A fun double sum involving binomial coefficients

Question

I came across the following double sum expression:

$$ S(k) := \sum_{l=0}^{2(k+1)-1} \sum_{i=0}^{l} (-1)^i\binom{2(k+1)-i}{2(k+1)-l} \left[ 3\binom{2k}{i-2}+ 3k\left(\binom{2k}{i-1} - \binom{2k-1}{i-2}\right) + \frac{1}{2}k(k-1)\binom{2k}{i} + \frac{1}{2}k(k+1)\binom{2k-2}{i-2} - k^2\binom{2k-1}{i-1}\right] $$

(Here I'm using the convention that $\binom{n}{m}=0$ for $m<0$.) My suspicion is that this expression is equal to the simple polynomial $S(k)=8k^2-12k+3$ for any natural number $k$. Indeed, I've been able to verify this computationally for $1\leq k\leq 100$. Simplification with Mathematica doesn't work and I've been unable to do an inductive argument. Also I've been unable to show that $S(k)$ is a polynomial of degree 2 (or of any constant degree...)

Any help would be greatly appreciated!

Answer 1

Remark a bit too long for a comment. The bracket $$3\binom{2k}{i-2}+ 3k\left(\binom{2k}{i-1} - \binom{2k-1}{i-2}\right) + \frac{1}{2}k(k-1)\binom{2k}{i} + \frac{1}{2}k(k+1)\binom{2k-2}{i-2} - k^2\binom{2k-1}{i-1}$$ can probably be simplified, by factorizing by $$\frac{k(2k-2)!}{i!(2k-i)!}.$$