,
$\begin{array}\\
\sum\limits_{n=0}^\infty \frac{(n+2)!}{(2!)(n!)}x^n
&=\sum\limits_{n=0}^\infty \binom{n+2}{2}x^n\\
&=\frac1{x^2}(1-x)^{-3}\\
\end{array}
$
by the generalized binomial theorem
(since
$\binom{-n}{k}
=(-1)^n \binom{n+k-1}{k}
$).
</s>