While solving a differential equation problem involving power series, I stumbled upon a sum (below) that seemed to be always equal to $0$, for any positive integer $s$.
$$ \sum_{k=0}^s \left( \frac{ \prod_{r=1}^k (-4r^2+10r-3) \prod_{r=1}^{s-k} (-4r^2+6r+1)}{2^s (2k)! (2s-2k+1)!} \times (2s-4k+1) \right) $$
Why is this sum always equal to $0$?
The simplified version of this equation would be: $$ \frac{1}{(-2)^s (2s+1)!} \sum_{k=0}^s \left( \binom{2s+1}{2k} (2s-4k+1) \prod_{r=1}^k (4r^2-10r+3) \prod_{r=1}^{s-k} (4r^2-6r-1) \right) $$ or $$ \frac{1}{(-2)^s (2s)!} \sum_{k=0}^s \left( \left( \binom{2s}{2k} - \binom{2s}{2k-1} \right) \prod_{r=1}^k (4r^2-10r+3) \prod_{r=1}^{s-k} (4r^2-6r-1) \right) $$
Update:
$4r^2-10r+3$ and $4r^2-6r-1$ were derived from $n^2-5n+3$ by replacing $n$ with $2r$ and $2r+1$, respectively, which means this could be rewritten as: $$ \frac{1}{(-2)^s (2s)!} \sum_{k=0}^s \left( \left( \binom{2s}{2k} - \binom{2s}{2k-1} \right) \prod_{r=1}^k a_{2r} \prod_{r=1}^{s-k} a_{2r+1} \right) $$ where $a_n=n^2-5n+3$
If the sum on the first line is indeed equal to $0$, then this would also be true: $$ \sum_{k=0}^s \left( \binom{2s}{2k} \prod_{r=1}^k a_{2r} \prod_{r=1}^{s-k} a_{2r+1} \right) = \sum_{k=0}^s \left( \binom{2s}{2k-1} \prod_{r=1}^k a_{2r} \prod_{r=1}^{s-k} a_{2r+1} \right) $$
Then, I changed the coefficients of $a_n$ into unknown constants to check whether this equality is true in other cases: $a_n=n^2-bn+c$
It seemed that this equality stands when $b$ is an odd integer greater than or equal to $5$ and $s$ is an integer greater than or equal to $(b-3)/2$; when these conditions are met, the value of $c$ does not seem to affect anything.
How could one mathematically derive or prove the above conclusion?