Let $M$ and $N$ be positive integers such that $2N-3M\geq 0$. I would like to know if the (finite) sum
$$
\sum_{i=0}^{\infty}{{2N-3M}\choose{N-3i}}{M\choose i}^3
$$
has a nice closed form (in terms of $M$ and $N$), or a significantly simpler form.
The generalized Vandermonde convolution gives the identity
$$
\sum_{i_1,i_2,i_3}{{2N-3M}\choose{N-i_1-i_2-i_3}}{M\choose{i_1}}{M\choose{i_2}}{M\choose{i_3}}={{2N}\choose N}
$$
My sum appears as the sum of the diagonal terms $i_1=i_2=i_3$ on the left hand side.
The methods I have found to sum products of binomial coefficients all seem to be limited to products of binomials in which every appearance of the summation variable has coefficient 1. But in my series, there is a $3i$ term. Note that the sum is equal (up to a constant depending on $M$ and $N$) to the generalized hypergeometric series
$$
\,_9F_2\left(\begin{align*}-\frac{N}{3},-\frac{N+1}{3},-\frac{N+2}{3},&-\frac{N-3M}{3},-\frac{N-3M+1}{3},-\frac{N-3M+2}{3},&-M,-M,-M\\
&1,1&\end{align*}\Bigg|-3^6\right)
$$
This is not "well-poised", and so the identities I have found in the literature do not help in simplifying it. This is related to the appearance of the $3i$ term.
Best Answer
$$\sum_{i=0}^{\infty}{{2N-3M}\choose{N-3i}}{M\choose i}^3=\binom{2 N-3 M}{N} \times$$ $$ _6F_5\left(-M,-M,-M,\frac{1-N}{3},\frac{2-N}{3},-\frac{N}{3} ;1,1,\frac{N-3M+1}{3},\frac{N-3M+2}{3},\frac{N-3M+3}{3};1 \right)$$ looks a "bit" nicer.