What is the closed form of the following binomial identities:
$$\sum_{k=0}^{m}\binom{n}{k}\binom{r}{k}k\tag{I}$$
I'm not sure if we can find a closed form using Vandermonde's identity.
$$\sum_{k=0}^{n}\binom{m-k-1}{m-n-1}\left(k \right)\tag{II}$$
If I knew a closed form for $\sum_{k=0}^{n}\binom{k}{n}k$ then I would handle that, but unfortunately I don't know.
$$\sum_{k=0}^{n}\binom{n+k}{k}\binom{n}{k}\frac{\left(-1 \right)^k}{k+1}\tag{III}$$
I tried some binomial transformation, but that was not helpful.
Source :
Concrete mathematics (second edition)
Best Answer
Let $\Gamma(x)$ denote the Gamma function. (In particular $n!=\Gamma(n+1)$.) Then Mathematica gives the following results by using first Zeilberger's algorithm and then algorithm Hyper (both described in this book):
$$\sum_{k=0}^{m}\binom{n}{k}\binom{r}{k}k=-(m+1) \binom{n}{m+1} \binom{r}{m+1} \, _3F_2(1,m-n+1,m-r+1;m+2,m+2;1)-\binom{n}{m+2} \binom{r}{m+2} \, _3F_2(2,m-n+2,m-r+2;m+3,m+3;1)+\frac{\Gamma (n+r)}{\Gamma (n) \Gamma (r)}$$
(what an evil first sum)
$$\sum_{k=0}^{n}\binom{m-k-1}{m-n-1} k=\frac{\Gamma (m+1)}{\Gamma (n) \Gamma (m-n+2)}$$
and $$\sum_{k=0}^{n}\binom{n+k}{k}\binom{n}{k}\frac{\left(-1 \right)^k}{k+1}=\frac{1}{\Gamma (1-n) \Gamma (n+2)}.$$