Binomial identities,combinatorial, equivalence

1. Feb 22, 2012

zmth

1. The problem statement, all variables and given/known data

To make it simpler just assume n is a positive even integer though it is also true when this is not the case but then the limits on s will be half an odd integer(s). We also
assume L is a non-negative integer and s goes by unit steps in the summation as usual.

sum_{s=0^{n/2} (2*s+1)^2 *(2*L+1)!*(2*L+2)! / ( (n/2+s+1)!*(2*L+1-n/2-s)! *(2*L+2-n/2+s)!(n/2-s)!) }

NOw this sum expression reduces to simply binom(4*L+2,n) = (4*L+2)!/n!/(4*L+2-n)!

2. The attempt at a solution

I have got it to the form (1/(2*L+2))*sum_{s=0^{n/2} (2*s+1)^2 * binom(2*L+2,n/2+s+1)*binom(2*L+2,n/2-s) } which the reader may verify is correct but seems
i can go no further with this as think this is the wrong way to go about it. Think we need to get the " s's " in the summation expressed as sum of products of binomials
where "s" is NOT in the numerator - that is only in the 2nd term of the binomials as...... binom(....,...s....)*.... I have tried in vain and cannot seem to reduce the sum
to closed form, that is with no summations, and equal to the correct answer binom(4*L+2,n) = (4*L+2)!/n!/(4*L+2-n)!. Can anyone use combinatorial methods to solve this ?