I need to prove the following statement.
Let $ n, g, m, a ,t$ be integers. Prove that the following statement is true for all $ n \geq g(1+2m)+1 $, $ g\geq 2t $, $ m\geq t $, $ 0\leq a <t $, and $ t>0 $:


$$\sum\_{l=0}^{m}(-1)^l\frac{\binom{m}{l}}{\binom{n/g-l}{m+1}}\left[\frac{\binom{n-2t}{gl-a}}{\binom{n}{gl}}-\frac{\binom{n-2t}{gl-2t+a}}{\binom{n}{gl}}\right]=0.$$


It works for multiple different values of the parameters. I tried induction or some combinatorial identities but could not come with a rigorous proof. Any idea might be helpful.



A summation involving fraction of binomial coefficients
First let's notice that
\begin{split}
\frac{\binom{m}l}{\binom{n/g-l}{m+1}} &= \binom{n/g}l\binom{n/g-l-m-1}{m-l} \frac{m!(m+1)!\Gamma(n/g-2m)}{\Gamma(n/g+1)} \\
&=(-1)^{m-l}\binom{n/g}l\binom{2m-n/g}{m-l} \frac{m!(m+1)!\Gamma(n/g-2m)}{\Gamma(n/g+1)}
\end{split}
and similarly
$$\frac{\binom{n-2t}{gl-a}}{\binom{n}{gl}} = \binom{gl}{a}\binom{n-gl}{2t-a} \frac{a!(2t-a)!(n-2t)!}{n!}.$$


Hence, the identity in question is equivalent to
$$\sum\_{l=0}^m \binom{n/g}l\binom{2m-n/g}{m-l} \binom{gl}{a}\binom{n-gl}{2t-a}$$
being invariant under replacement of $a$ with $2t-a$.


Multiplying the last expression by $x^a$ and summing over $a=0..2t$, we get the coefficient of $y^{2t}$ in
$$\sum\_{l=0}^m \binom{n/g}l\binom{2m-n/g}{m-l} (1+xy)^{gl} (1+y)^{n-gl},$$
and doing same after replacement of $a$ with $2t-a$, we get the coefficient of $y^{2t}$ in
$$\sum\_{l=0}^m \binom{n/g}l\binom{2m-n/g}{m-l} (1+y)^{gl} (1+xy)^{n-gl}.$$
Since $a$ and $t\leq m$ are arbitrary, we need to prove that the last two expressions as series in $x,y$ are equal modulo $y^{2m+1}$. Equivalently, we need to show that $F(x,y)\equiv G(x,y)\pmod{y^{2m+1}}$, where:
\begin{split}
F(x,y) &:= [z^m]\ (1+z(1+xy)^g)^{n/g} (1+z(1+y)^g)^{2m-n/g} (1+y)^{n-mg}\\
&= [z^m]\ \left(\frac{(1+y)^g+z(1+xy)^g}{1+z}\right)^{n/g} (1+z)^{2m},
\end{split}
and
\begin{split}
G(x,y) &:= [z^m]\ (1+z(1+y)^g)^{n/g} (1+z(1+xy)^g)^{2m-n/g} (1+xy)^{n-mg} \\
&=[z^m]\ \left(\frac{(1+xy)^g+z(1+y)^g}{1+z}\right)^{n/g} (1+z)^{2m}.
\end{split}




---


Using [Lagrange–Bürmann formula](https://en.wikipedia.org/wiki/Lagrange_inversion_theorem#Lagrange%E2%80%93B%C3%BCrmann_formula), we derive the expession
$$F(x,y) = [w^m]\ \left( \frac{A+B}2 + \frac{A-B}2\sqrt{1-4w} \right)^{n/g}\frac1{\sqrt{1-4w}},$$
where $A:=(1+y)^g$ and $B:=(1+xy)^g$, and the expression for $G(x,y)$ is obtained by exchanging $A$ and $B$.
It remains to notice that $A-B$ is a multiple of $y$ and in the expansion
$$F(x,y) = (-4)^m\left( \frac{A+B}2 \right)^{n/g} \sum\_{j=0}^{2m} \binom{n/g}j \left( \frac{A-B}{A+B} \right)^j \binom{j/2-1/2}{m} + O(y^{2m+1})$$
the terms with odd $j$ are zero, while for even $j$ the corresponding terms in $F$ and $G$ coincide.
QED


