Summation #1

$\begin{aligned} S & = \sum_{n=0}^\infty \frac {(n+2)!}{2^n n!} \\ & = \sum_{\color{#3D99F6}n=0}^\infty \frac {(n+1)(n+2)}{2^n} \\ S & = \frac 2{2^0} + \frac 6{2^1} + \frac {12}{2^2} + \frac {20}{2^3} + \cdots + \frac {(n+1)(n+2)}{2^n} + \cdots & \small \color{#3D99F6} ... (1) \\ \frac S2 & = \frac 2{2^1} + \frac 6{2^2} + \frac {12}{2^3} + \frac {20}{2^4} + \cdots + \frac {n(n+1)}{2^n} + \cdots & \small \color{#3D99F6} ... \frac {(1)}2 \to (2) \\ \frac S2 & = \frac 2{2^0} + \frac 4{2^1} + \frac 6{2^2} + \frac 8{2^3} + \cdots + \frac {2(n+1)}{2^n} + \cdots & \small \color{#3D99F6} ... (1)-(2) \to (3) \\ \frac S4 & = \frac 2{2^1} + \frac 4{2^2} + \frac 6{2^3} + \frac 8{2^4} + \cdots + \frac {2n}{2^n} + \cdots & \small \color{#3D99F6} ... \frac {(3)}2 \to (4) \\ \frac S4 & = \frac 2{2^0} + \frac 2{2^1} + \frac 2{2^2} + \frac 2{2^3} + \cdots + \frac 2{2^n} + \cdots & \small \color{#3D99F6} ... (3)-(4) \to (5) \\ & = 2 + \color{#3D99F6} \frac 1{2^0} + \frac 1{2^1} + \frac 1{2^2} + \cdots & \small \color{#3D99F6} \text{Infinite geometric progression} \\ & = 2 + {\color{#3D99F6}\frac 1{1-\frac 12}} = 4 \end{aligned}$

Therefore $S = \boxed{16}$

$\sum_{n=0}^{\infty}\frac{(n+2)!}{2^nn!}=\sum_{n=0}^{\infty}\frac{1}{2^{n-1}}\binom{n+2}{2}$

with some help form Pascal's formula, which says

$\binom{n+1}{k+1}=\binom{n}{k+1}+\binom{n}{k}$

$\sum_{n=0}^{\infty}\frac{1}{2^{n-1}}\binom{n+2}{2}=\sum_{n=0}^{\infty}\frac{1}{2^{n-1}} \big( \binom{n+1}{2}+\binom{n+1}{1} \big)$

$\sum_{n=0}^{\infty}\frac{1}{2^{n-1}} \big( \binom{n+1}{2}+\binom{n+1}{1} \big)=\sum_{n=0}^{\infty}\frac{1}{2^{n-1}} \big( \binom{n+1}{2} \big) +\sum_{n=0}^{\infty} \big( \frac{n+1}{2^{n-1}}\big)$

Generating functions can be utilised, if the above summations are written as below.

$f(x)=\sum_{n=0}^{\infty}\frac{1}{2^{n-1}}\binom{n+2}{2}x^n=\sum_{n=0}^{\infty}\frac{1}{2^{n-1}} \binom{n+1}{2} x^n +\sum_{n=0}^{\infty} \frac{n+1}{2^{n-1}} x^n$

can be further simplified as

$f(x)=\frac{1}{2}f(x)+4+x \frac{d}{dx}\big( \frac{1}{2-x} \big)+\frac{3}{2-x}$

the rest is a simple derivative calculation and solving for $f(x)$ . Finally evaluate $f(x)$ for $x=1$ .