<s>[INST] 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?
 [/INST] $$\frac{1}{(1-\frac{x}{3})^2}=\frac{1}{1-\frac{x}{3}}\cdot\frac{1}{1-\frac{x}{3}}$$
$$=\sum\limits_{n=0}^\infty(\frac{x}{3})^n\cdot\sum\limits_{n=0}^\infty(\frac{x}{3})^n $$
$$=(1+\frac{x}{3}+(\frac{x}{3})^2+\ldots)(1+\frac{x}{3}+(\frac{x}{3})^2+\ldots).$$
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:
$$\sum\limits_{n=0}^\infty (n+1)\frac{x^n}{3^n}.$$
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