Finding a coefficient using generating functions I need to find a coefficient of $x^{21}$ inside the following expression:
$$(x^{2} + x^{3} + x^{4} + x^{5} + x^{6})^{8}$$
I think the only way (using generating functions) is to express the parentheses content with a generating function.
The generating function for the $(1, x, x^{2}, x^{3}, ...)$ sequence is equal to $\frac{1}{1-x}$.
However, well...the expression at the top is not infinite, so I can't really express it as a generating function.
How can I do this?
 A: $$(x^{2} + x^{3} + x^{4} + x^{5} + x^{6})^{8}=(x^2(1+x+x^2+x^3+x^4))^8=$$
$$=x^{16}\left(\frac{1-x^5}{1-x}\right)^8=x^{16}(1-x^5)^8(1-x)^{-8}=$$
$$=x^{16}\sum_{k=0}^{8}\binom{8}{k}(-x)^{5k}\sum_{l=0}^{\infty}\binom{8+l-1}{l}x^l=$$
$$=\sum_{l=0}^{\infty}\sum_{k=0}^{8}(-1)^{5k}\binom{8+l-1}{l}\binom{8}{k}x^{16+5k+l}$$
$$16+5k+l=21\Rightarrow k=1,l=0 \text{and}k=0,l=5$$
coefficient next $x^{21}$ is
$$(-1)^0\binom{8+5-1}{5}\binom{8}{0}+(-1)^5\binom{8+0-1}{0}\binom{8}{1}=\binom{12}{5}-8$$
A: Hint Use a finite geometric series:
$$x^2 + \ldots + x^6 = x^2 \cdot \frac{1-x^5}{1-x}$$
and now you are looking for 
$$\left[x^{13}\right] \left(1-x^5\right)^8 \cdot \frac{1}{(1-x)^8}$$
A: $(x^{2} + x^{3} + x^{4} + x^{5} + x^{6})^{8} = x^{16}(1+x+x^2+x^3+x^4)^8$, so you should find the coefficient of $x^5$ in $(1+x+x^2+x^3+x^4)^8$. If you write $5$ as sum of $8$ terms $4\geq k_i\geq 0$ in all possible ways, you'll to see  in how many ways you can get $x^5$. It is also known what is the number of sums of $k$ nonegative terms which sum up in $n$: $n+k-1 \choose k-1$. In your case, there are $8$ sums $5=0+0+...+5=0+...+0+5+0=...=5+0+..+0$,which you cannot get in the products, because $k_i\leq 4$. So the answer is ${5+8-1} \choose{8-1}$$-8$.
