Return of the sums

$\begin{aligned} S & = \sum_{n \ge 0} \frac {n^2(n-1)}{n!} & \small \color{#3D99F6} \text{Summand }= 0 \text{, when }n=0, 1 \\ & = \sum_{n=2}^\infty \frac {n^2(n-1)}{\color{#3D99F6}n!} & \small \color{#3D99F6} \text{Note that }n! = n(n-1)(n-2)! \\ & = \sum_{\color{#3D99F6}n=2}^\infty \frac n{(n-2)!} \\ & = \sum_{\color{#D61F06}n=0}^\infty \frac {n+2}{n!} \\ & = {\color{#3D99F6}\sum_{n=0}^\infty \frac n{n!}} + 2 \sum_{n=0}^\infty \frac 1{n!} & \small \color{#3D99F6} \text{Summand }= 0 \text{, when }n=0 \\ & = {\color{#3D99F6}\sum_{\color{#D61F06}n=1}^\infty \frac n{n!}} + 2 \sum_{n=0}^\infty \frac 1{n!} \\ & = \sum_{\color{#D61F06}n=1}^\infty \frac 1{(n-1)!} + 2 \sum_{n=0}^\infty \frac 1{n!} \\ & = \sum_{\color{#3D99F6}n=0}^\infty \frac 1{n!} + 2 \sum_{n=0}^\infty \frac 1{n!} \\ & = 3 \sum_{n=0}^\infty \frac 1{n!} \\ & = 3e \end{aligned}$

$\implies a = \boxed{3}$ .

$\begin{aligned} \sum_{n = 0}^{\infty} \frac{n^2(n-1)}{n!} &= \sum_{n = 0}^{\infty} \frac{n(n-1)(n - 2 + 2)}{n!} \\ &= \sum_{n = 0}^{\infty} \frac{n(n-1)(n-2)}{n!}+ 2\sum_{n = 0}^{\infty} \frac{n(n-1)}{n!} \\ &= \frac{d^3}{dx^3}e^x|_{x = 1} + 2\frac{d^2}{dx^2}e^x |_{x=1} \\ &= e + 2e \\ &= 3e \end{aligned}$

We are given $\displaystyle \sum_{n \ge 0} \dfrac{n^2(n-1)}{n!}$ .

We can rewrite this as $\displaystyle \sum_{n = 2}^{\infty} \dfrac{n}{(n-2)!} = \sum_{n \ge 0} \dfrac{n+2}{n!} \\ \displaystyle = 2\sum_{n \ge 0} \dfrac{1}{n!} + \sum_{n \ge 0} \dfrac{n}{n!} = 2e + \sum_{n \ge 0} \dfrac{1}{(n-1)!} = 3e.$

Hence, we see that $a = \boxed{3}$ .

$\begin{aligned}\text{Let, } S = \sum_{n=0}^{\infty}{\dfrac{n^2(n-1)}{n!}} \\ \text{Evaluating the first 2 terms, } \\ S = 0 + 0 + \sum_{n=2}^{\infty}{\dfrac{n}{(n-2)!}} \\ S = 2 + \frac{3}{1!} + \frac{4}{2!} + \frac{5}{3!} + \frac{6}{4!} + \cdots \\ S = 2 + 3 + 4\times\frac{1}{2!} + 5\times\frac{1}{3!} + 6\times\frac{1}{4!} + \cdots \\ S = e + 3 + 3\times\frac{1}{2!} + 4\times\frac{1}{3!} + 5\times\frac{1}{4!} + \cdots \\ S = e + e + 1 + 2\times\frac{1}{2!} + 3\times\frac{1}{3!} + 4\times\frac{1}{4!} + \cdots \\ S = e + e + 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \cdots \\ S = e + e + e = 3e \\ \therefore \boxed{a = 3} \end{aligned}$