Think!

$\begin{aligned} \displaystyle \sum^{\infty}_{n=0}x^n & = \frac{1}{1-x} \\ \displaystyle \sum^{\infty}_{n=0} nx^{n-1} & = \frac{d}{dx} (1-x)^{-1} \\ \displaystyle \sum^{\infty}_{n=0} nx^{n-1} & = \frac{1}{(1-x)^2}\Rightarrow \displaystyle \sum^{\infty}_{n=0} nx^{n} = \frac{x}{(1-x)^2} \\ \displaystyle \sum^{\infty}_{n=0}n^2x^{n-1} & = \frac{d}{dx} x(1-x)^{-2} \\ \displaystyle \sum^{\infty}_{n=0}n^2x^{n-1} & =\frac{x+1}{(1-x)^3} \Rightarrow \displaystyle \sum^{\infty}_{n=0}n^2x^n=\frac{x(x+1)}{(1-x)^3} \end{aligned}$

Placing $x=\frac{1}{e}$ ,

$\displaystyle \sum^{\infty}_{n=0}\frac{n^2}{e^n}=\frac{ \frac{1}{e}\left(\frac{1}{e}+1\right) }{\left(1-\frac{1}{e}\right)^3}=\frac{e(e+1)}{(e-1)^3} \\ \Rightarrow 1+1+3=\boxed{5}$

Consider:

$\begin{aligned} \sum_{n=0}^\infty n^2e^{-n\color{#3D99F6}{x}} & = \sum_{n=0}^\infty \color{#3D99F6}{\frac{\partial^2}{\partial x^2}} e^{-n\color{#3D99F6}{x}} \quad \quad \small \color{#3D99F6}{\text{See note.}} \\ & = \frac{\partial^2}{\partial x^2} \color{#3D99F6}{\sum_{n=0}^\infty \left(e^{-x}\right)^n \quad \quad \small \text{Sum of infinite GP}} \\ & = \frac{\partial^2}{\partial x^2} \color{#3D99F6}{\left( \frac{1}{1-e^{-x}} \right)} = \frac{\partial}{\partial x} \left( \frac{- e^{-x}}{(1-e^{-x})^2} \right) \\ & = \frac{e^{-x}}{(1-e^{-x})^2} + \frac{2e^{-2x}}{(1-e^{-x})^3} \quad \quad \small \color{#3D99F6}{\text{Putting }x = 1} \\ \Rightarrow \sum_{n=0}^\infty \frac{n^2}{e^n} & = \frac{e^{-1}}{(1-e^{-1})^2} + \frac{2e^{-2}}{(1-e^{-1})^3} = \frac{e^{-1}+e^{-2}}{(1-e^{-1})^3} = \frac{e(e+1)}{(e-1)^3} \end{aligned}$

$\Rightarrow A+B+C = 1+1+3 = \boxed{5}$

$\color{#3D99F6}{\text{Note: }}$ I learned this trick from Isaac Buckley . It is surely handy here.

Let the summation be equal to S then

S=0+ 1/e + 2^2 / e^2 + 3^3/e^3 + .......so on -(1) S/e= 1/e^2 + 2^2 /e^3 + .........so on - (2)

Subtract 1 from 2

S(1-1/e)= T = 1/e + 3/e^2 + 5/e^3........ (3)

Take LHS =T (new variable)

T/e = 1/e^2 + 3/e^3 ....... (4)

Subtract 3 from 4

T(1-1/e)= 1/e + 2[1/e^2 + 1/e^3 +........]

Then terms in square brackets can be solved by infinite GP formula ie a/1-r

====> (1/e^2)/(1-1/e) which is equal to 1/e(e-1)

Putting back and expressing T in terms of S we get on LHS S((e-1)^2/e^2)

And solving RHS we get (e+1)/(e.(e-1)) Equating We get

S= e(e+1)/(e-1)^3

A=1 B=1 C=3 A+B+C= 5