# Two combinatorial identities

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.

• In the RHS of the first identity, do you have $\alpha kl$ or rather $\beta kl$? I would expect something depending only on ${\alpha+\beta\over\beta}$, not on $\alpha$ or $\beta$, since so is the LHS. – Pietro Majer Feb 5 '17 at 8:32
• I guess dividing by ${ {\alpha+\beta\over\beta} l \choose l}$ both sides in the first identity should give a Vandermonde's identity en.wikipedia.org/wiki/Vandermonde%27s_identity – Pietro Majer Feb 5 '17 at 9:09
• @PietroMajer I doubt. This identity is telescoping and Vandermonde's is not (or at least I do not know how it is). – Fedor Petrov Feb 5 '17 at 17:54
• Yes; but I suspected the wanted identity may be different (see my first comment) – Pietro Majer Feb 5 '17 at 18:59

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 $l$): $$m\frac{\binom{z}{m}\binom{xy/z}{m}} {\binom{-x-1}m \binom{-y - 1}{ m}}=h(m-1)-h(m),\\ h(m):=\frac{z(m+x+1)(m+y+1)}{(z+x)(z+y)}\cdot \frac{\binom{z}{m+1}\binom{xy/z}{m+1}}{\binom{-x-1}{m+1}\binom{-y-1}{m+1}}.$$

• Maple 2016 outputs $${\frac {ABCD{\mbox{_3F_2}(2,-AB+1,-CD+1;\,-AD+2,-BC+2;\,1)}}{ \left( AD-1 \right) \left( BC-1 \right) }}$$ for the RHS of your supposed identity. – user64494 Feb 5 '17 at 16:20

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.

• I did not see the update to Fedor's post which appeared as I was typing the formulas; I'll leave this answer here anyway. – Vladimir Dotsenko Feb 5 '17 at 18:04

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

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