Calculate the sum of infinite series with general term $\frac{n^2}{2^n}$. Please explain different methods to calculate the sum of infinite series with $\dfrac{n^2}{2^n}$ as it's general term i.e. Calculate 
$$\sum_{n=0}^\infty \dfrac {n^2}{2^n}$$
Please avoid the method used for summation of arithmetic geometric series. It is very tedious approach. Does any other simpler approach exist?
 A: For $|x|<1,$ $$\sum_{r=0}^\infty x^r=\frac1{1-x}$$ 
Differentiate wrt $x,$   $$\sum_{r=0}^\infty rx^{r-1}=\frac1{(1-x)^2}$$ 
Multiply by $x$
Differentiate wrt $x$
Multiply by $x$
Can you recognize $x$ here?
A: The answer below is (perhaps) not given with perfect seriousness, since in a sense it is on the complicated side. However, if one thinks in probabilistic terms, it is not completely unreasonable.   
Imagine tossing a fair coin until we obtain a head. Let $X$ be the number of tosses. Then 
$$E(X^2)=\sum_1^\infty \frac{n^2}{2^n}.$$
 Condition on the result of the first toss. The conditional expectation of $X^2$, given that we got a head on the first toss, is $1$. 
The conditional expectation of $X^2$, given that we got a tail on the first toss, is $E(1+X)^2$, which by the linearity of expectation is equal to $1+2E(X)+E(X^2)$.
It follows that
$$E(X^2)=\frac{1}{2}\cdot 1+\frac{1}{2}(1+2E(X)+E(X^2)).$$
We conclude that
$$E(X^2)=2+2E(X).$$
It remains to find $E(X)$. By a conditioning argument similar to the one above, but quite a bit simpler, we get that $E(X)=2$, giving $E(X^2)=6$.  
