<s>[INST] Find close expression for the sum $S_{n,k}=\sum\limits_{i=0}^{2n} (-1)^i \binom{n-1}{i} \binom{n+1}{k-i}.$ Find close expression for the sum $$S_{n,k}=\sum_{i=0}^{2n} (-1)^i \binom{n-1}{i} \binom{n+1}{k-i}.$$
For small $k$ I have got following
\begin{align}
&S_{n,0}=1,\\
&S_{n,1}=2,\\
&S_{n,2}=-n+2,\\
&S_{n,3}=-2n+2,\\
&S_{n,4}=\frac{(n-1)(n-4)}{2},\\
&S_{n,5}=(n-1)(n-2),\\
&S_{n,6}=-\frac{(n-1)(n-2)(n-6)}{6}
\end{align}
What is the expression for $S_{n,k}$ for arbitrary $n,k$?
 [/INST] $$S_{n,k}=\sum_{i=0}^{2n} (-1)^i \binom{n-1}{i} \binom{n+1}{k-i}=\sum_{i=0}^{k} (-1)^i \binom{n-1}{i} \binom{n+1}{k-i}$$
Let $[x^j] f(x)$ denote the coefficient of $x^j$ in the expansion of $f(x)$.
$$\large\therefore S_{n,k}=[x^k]\ \sum_{k=0}^{\infty}\sum_{i=0}^{k} (-1)^i \binom{n-1}{i} \binom{n+1}{k-i}x^k$$$$\large=[x^k]\ \bigg(\sum_{j=0}^{\infty}(-1)^j \binom{n-1}{j}x^j\bigg)\bigg(\sum_{j=0}^{\infty}\binom{n+1}{j}x^j\bigg)$$$$\large=[x^k]\ (1-x)^{n-1}(1+x)^{n+1}=[x^k]\ (1-x^2)^{n-1}(x^2+2x+1)$$
$$\LARGE \therefore S_{n,k}=\begin{cases} 
      (-1)^{\frac{k}{2}}\Bigg(\binom{n-1}{\frac{k}{2}} - \binom{n-1}{\frac{k}{2}-1}\Bigg)& k\text{ is even} \\
      2(-1)^{\frac{k-1}{2}}\huge\binom{n-1}{\frac{k-1}{2}} & k\text{ is odd}
   \end{cases}$$
$\blacksquare$
Also see Cauchy product.
</s>