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?
 A: $S=\sum_{i=1}^{\infty}\frac{i^2}{4^i} \rightarrow 4S=\sum_{i=1}^{\infty}\frac{i^2}{4^{i-1}}=1+\sum_{i=1}^{\infty}\frac{(i+1)^2}{4^i}$
Thus we have $3S=1+\sum_{i=1}^{\infty}\frac{2i+1}{4^i}=1+\frac{1}{3}+2\sum_{i=1}^{\infty}\frac{i}{4^i}$
$T=\sum_{i=1}^{\infty}\frac{i}{4^i} \rightarrow 4T=\sum_{i=1}^{\infty}\frac{i}{4^{i-1}}=1+\sum_{i=1}^{\infty}\frac{i+1}{4^i}$
Thus we have $3T=1+\sum_{i=1}^{\infty}\frac{1}{4^i}=\frac{4}{3}$, or $T=\frac{4}{9}$. Substituting this leads to $3S=\frac{20}{9}$, or $S=\frac{20}{27}$.
A: 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.
A: \begin{eqnarray*}
\sum_{i=0}^{\infty} i^2 x^i = \frac{x(1+x)}{(1-x)^3} \mid_{x=\frac{1}{4}} = \frac{\frac{1}{4} \left( 1+ \frac{1}{4} \right)}{\left(1- \frac{1}{4} \right)^3} = \color{red}{ \frac{20}{27}} .
\end{eqnarray*}
EDIT:
You are happy with a geometric sum ?
\begin{eqnarray*}
\sum_{i=0}^{\infty} x^i = \frac{1}{1-x} 
\end{eqnarray*}
Now differentiate this equation with respect to $x$ ( use $ \frac{d}{dx} x^n = nx^{n-1}$)
\begin{eqnarray*}
\sum_{i=0}^{\infty} i x^{i-1} = \frac{1}{(1-x)^2} 
\end{eqnarray*}
multiply by $x$
\begin{eqnarray*}
\sum_{i=0}^{\infty} i x^{i} = \frac{x}{(1-x)^2} 
\end{eqnarray*}
differentiate again (The RHS is tricky & I will supply more detail if needs be)
\begin{eqnarray*}
\sum_{i=0}^{\infty} i^2 x^{i-1} = \frac{1+x}{(1-x)^3} 
\end{eqnarray*}
finally multiply by $x$ and we have the result
\begin{eqnarray*}
\sum_{i=0}^{\infty} i^2 x^{i} = \frac{x(1+x)}{(1-x)^3}. 
\end{eqnarray*}
