How to prove $\sum_{n=0}^{\infty} \frac{n^2}{2^n} = 6$?

First observe that the sum converges (by, say, the root test).

We already know that $\displaystyle R := \sum_{n=0}^\infty \frac{1}{2^n} = 2$.

Let $S$ be the given sum. Then $\displaystyle S = 2S - S = \sum_{n=0}^\infty \frac{2n+1}{2^n}$.

Now use the same trick to compute $\displaystyle T := \sum_{n=0}^\infty \frac{n}{2^n}$: we have $\displaystyle T = 2T - T = \sum_{n=0}^\infty \frac{1}{2^n} = R = 2$. Hence $S = 2T + R = 6$.

One can continue like this to compute $\displaystyle X:= \sum_{n=0}^\infty \frac{n^3}{2^n}$. We have $\displaystyle X = 2X-X = \sum_{n=0}^\infty \frac{3n^2+3n+1}{2^n} = 3S+3T+R = 26$. Sums with larger powers can be computed in the same way.

If Calculus is allowed, using infinite geometric series formula for $|r|<1$ $$\sum_{0\le n<\infty}r^n=\frac1{1-r}$$

Differentiating wrt $r$ $$\sum_{0\le n<\infty}nr^{n-1}=\frac1{(1-r)^2}$$

$$\implies \sum_{0\le n<\infty}nr^n=\frac r{(1-r)^2}=\frac{1-(1-r)}{(1-r)^2}=\frac1{(1-r)^2}-\frac1{(1-r)}$$

Differentiating wrt $r$ (fixed sign error!)

$$\implies \sum_{0\le n<\infty}n^2r^{n-1}=\frac2{(1-r)^3}-\frac1{(1-r)^2}$$

$$\implies \sum_{0\le n<\infty}n^2r^n=\frac{2r}{(1-r)^3}-\frac r{(1-r)^2}$$

Here $\displaystyle r=\frac12$

Consider $$ \frac{1}{1-z} = \sum_{n=0}^{\infty} z^n $$ Differentiating and multiplying by $z$, one has $$ \frac{z}{(1-z)^2} = z\sum_{n=1}^{\infty} nz^{n-1} = \sum_{n=1}^{\infty} nz^n $$ Repeating the above process, $$ \sum_{n=1}^{\infty} n^2z^n = \frac{z(1+z)}{(1-z)^3} $$ Now plug in $z=1/2$

Let me show you a slightly different approach; this approach is very powerful, and can be used to compute values for a large number of series.

Let's think of this in terms of power series. You noticed that you can write $$ \sum_{n=0}^{\infty}\frac{n^2}{2^n}=\sum_{n=0}^{\infty}n^2\left(\frac{1}{2}\right)^n; $$ so, let's consider the power series $$ f(x)=\sum_{n=0}^{\infty}n^2x^n. $$ If we can find a simpler expression for the function $f(x)$, and if $\frac{1}{2}$ lies within its interval of convergence, then your series is exactly $f(\frac{1}{2})$.

Now, we note that $$ f(x)=\underbrace{0}_{n=0}+\sum_{n=1}^{\infty}n^2x^n=x\sum_{n=1}^{\infty}n^2x^{n-1}. $$ But, notice that $\int n^2x^{n-1}\,dx=nx^n$; so, $$\tag{1} \sum_{n=1}^{\infty}n^2 x^{n-1}=\frac{d}{dx}\left[\sum_{n=0}^{\infty}nx^n\right]. $$ Now, we write $$\tag{2} \sum_{n=0}^{\infty}nx^n=\underbrace{0}_{n=0}+x\sum_{n=1}^{\infty}nx^{n-1}=x\frac{d}{dx}\left[\sum_{n=0}^{\infty}x^n\right]. $$ But, this last is a geometric series; so, as long as $\lvert x\rvert<1$, $$ \sum_{n=0}^{\infty}x^n=\frac{1}{1-x}. $$ Plugging this back in to (2), we find that for $\lvert x\rvert<1$, $$ \sum_{n=0}^{\infty}nx^n=x\frac{d}{dx}\left[\frac{1}{1-x}\right]=x\cdot\frac{1}{(1-x)^2}=\frac{x}{(1-x)^2}. $$ But, plugging this back in to (1), we find that $$ \sum_{n=1}^{\infty}n^2x^{n-1}=\frac{d}{dx}\left[\frac{x}{(1-x)^2}\right]=\frac{x+1}{(1-x)^3} $$ So, finally, $$ f(x)=x\cdot\frac{x+1}{(1-x)^3}=\frac{x(x+1)}{(1-x)^3}. $$ Plugging in $x=\frac{1}{2}$, we find $$ \sum_{n=0}^{\infty}\frac{n^2}{2^n}=f\left(\frac{1}{2}\right)=6. $$