Closed form of the sum of the product of three binomial coefficients

Question

I encountered with this kind of series from the calculation in quantum optics: $$\sum_{n,m=0}^\infty \sum_{k,l=0}^{\min(n,m)}\binom{n}{k}\binom{m}{l}\binom{n+m-k-l}{m-k}A^{n+m}B^kC^l$$ Provided that the choice of $A,B$ and $C$ are made to make the series converge, can one find a closed form of this series? Any kind of help will be appreciated.

edit : I have reduced the expression into the following one: $$ \sum_{n=0}^\infty A^{2n}(1+B)^n\frac{1}{n!}\frac{d^n}{dx^n}\left[\frac{1-x+A(1+Cx)}{1-x-A(1+Cx)}\frac{(1+Cx)^n}{(1-x)^{n+1}}\right]_{x=0}$$ Although it has a similar form with the Taylor series, I can't find a way to sort this.

Answer 1

I found a closed form for a more general series: $$\sum_{n,m=0}^\infty \sum_{k,l=0}^{\min(n,m)}\binom{n}{k}\binom{m}{l}\binom{n+m-k-l}{m-k}A^{n-k}B^{m-l}C^kD^l,$$

which is: $$ =\frac{1}{1-A-B-AC-BD-CD}.$$

One can find this closed form by using integral transform appropriately.