A summation similar to Vandermonde identity

We seek to show that

$$\sum_{j=0}^k {k\choose j} {j/2\choose n} (-1)^{n+k-j} = \frac{k}{n} (-1)^k 2^{k-2n} {2n-k-1\choose n-1}$$

where $n\ge k\ge 0.$ We get for the even component

$$\sum_{p=0}^{\lfloor k/2 \rfloor} {k\choose 2p} {p\choose n} (-1)^{n+k} = 0$$

because $n\gt p$ and $p\ge 0.$ This leaves the odd component

$$- (-1)^{n+k} \sum_{p=0}^{\lfloor (k-1)/2 \rfloor} {k\choose 2p+1} {p+1/2\choose n}.$$

Now we have

$${p+1/2\choose n} = \frac{1}{n!} \prod_{q=0}^{n-1} (p+1/2-q) = \frac{1}{2^n n!} \prod_{q=0}^{n-1} (2p+1-2q) \\ = \frac{1}{2^n n!} \prod_{q=0}^{p} (2p+1-2q) \prod_{q=p+1}^{n-1} (2p+1-2q) \\ = \frac{1}{2^n n!} \frac{(2p+2)!}{2^{p+1} (p+1)!} (-1)^{n-p-1} \prod_{q=p+1}^{n-1} (2q-2p-1) \\ = \frac{1}{2^n n!} \frac{(2p+2)!}{2^{p+1} (p+1)!} (-1)^{n-p-1} \frac{(2n-2p-2)!}{2^{n-p-1} (n-p-1)!} \\ = \frac{(-1)^{n-p-1} (2n)!}{2^{2n} n!^2} {2n\choose 2p+2}^{-1} {n\choose p+1} \\ = \frac{(-1)^{n-p-1}}{2^{2n}} {2n\choose n} {2n\choose 2p+2}^{-1} {n\choose p+1}.$$

where $p\lt n.$ It will be helpful to re-write this as

$$\frac{p+1}{n} \frac{(-1)^{n-p-1}}{2^{2n}} {2n\choose n} {2n-1\choose 2p+1}^{-1} {n\choose p+1} \\ = \frac{(-1)^{n-p-1}}{2^{2n}} {2n\choose n} {2n-1\choose 2p+1}^{-1} {n-1\choose p}.$$

We thus get for our sum

$$\frac{(-1)^{k}}{2^{2n}} {2n\choose n} \sum_{p=0}^{\lfloor (k-1)/2 \rfloor} (-1)^p {k\choose 2p+1} {2n-1\choose 2p+1}^{-1} {n-1\choose p}.$$

Now observe that

$${k\choose 2p+1} {2n-1\choose 2p+1}^{-1} = \frac{k!}{(k-2p-1)!} \frac{(2n-2p-2)!}{(2n-1)!} \\ = {2n-1\choose k}^{-1} {2n-2p-2\choose k-2p-1}.$$

This yields for the sum

$$\frac{(-1)^{k}}{2^{2n}} {2n\choose n} {2n-1\choose k}^{-1} \sum_{p=0}^{\lfloor (k-1)/2 \rfloor} (-1)^p {2n-2p-2\choose k-2p-1} {n-1\choose p}.$$

Now to treat the remaining sum we have

$$[z^{k}] (1+z)^{2n-2} \sum_{p=0}^{\lfloor (k-1)/2 \rfloor} (-1)^p z^{2p+1} (1+z)^{-2p} {n-1\choose p}.$$

The coefficient extractor enforces the upper limit $\lfloor (k-1)/2\rfloor \ge p$ so we may continue with

$$[z^{k}] (1+z)^{2n-2} \sum_{p\ge 0} (-1)^p z^{2p+1} (1+z)^{-2p} {n-1\choose p} \\ = [z^{k}] (1+z)^{2n-2} z \left(1-\frac{z^2}{(1+z)^2}\right)^{n-1} \\ = [z^{k}] z (1+2z)^{n-1}.$$

This means for $k=0$ the sum is zero. For $k\ge 1$ we get including the factor in front

$$\bbox[5px,border:2px solid #00A000]{ \frac{(-1)^{k}}{2^{2n}} {2n\choose n} {2n-1\choose k}^{-1} {n-1\choose k-1} 2^{k-1}.}$$

To simplify this we expand the binomial coefficients

$$\frac{(-1)^k}{2^{2n-k+1}} \frac{(2n)!\times k! \times (2n-1-k)! \times (n-1)!} {n! \times n! \times (2n-1)! \times (k-1)! \times (n-k)!} \\ = \frac{(-1)^k}{2^{2n-k+1}} \frac{(2n)\times k \times (2n-1-k)!} {n \times n! \times (n-k)!} \\ = \frac{(-1)^k}{2^{2n-k}} \frac{k \times (2n-1-k)!} {n! \times (n-k)!}.$$

This yields at last

$$\bbox[5px,border:2px solid #00A000]{ \frac{(-1)^k}{2^{2n-k}} \frac{k}{n} {2n-1-k\choose n-1}.}$$