A combinatorial identity in proving the Taylor expansion of the Spread Polynomials In trying to prove the Taylor expansion of the Spread Polynomials as given ( also in Wikipedia ) by S Goh in a new way I miss a final decisive step. How to prove a combinatorial simplification for this sum ?
$$\sum_{k+l=i}\frac{2m+1}{2k+1}\frac{2m+1}{2l+1}\binom{m+k}{2k}\binom{m+l}{2l}$$
If I compare this expression with the one of Goh the result should be
$$\frac{2m+1}{i+1}\binom{2m+1+i}{2m-i}$$
Please rely not 100% on this expression of the result because I might have done an error in my modification of Goh result when making a comparison. But the form of the result should be correct : a factor times a binomial both dependent on the integer $i$. $m$ is also an integer. 
 A: $$\sum_{k+l=i}\frac{2m+1}{2k+1}\frac{2m+1}{2l+1}\binom{m+k}{2k}\binom{m+l}{2l}$$
We have the obvious constraints $k, l \ge 0$, which we can rewrite (substituting $l = i - k$) as $0 \le k \le i$. The numerators of the two fractions can be pulled out as independent of $k$, so we have
$$(2m+1)^2 \sum_{k=0}^i\frac{(m+k)!(m+i-k)!}{(m-k)!(2k+1)!(m-i+k)!(2i-2k+1)!}$$
Wolfram Alpha (probably using Zeilberger's method) evaluates that sum as $$-\frac{(-1)^i 4^{1-i} m(m+1)(1-2m)_{i-1} (2m+3)_{i-1}}{3\left(\frac52\right)_{i-1}(3)_{i-1}}$$
where $(a)_n$ is the Pochhammer symbol: $(a)_n = a(a+1)\ldots(a+n-1)$. For positive $a$ and $n$ we have $(a)_n = \frac{(a+n-1)!}{(a-1)!}$, but we need to tread a bit more carefully with the first one in the numerator:
$$\begin{eqnarray}(-1)^{i-1} (-b)_{i-1} & = & (-1)^{i-1} (-b)(-b+1)\ldots(-b+i-2) \\
& = & (b)(b-1)\ldots(b-i+2)
\end{eqnarray}$$
So putting back in the $(2m+1)^2$ and expanding to factorials we have the original sum as
$$\begin{eqnarray}&\frac{4^{1-i} m(m+1)(2m+1)^2 \frac{(2m-1)!}{(2m-i)!}  \frac{(2m+i+1)!}{(2m+2)!}}{3 \left(\frac52\right)\ldots\left(\frac12 + i\right)  \frac{(i+1)!}{2!}}\\
= & \frac{(2m+1) (2m+i+1)!}{3 \cdot 2^i (2m-i)! \left(5 \cdot 7 \cdots(1 + 2i)\right) (i+1)!} \\
= & \frac{(2m+1) (2m+i+1)!}{2^i (2m-i)! (2i + 1)!! (i+1)!} \\
= & \frac{2m+1}{i+1} \binom{2m+i+1}{2m-i}
\end{eqnarray}$$
as desired.
