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I am working with the equation $$ \sum^{2n+1}_{k=0} \binom{2n+1}{k}(x^k -(-x)^k), \ n = 0,1,2,..$$ and want to rewrite it in terms of rising Pochhammer symbols.

I am aware of the relation $$ \frac{(x)_n}{n!} = \binom{x+n-1}{n}.$$ But how could I manipulate my binomial coefficient to get it in to this form?

Any help is greatly appreciated!

Edit: Just to clarify, here we denote $(x)_n$ as the rising factorial i.e $x(x+1)(x+2)\dots(x+n-1)$

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  • $\begingroup$ Since $$\frac{(x)_k}{k!} = \binom{x+k-1}{k}$$ then just solve for $x$ in $x+k-1=2n+1$ to get the argument of your Pochhammer symbol. $\endgroup$ – Tito Piezas III Jan 31 '18 at 15:42
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We can write \begin{align*} \binom{2n+1}{k}=\frac{(2n+1)(2n)\cdots(2n-k+2)}{k!}=\frac{(2n-k+2)_k}{k!} \end{align*}

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  • $\begingroup$ @TitoPiezasIII: Sorry for this inconvenience and thanks for upvote. :-) $\endgroup$ – Markus Scheuer Jan 31 '18 at 15:37

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