<s>[INST] Find the coefficient of $x^{100}$ in the power series representing the function $\dots$ Find the coefficient of $x^{100}$ in the power series representing the function:
$$f(x)=(x+x^2+x^3+ \cdots) \cdot (x^2+x^3+x^4 \cdots) \cdot (x^3+x^4+x^5 \cdots)$$
Here is what I have so far:
$x+x^2+x^3+ \cdots = x(1+x+x^2+x^3+ \cdots)$ and this equals to the geometric series $\frac{x}{1-x}$
$x^2+x^3+x^4 \cdots = x^2(1+x+x^2+x^3+ \cdots)$ and this equals $\frac{x^2}{1-x}$
$x^3+x^4+x^5 \cdots = x^3(1+x+x^2+x^3+ \cdots)$ and this equals $\frac{x^3}{1-x}$
So now we have $F(x)=\left(\frac{x}{1-x}\right)\left(\frac{x^2}{1-x}\right)\left(\frac{x^3}{1-x}\right)$
This is the same as $F(x)=x^6(1-x)^{-1}$  
Now we can use the following way to expand and find the coefficient in front of $x^{100}$, $(1+x)^{\alpha}=1+\alpha x + {\alpha \choose 2}x^2 + {\alpha \choose 3}x^3 + \dots$  
For our function we get $x^6(1+(-1)(-x) + {-1 \choose 2}(-x)^2 + {-1 \choose 3}(-x)^3 + \dots)$, but $-1 \choose \beta$ where $\beta \ge 2$ will always equal $1$ or $-1$ and I do not think I am doing this correct because according to this the coefficient of $x^{100}$ will be either $1$ or $-1$?
What am I doing wrong?
 [/INST] A different conceptual interpretation: We have
$$
F(x) = x^6 (1+x+x^2+\cdots)^3
$$
We can interpret this as a generating function representing a fixed prefix of six balls, plus three (distinguishable) sacks of zero or more balls each.  The coefficient of $x^{100}$ is then the number of ways to total $100$ balls.  Factoring out the six-ball prefix, that leaves the number of ways to obtain $94$ balls in the three sacks, which we can determine using stars-and-bars as
$$
\binom{96}{2} = 4560
$$
</s>