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.
 A: $$\begin{align}
\frac{1-x^{11}}{(1-x)^4}&=(1-x^{11})\sum_{n\ge0}\frac{(n+1)(n+2)(n+3)}{6}x^n\\
&=\sum_{n\ge0}\frac{(n+1)(n+2)(n+3)}{6}x^n-\sum_{n\ge0}\frac{(n+1)(n+2)(n+3)}{6}x^{n+11}\\
&=\sum_{n\ge0}\frac{(n+1)(n+2)(n+3)}{6}x^n-\sum_{n\ge11}\frac{(n-10)(n-9)(n-8)}{6}x^{n}\\
&=\sum_{n\ge0}\frac{1}{6}\bigg((n+1)(n+2)(n+3)-(n-10)(n-9)(n-8)[n\ge11]\bigg)x^n,
\end{align}$$
where $[n\ge11]$ is the Iverson Bracket.
A: 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.
A: It is convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ of a series.

We obtain
\begin{align*}
\color{blue}{[x^{31}]\frac{1-x^{11}}{(1-x)^4}}
&=\left([x^{31}]-[x^{20}]\right)\sum_{j=0}^{\infty}\binom{-4}{j}(-x)^j\tag{1}\\
&=\left([x^{31}]-[x^{20}]\right)\sum_{j=0}^{\infty}\binom{j+3}{j}x^j\tag{2}\\
&=\binom{34}{31}-\binom{23}{20}\tag{3}\\
&\,\,\color{blue}{=4\,213}
\end{align*}

Comment:

*

*In (1) we use the rule $[x^p]x^qA(x)=[x^{p-q}]A(x)$ and make a binomial series expansion.


*In (2) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.


*In (3) we select the coefficients accordingly.
