How do you evaluate this sum?

Question

$$\sum_{n=2}^{\infty} n(n-1)(1/2)^n $$

I believe the answer is 4 but I am unable to understand how to work it out.

Thanks for any help.

Answer 1

HINT: $\sum\limits_{n=2}^\infty n(n-1)x^n = x^2 \sum\limits_{n=2}^\infty n(n-1)x^{n-2}$. Can you figure out $\sum\limits_{n=2}^\infty n(n-1)x^{n-2}$ by differentiating a well-known power series?

Answer 2

Ted's answer is definitely the answer that must be given first of all and this is just given for diversify using probabilistic language:

Let $X$ a random variable that follows geometric distribution $\mathcal G(p)$:

$$\Bbb P(X=n)=q^{n-1}p,\quad n\in\Bbb N,\; q=1-p$$ We know that

$$\Bbb E(X)=\sum_{n=1}^\infty n\Bbb P(X=n)=\frac1p$$ and that $$\Bbb V(X)=\Bbb E(X^2)-\Bbb E^2(X)=\frac q{p^2}$$ so with $p=\frac12$ you look for this sum $$\Bbb E(X^2)-\Bbb E(X)=\Bbb V(X)+\Bbb E^2(X)-\Bbb E(X)=2+4-2=4$$