Partial Fraction of $\frac{1-x^{11}}{(1-x)^4} $ for Generating Function The original question involves using generating functions to solve for the number of integer solutions to the equation $c_1+c_2+c_3+c_4 = 20$ when $-3 \leq c_1, -3 \leq c_2, -5 \leq c_3 \leq 5, 0 \leq c_4$.
Using generating functions I was able to get it into the rational polynomial form:
$$f(x) = {\left(\frac{1}{1-x}\right)}^3\left(\frac{1-x^{11}}{1-x}\right) = \frac{1-x^{11}}{{(1-x)}^4}$$
I was also able to determine that the sequence could be represented in two factors:
$${\left(1+x^1+x^2+x^3+\cdots\right)}^3\left(1+x^1+x^2+\cdots+x^{10}\right)$$
However, to find the coefficient on $x^{31}$ to solve the problem, I figured I would have to get the term $\frac{1-x^{11}}{{(1-x)}^4}$ into a more typical generating function summation form. Thus, I endeavored to find the partial fraction decomposition of the term, however, I can't seem to do it at all.
How would I find the partial fraction decomposition of $\frac{1-x^{11}}{{(1-x)}^4}$?
Or is there a better method in using ${(1+x^1+x^2+x^3+\ldots)}^3(1+x^1+x^2+\ldots+x^{10})$ in order to find the coefficient on $x^{31}$?
Thank you very much for your help, I've been trying this partial fraction for a while now and Wolfram alpha doesn't seem to be giving me an answer that is of much value.

I would use the
generalized binomial theorem
in the form
$
\dfrac1{(1-x)^s}
=\sum_{k=0}^{\infty}
\binom{s+k-1}{s-1}x^k
$.
Then
$\begin{array}\\
\dfrac{1-x^a}{(1-x)^s}
&=\sum_{k=0}^{\infty}
\binom{s+k-1}{s-1}x^k
-x^a\sum_{k=0}^{\infty}
\binom{s+k-1}{s-1}x^k\\
&=\sum_{k=0}^{\infty}
\binom{s+k-1}{s-1}x^k
-\sum_{k=0}^{\infty}
\binom{s+k-1}{s-1}x^{k+a}\\
&=\sum_{k=0}^{\infty}
\binom{s+k-1}{s-1}x^k
-\sum_{k=a}^{\infty}
\binom{s+k-a-1}{s-1}x^{k}\\
&=\sum_{k=0}^{a-1}
\binom{s+k-1}{s-1}x^k
+\sum_{k=a}^{\infty}
\binom{s+k-1}{s-1}x^k
-\sum_{k=a}^{\infty}
\binom{s+k-a-1}{s-1}x^{k}\\
&=\sum_{k=0}^{a-1}
\binom{s+k-1}{s-1}x^k
+\sum_{k=a}^{\infty}
\left(\binom{s+k-1}{s-1}-\binom{s+k-a-1}{s-1}\right)x^k
\\
&=\sum_{k=0}^{a-1}
\binom{s+k-1}{s-1}x^k
+\sum_{k=a}^{\infty}
\left(\dfrac{(s+k-1)!}{(s-1)!k!}-\dfrac{(s+k-a-1)!}{(s-1)!(k-a)!}\right)x^k
\\
&=\sum_{k=0}^{a-1}
\binom{s+k-1}{s-1}x^k
+\sum_{k=a}^{\infty}
\left(\dfrac{(s+k-1)!}{(s-1)!k!}-\dfrac{(s+k-a-1)!}{(s-1)!(k-a)!}\right)x^k
\\
\end{array}
$
You can do more manipulation if you want.
