<s>[INST] A closed formulae for the coefficient of $x^k$ in $(x-1)^a(x+1)^b$ Let a,b positive integer 
Do you know any closed formulae for the coefficient of $x^k$ 
in $(x-1)^a(x+1)^b=\sum_{k=0}^{a+b}u(k;a,b)x^k$ ?
I look for an a closed expression of $u(k;a,b)$ involving maybe integral , special function
not the symbol $\sum$
thank's for your help
 [/INST] The expansion follows as:
\begin{align}
(x-1)^{a} (x+1)^{b} &= (-1)^{a} (1-x)^{a} (1+x)^{b} \\
&= (-1)^{a} \sum_{r=0}^{a} \binom{a}{r} (-1)^{r} x^{r} \cdot \sum_{s=0}^{b} \binom{b}{s} x^{s} \\
&= (-1)^{a} \sum_{r=0} \sum_{s=0} (-1)^{r} \binom{a}{r} \binom{b}{s} x^{r+s} \\
&= (-1)^{a} \sum_{r=0}^{a|b} (-1)^{r} \left( \sum_{s=0}^{r} \binom{a}{r-s} \binom{b}{s} (-1)^{s} \right) x^{r}.
\end{align}
Now
\begin{align}
\sum_{s=0}^{r} \binom{a}{r-s} \binom{b}{s} (-1)^{s} &= \sum_{s=0}^{r} \frac{a!}{(r-s)!(a-r+s)!} \frac{b!}{s! (b-s)!} (-1)^{s} \\
&= \binom{a}{r} \sum_{s=0}^{r} \frac{1}{(r+1)_{-s} (a+1-r)_{s}} \frac{(-1)^{s}}{s! (b+1)_{-s}} \\
&= \binom{a}{r} \sum_{s=0}^{r} \frac{(-r)_{s} (-b)_{s} (-1)^{s}}{s! (a+1-r)_{s}} \\
&= \binom{a}{r} {}_{2}F_{1}(-r, -b; a-r+1; -1).
\end{align}
From this the expansion becomes
\begin{align}
(x-1)^{a} (x+1)^{b} &= \sum_{r=0}^{a|b} \left( (-1)^{r+a} \binom{a}{r} {}_{2}F_{1}(-r, -b; a-r+1; -1) \right) x^{r}.
\end{align}
The use of $a|b$ is to mean min(a,b), ie the minimum of $a$ or $b$.
</s>