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How can I find the closed form (or product form) expression for $$ \sum_{k=n}^p \frac{(-)^k k!}{(p-k)!(k-n)!(k+n+1)!}x^k \quad ?$$ Any help is appreciated.

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    $\begingroup$ I did not think $(-)^k$ is still a notation people use. Am I mistaken? It reminds of early 20th century British math (or perhaps I should say maths) textbooks like Whittaker and Watson. $\endgroup$ – KConrad Sep 8 '16 at 4:49
  • $\begingroup$ Ok why is it not in use? I just use it because its shorter and looks better $\endgroup$ – Matt Majic Sep 8 '16 at 4:54
  • $\begingroup$ It was tried a long time ago and failed to catch on. I can't say for sure why, but I personally think it looks like a typographical error. After all, $-$ is not a number. $\endgroup$ – KConrad Sep 8 '16 at 13:17
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Maple writes it as a hypergeometric:

$$ {\frac { \left( -1 \right) ^{n}n!\,{x}^{n} {\mbox{$_2$F$_1$}(n+1,-p+n;\,2\,n+2;\,x)}}{ \left( p-n \right) !\, \left( 1+2\,n \right) !}} $$

I doubt that you'll get anything more "closed form" than that in general.

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I agree there is no closed formula for this, however, one may offer a recurrence. Here are two of them, one in terms of $p$ and the other in terms of $n$. To this end, denote the given sum by $S(p,n)$ after suppressing $x$. Then,

$$(p+n+3)(p-n+2)S(p+2,n)+(p+2)(x-2)S(p+1,n)-(x-1)S(p,n)=0$$ and $$x(p+n+3)S(p,n+2)+(x-2)(2n+3)S(p,n+1)-x(p-n)S(p,n)=0.$$ Just add initial conditions.

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