Two combinatorial identities

Question

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.

Answer 1

At first, I reformulate your identity (the first identity, the second is the same up to change of variables, as you note in the post).

Denote $(\alpha+\beta)/\beta=\lambda$ and divide both parts by $\binom{\lambda k}k \binom{\lambda l}l$. We get an equivalent identity $$ \sum_{m=0}^k m\frac{k(k-1)\dots (k-m+1)\cdot (\lambda l-l)\dots (\lambda l-l-m+1)}{(\lambda k-k+1)\dots (\lambda k-k+m)(l+1)\dots(l+m)}=\left(1-\frac1{\lambda}\right)\frac{kl}{k+l}. $$ Denote $\lambda k-k=x$, then $\lambda l-l=xl/k$, $\left(1-\frac1{\lambda}\right)\frac{kl}{k+l}=\frac{kxl}{(k+l)(k+x)}$ and we rewrite your identity as $$ \sum_{m=0}^km\frac{\binom{k}m\binom{xl/k}m}{\binom{-l-1}m\binom{-x-1}m}=\frac{kxl}{(k+l)(k+x)}. $$ Both parts are rational functions in $x$, $l$. The more general hypergeometric identity is $$ \sum_{m=0}^\infty m \frac{\binom{AB}m\binom{CD}m}{\binom{AD-1}m\binom{BC-1}m}=\frac{ABCD}{(A-C)(B-D)} $$ whenever LHS is well-defined, our case correspond to $A=-1,B=-k,C=x/k,D=l$.

UPDATE I proved it with the help of Wolframalpha. This is simply telescoping (here $z$ is old $k$ and $y$ is old $l$): $$m\frac{\binom{z}{m}\binom{xy/z}{m}} {\binom{-x-1}m \binom{-y - 1}{ m}}=h(m-1)-h(m),\\ h(m):=\frac{z(m+x+1)(m+y+1)}{(z+x)(z+y)}\cdot \frac{\binom{z}{m+1}\binom{xy/z}{m+1}}{\binom{-x-1}{m+1}\binom{-y-1}{m+1}}.$$

Answer 2

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.

Answer 3

In addition to the Fedor's answer: Maple 2016 code

sum(m*binomial(k, m)*binomial(x*l/k, m)/(binomial(-l-1, m)*binomial(-x-1, m)), m = 0 .. k);

produces $${\frac {lxk}{{k}^{2}+lk+xk+xl}}. $$