Sequence from generating function $\frac{1}{(1 - \frac{x}{3})^2}$ I know that for 
$$ \frac{1}{1 - \frac{x}{3}} $$
sequence would be $a_n = \frac{1}{3^n}$ for $n \geq 0$ ($\sum_{n \geq 0} \frac{1}{3^n} x^n$ ).
How should I approach with that power of two?
 A: In general, by computing the Taylor series expansion. 
In this case it happens to work to use $\frac{1}{(1-z)^2}=\frac{\partial}{\partial z}\frac{1}{1-z}$.
But a more widely applicable bit of knowledge is that if $f(z)=\sum_{n=0}^\infty a_nz^n$, then 
$$\frac{1}{1-z}\,f(z)=\sum_{n=0}^\infty\left(\sum_{k=0}^n a_k\right)z^k.$$
And $\frac{1}{(1-z)^2}=\frac{1}{1-z}\frac{1}{1-z}$.
Lastly, set $z=\frac{x}{3}$ for a cleaner computation.
A: $$\frac{1}{(1-\frac{x}{3})^2}=\frac{1}{1-\frac{x}{3}}\cdot\frac{1}{1-\frac{x}{3}}$$
$$=\sum\limits_{n=0}^\infty(\frac{x}{3})^n\cdot\sum\limits_{n=0}^\infty(\frac{x}{3})^n $$
$$=(1+\frac{x}{3}+(\frac{x}{3})^2+\ldots)(1+\frac{x}{3}+(\frac{x}{3})^2+\ldots).$$
Now we want a single sum in the form of $\sum a_nx^n$, and so we're going to find all of the terms resulting from the multiplication above that correspond to $x^0, x^1, x^2,...$ and match those coefficients to $a_0, a_1, a_2,...$.  For example, the only way to get a constant term is from multiplying the first terms of both sequences (1) together, so $a_0=1$.  Then, there are two ways to get $x^1$, either from multiplying the 1 in the first sequence to $\frac{x}{3}$ in the second or vice versa, so we get $a_1=2\cdot\frac{1}{3}$.  In general, when we add 1 to our index $n$, there is one more way to get an $x^n$ term. And so $a_n=(n+1)\cdot(\frac{1}{3})^n$, for $n\ge0$.  Thus, your series would be:
$$\sum\limits_{n=0}^\infty (n+1)\frac{x^n}{3^n}.$$
A: For yet another answer,
$$ \frac1{1-x} = \sum_{n=0}^\infty x^n,$$
and
$$ \frac1{1-\frac13 x} = \sum_{n=0}^\infty \left(\frac13\right)^n x^n.$$
Differentiating we have
$$ \frac{\frac13}{\left(1-\frac13 x\right)^2} = \sum_{n=0}^\infty(n+1)\left(\frac13\right)^{n+1}x^n.  $$
Hence
$$\frac1{\left(1-\frac13 x\right)^2} = \sum_{n=0}^\infty(n+1)\left(\frac13\right)^nx^n. $$
A: Same, just use the binomial coefficient $\binom{-2}{k}$. If you feel scared, just set $\alpha=-2$ and do the Binomial expansion. You should get something like $(-1)^k k(k+1)$ I think. 
