TITLE: Two combinatorial identities
QUESTION [8 upvotes]: In my research, I found two combinatorial identities. Mathematica can give the answers immediately, but I don't know how to prove them. Could someone help me? Thank you!
Here are they:
Let $\alpha$, $\beta$ be two arbitrary complex numbers, and $k$, $l$ be two positive integers, then:
\begin{align*}
\sum_{m=1}^k m \binom{\frac{\alpha+\beta}{\beta}k}{k-m}\binom{\frac{\alpha+\beta}{\beta}l}{l+m}
=&\frac{\alpha\,k\,l}{(k+l)(\alpha+\beta)}\binom{\frac{\alpha+\beta}{\beta}k}{k}\binom{\frac{\alpha+\beta}{\beta}l}{l}\\
\sum_{m=1}^k m \binom{\frac{\alpha+\beta}{\beta}k}{k-m}\binom{\frac{\alpha+\beta}{\alpha}l}{l-m}
=&\frac{\alpha\,\beta\,k\,l}{(k\,\alpha+l\,\beta)(\alpha+\beta)}\binom{\frac{\alpha+\beta}{\beta}k}{k}\binom{\frac{\alpha+\beta}{\alpha}l}{l}
\end{align*}
Actually, the positive integer $l$ can be also a complex number. Then, if one replaces $\frac{\alpha}{\beta}l$ by $\tilde{l}$ in the first identity, it becomes equivalent to the second identity. Besides this observation, I can only prove some trivial cases.

REPLY [5 votes]: This is inspired by Fedor's answer:
Consider
 $$
f(m):=-\frac{\binom{ab}{m}\binom{cd}{m}(m-bc)(m-ad)}{\binom{ad-1}{m}\binom{bc-1}{m}(a-c)(b-d)}.
 $$
Then 
 $$
f(m+1)-f(m)=-\frac{\binom{ab}{m+1}\binom{cd}{m+1}(m+1-bc)(m+1-ad)}{\binom{ad-1}{m+1}\binom{bc-1}{m+1}(a-c)(b-d)}+\frac{\binom{ab}{m}\binom{cd}{m}(m-bc)(m-ad)}{\binom{ad-1}{m}\binom{bc-1}{m}(a-c)(b-d)}=-
\frac{\binom{ab}{m}\binom{cd}{m}}{\binom{ad-1}{m}\binom{bc-1}{m}(a-c)(b-d)}
\left(\frac{\frac{ab-m}{m+1}\frac{cd-m}{m+1}}{\frac{ad-1-m}{m+1}\frac{bc-1-m}{m+1}}(m+1-bc)(m+1-ad)-(m-bc)(m-ad)\right)=-
\frac{\binom{ab}{m}\binom{cd}{m}}{\binom{ad-1}{m}\binom{bc-1}{m}(a-c)(b-d)}(m^2-m(ab+cd)+abcd-m^2+m(ad+bc)-abcd)=m\frac{\binom{ab}{m}\binom{cd}{m}}{\binom{ad-1}{m}\binom{bc-1}{m}}.
 $$
Thus, the sum
 $$
\sum_{m=0}^n m\frac{\binom{ab}{m}\binom{cd}{m}}{\binom{ad-1}{m}\binom{bc-1}{m}}
 $$
is equal $f(n+1)-f(0)$, which implies the identity you want.