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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?

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    $\begingroup$ Tiny correction: I think it should be "... for any positive integer $s$", since for $s=0$ the result is $1$. $\endgroup$ – Théophile Nov 12 '15 at 23:52
  • $\begingroup$ @Théophile Thanks for catching that. $\endgroup$ – JungHwan Min Nov 13 '15 at 3:17
  • $\begingroup$ This sum is indeed 0. This can be checked by using Dougall-Ramanujan summation formula. $\endgroup$ – Nemo Nov 14 '15 at 0:10
  • $\begingroup$ @MartinNicholson: Your hint is not clear to me. Could you be somewhat more specific? $\endgroup$ – Markus Scheuer Nov 15 '15 at 8:42
  • $\begingroup$ @MartinNicholson: Yes, it is! :-) Thanks a lot! (+1) $\endgroup$ – Markus Scheuer Nov 15 '15 at 9:46
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Let $$ \alpha=\frac{5+\sqrt{13}}{4},\beta=\frac{5-\sqrt{13}}{4} $$ be the roots of equation $4r^2-10r+3=0$, and $$ \gamma=\frac{3+\sqrt{13}}{4},\delta=\frac{3-\sqrt{13}}{4} $$ be the roots of equation $4r^2-6r-1=0$. Note that $\alpha-\gamma=\beta-\delta=\frac{1}{2}$,$\ \alpha+\beta=\frac{5}{2}$ (this will be used below). Using the Pochhammer symbol $$ (x)_n=x(x+1) ... (x+n-1) $$ and the identities $$ (x)_{n-k}=(-1)^k\frac{(x)_n}{(-x-n+1)_k}, $$ $$ (2k)!=4^kk!\left(\frac{1}{2}\right)_k,\quad (2s-2k+1)!=2^{2s-2k}\left(1\right)_{s-k}\left(\frac{3}{2}\right)_{s-k} $$ one can rewrite the sum as $$ A(s)\cdot\sum_{k=0}^s\frac{(1-\alpha)_k(1-\beta)_k(-s)_k\left(-s-\frac{1}{2}\right)_k}{k!\left(\frac{1}{2}\right)_k(\gamma-s)_k(\delta-s)_k}\left(s-2k+\frac{1}{2}\right) $$ where $A(s)$ is a non-essential constant that had absorbed all prefactors that depend only on $s$ (but not on $k$). Up to a prefactor this is equal to $$ _7F_6\left({\textstyle a,1+\frac{a}{2},b,c,d,e,-s \atop \textstyle \frac{a}{2},1+a-b,1+a-c,1+a-d,1+a-e,1+a+s}\,;1\right) $$ where $a=-s-\frac{1}{2}$,$\ b=1-\alpha$,$\ c=1-\beta$,$\ d=e=\frac{1+a}{2}$. One can check that $1+2a=b+c+d+e-s$, so the series can be summed using Dougall's formula

To show that this series under consideration is $0$ it is enough to note that (one can see this from the link above) this sum is proportional to the Pochhammer symbol $(1+a-b-c)_s=(1-s)_s$ which is $0$ for positive $s$.

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    $\begingroup$ Very instructive answer! $\endgroup$ – Markus Scheuer Nov 15 '15 at 9:49

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