Sum $ \sum\limits_{k=1}^{\infty} \frac{k^2}{2^k}$

Observe that by just differentiating

$$ 1+x+x^2+...+x^n=\frac{1-x^{n+1}}{1-x}, \quad |x|<1, \tag2 $$ with respect to $x$ and by multiplying by $x$, we get the identity $$ \sum_{k=1}^{n}kx^k=\frac{1-x^{n+1}}{(1-x)^2}+\frac{-(n+1)x^{n}}{1-x} , \quad |x|<1,\tag3$$ differentiating once more and multiplying by $x$ gives, as $n \to \infty$, using $|x|<1$: $$ \sum_{k=1}^{\infty}k^2x^k=\frac{x (1+x)}{(1-x)^3} , \quad |x|<1,\tag4$$ then put $x:=\dfrac12$ to obtain $$ \sum_{k=1}^{n}k^2/2^k \longrightarrow 6, \quad \text{as} \, n \to \infty. $$

Let $$ S = \sum_{k\geq 1}\frac{k^2}{2^k}. $$ Then: $$ 2S = \sum_{k\geq 1}\frac{k^2}{2^{k-1}} = 1+\sum_{k\geq 1}\frac{(k+1)^2}{2^k},$$ hence: $$ S = 2S-S = 1+\sum_{k\geq 1}\frac{2k+1}{2^k} = 2+2\sum_{k\geq 1}\frac{k}{2^k}$$ but with the same technique, by letting $T=\sum_{k\geq 1}\frac{k}{2^k}$, we have: $$ T=2T-T = 1+\sum_{k\geq 1}\frac{1}{2^k} = 2$$ from which $S=\color{red}{6}$ follows.