I want to calculate
$$ \sum_{k=1}^{\infty} \frac{k^2}{2^k} $$
Usually (assuming it was $k$ instead of $k^2$) I would consider this as a special case of the series $ f(x) =\sum_{k=1}^{\infty} k \cdot z^k $ (i.e when $z=1/2$)
$$ \sum_{k=1}^{\infty} z^k \quad=\quad \frac{z}{1-z} $$ Derivation gives $$ \sum_{k=1}^{\infty} k \cdot z^{k-1} \quad=\quad \frac{1}{(1-z)^2} $$
Multiplying by z $$ \sum_{k=1}^{\infty} k \cdot z^k \quad=\quad \frac{z}{(1-z)^2} $$
And if $z=1/2$ it would be 2. But I have no idea how to handle the case when we have $k^2$, any tips?
