If $S= \sum_{n=0}^{\infty} \frac{n^2}{2^n} $ then $2S-S$ is

Question

I was trying to understand the accepted answer for How to prove $\sum_{n=0}^{\infty} \frac{n^2}{2^n} = 6$?
$$S= \sum_{n=0}^{\infty} \frac{n^2}{2^n} $$

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

I can't understand how this works.

Answer 1

Note that$$2S=\sum_{n=0}^\infty\frac{n^2}{2^{n-1}}=\sum_{n=1}^\infty\frac{n^2}{2^{n-1}}=\sum_{n=0}^\infty\frac{(n+1)^2}{2^n}.$$Can you take it from here?