<s>[INST] Infinite power summation where no common ratio can be seen Problem in book states to evaluate the sum:
$$\sum_{i=0}^\infty \frac{i^2}{4^i}$$
I can come up with the series S = $\frac{1^2}{4^1} + \frac{2^2}{4^2} + \frac{3^2}{4^3} + \frac{4^2}{4^4} + \frac{5^2}{4^5} + ...$
And 4S = $ 1 + \frac{2^2}{4^1} + \frac{3^2}{4^2} + \frac{4^2}{4^3} + \frac{5^2}{4^4} + \frac{6^2}{4^5} + ...$
If I subtract S from 4S I get 3S = $ 1 + \frac{3}{4^1} + \frac{5}{4^2} + \frac{7}{4^3} + \frac{9}{4^4} + \frac{11}{4^5} + ...$
I cannot find a common ratio to plug in to $\frac{1}{1-r}$ in order to figure this out.  I asked my professor for some assistance, and he said the answer should be $\frac{20}{27}$, but I cannot figure out what steps he took to get that.  He didn't explain it very well, and all other resources I've checked only show how to compute a geometric sum when there is a common ratio like $\frac{1}{2}$.
How do I determine a common ratio to figure out the sum?  I tried division by sequential terms, and that gave me a common ratio of $\frac{5}{16}$, but that left me with an answer of S = $\frac{16}{27}$, which isn't correct.
What am I doing wrong? Or what am I missing?
 [/INST] Consider $$\sum_{i=0}^\infty {i^2}{x^i}=\sum_{i=0}^\infty {[i(i-1)+i]}{x^i}=\sum_{i=0}^\infty {i(i-1)}{x^i}+\sum_{i=0}^\infty {i}{x^i}=x^2\sum_{i=0}^\infty {i(i-1)}{x^{i-2}}+x\sum_{i=0}^\infty {i}{x^{i-1}}$$ that is to say 
$$\sum_{i=0}^\infty {i^2}{x^i}=x^2\left(\sum_{i=0}^\infty {x^{i}} \right)''+x\left(\sum_{i=0}^\infty {x^{i}} \right)'$$ and you know that
$$\sum_{i=0}^\infty {x^{i}} =\frac{1}{1-x}\qquad \text{for}\qquad x <1$$ Now, things look to be simple.
When finished, replace $x$ by $\frac 14$ to get the final result.
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