Simplify sum $\sum_{i=0}^k(-1)^ii\binom{n}{i}\binom{n}{k-i}$ for $n\geq k\geq 0$ The problem asks us to simplify the following sum:
$$\sum_{i=0}^k(-1)^ii\binom{n}{i}\binom{n}{k-i}$$ for  $n\geq k\geq 0$. I've tried the following:
$\sum_{i=0}^k(-1)^ii\binom{n}{i}\binom{n}{k-i}$=$n\sum_{i=0}^k(-1)^i\binom{n-1}{i-1}\binom{n}{k-i}$=$n\sum_{i=0}^{k-1}(-1)^i\binom{n-1}{i}\binom{n}{k-1-i}$
But I have no idea whether it's simpler or not, it's everything I came up with. Is there anything better that can be done?
 A: Suppose we seek to evaluate
$$\sum_{q=0}^k (-1)^q q {n\choose q} {n\choose k-q}$$
which is (here $n\ge k$)
$$n\sum_{q=1}^k (-1)^q {n-1\choose q-1} {n\choose k-q}.$$
Introduce
$${n\choose k-q} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{k-q+1}} (1+z)^n \; dz.$$
Observe that  this controls the range  being zero when $q\gt  k$ so we
may extend $q$ to $n$ to obtain for the sum
$$\frac{n}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{k+1}} (1+z)^n 
\sum_{q=1}^n (-1)^q {n-1\choose q-1} z^q\; dz
\\ = -\frac{n}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{k}} (1+z)^n
\sum_{q=0}^{n-1} (-1)^q {n-1\choose q} z^q\; dz
\\ = -\frac{n}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{k}} (1+z)^n
(1-z)^{n-1} \; dz
\\ = -\frac{n}{2\pi i}
\int_{|z|=\epsilon}
\frac{1+z}{z^{k}} (1-z^2)^{n-1} \; dz.$$
This is
$$-n[z^{k-1}] (1-z^2)^{n-1} 
- n[z^{k-2}] (1-z^2)^{n-1}.$$ 
Hence we get for $k$ even
$$n(-1)^{1+(k-2)/2} {n-1\choose (k-2)/2}$$
and for $k$ odd
$$n(-1)^{1+(k-1)/2} {n-1\choose (k-1)/2}.$$
Join these two to obtain
$$n(-1)^{1+\lfloor (k-1)/2\rfloor} 
{n-1\choose \lfloor (k-1)/2\rfloor}
\\ = n(-1)^{\lfloor (k+1)/2\rfloor} 
{n-1\choose \lfloor (k-1)/2\rfloor}.$$
A: The generating function for this summation is 
\begin{align}
\sum_{k=0}^{\infty} S_{k} t^{k} = -n \, \frac{(1-t^{2})^{n}}{t(1-t)}
\end{align}
where
\begin{align}
S_{k} = \sum_{i=0}^{k} (-1)^{i} i \, \binom{n}{i} \binom{n}{k-i}.
\end{align}
In terms of "reduction" it is seen that
\begin{align}
S_{k} = - n \, \binom{n}{k-1} \, {}_{2}F_{1}(-n +1, -k+1; n-k+2; -1)
\end{align}
where ${}_{2}F_{1}$ is the hypergeometric function. 
A: Hint
If $\quad k=2m,\;(m\ge 1)\quad $ then $\displaystyle \quad S_{2m}=\sum_{i=0}^{2m}(-1)^ii{n\choose i}{n\choose 2m-i}$
If $\quad k=2m-1,\;(m\ge 1)\quad $ then $\displaystyle \quad S_{2m-1}=\sum_{i=0}^{2m-1}(-1)^ii{n\choose i}{n\choose 2m-1-i}$
Now, You can prove that 
$$S_{2m-1}=S_{2m}=(-1)^m.n{n-1\choose m-1}$$
So, for $k\ge 1$ $$\Large S_k=\sum_{i=0}^k(-1)^ii{n\choose i}{n\choose k-i}= n(-1)^{\left\lfloor\frac{k+1}{2}\right\rfloor} {n-1\choose \left\lfloor\frac{k-1}{2}\right\rfloor}$$
