<s>[INST] Coefficients of this generating function For the first part of a problem, I solved the generating function to be $F(x) = \frac{x^3}{(1-x)^2}$
Now it's the easy part that has me a little confused. What would the coefficients be in this case? 
I have computed so far:
$$\frac{x^3}{(1-x)^2} = x^3 (1+2x+3x^2 + \dots) =x^3 + 2x^4 + 3x^5 + \dots$$
Which leads me to believe that $f(n)=(n-2)$. But I know this only holds true for $ \ge 2$. What am i missing?
 [/INST] There is the binomial theorem, which says that 
$$(1+x)^n = \sum_{k \ge 0} \binom{n}{k} x^k$$
This is true even when $n$ is not necessarily a positive integer; we just have to define the binomial coefficient properly: for nonnegative integer $k$ and any $n$, we define
$$\binom{n}{k} = \frac{n(n-1)\cdots(n-k+1)}{k!}$$
With that definition, we have 
$$\binom{-2}{k} = \frac{(-2)(-3)\cdots(-2-k+1)}{k!} = (-1)^k \frac{2\cdot 3 \cdots (k+1)}{k!} = (-1)^k (k+1).$$
Thus, 
$$\frac{x^3}{(1-x)^2} = x^3(1-x)^{-2} = x^3\sum_{k \ge 0} \binom{-2}{k} x^k = x^3\sum_{k \ge 0} (k+1) x^k = \sum_{k \ge 0} (k+1) x^{k+3}$$
As the exponent is $k+3$, we can make the substitution $m = k+3$ (and so $k +1 = m - 2$) to get the above to be equal to
$$\sum_{k \ge 0} (k+1)x^{k+3} = \sum_{m \ge 3} (m-2) x^m$$
This means that the coefficient of any $x^m$ is 
$$[x^m]\frac{x^3}{(1-x)^2} = \begin{cases}
(m-2) & \text{if $m \ge 3$} \\
0 & \text{otherwise}
\end{cases}$$
</s>