It is many times that we need to compute discrete convolutions. Driven by this need we have discovered a following formula: \begin{equation} \sum\limits_{l=0}^k \binom{l+A_1}{A_2} \binom{-2 \beta (k-l)}{A_3} = \sum\limits_{L=0}^{A_3} \frac{1}{A_3!} \left.\frac{d^{A_3}}{d x^{A_3}} \left(x^{2 \beta} (1-x^{2 \beta})^L \right) \right|_{x=1} \cdot \binom{1+k+A_1+L}{1+A_2+L} \end{equation} Here $A_1,A_3$ are positive integers, $A_2 \ge A_1$ and $\beta$ is a parameter.
Note 1: When $A_3=0$ the right hand side reduces to one term only namely $\binom{1+k+A_1}{1+A_2}$ which is exactly what we get by using the telescoping property of the binomial factor (Pascal's triangle).
Note 2:If $\beta=-1/2$ the coefficients in the sum on the right hand side are equal to $\binom{A_3}{L} (-1)^L$ and therefore (using the Gauss' formula for $F_{2,1}$ at unity) the right hand side is equal to $\binom{k+A_1+1}{1+A_2+A_3}$. This is exactly what we should be getting in this case as we can prove by summation by parts for example.
Now, there are clearly two questions. Firstly can this neat formula be proven using some combinatorial reasoning and secondly are there other values of $\beta$ such that the right hand side simplifies even further?