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I have two questions regarding the binomial coefficient and Pochhammer’s Symbol when they contain negative value;

In the following example

$\sum\limits_{k=0}^{-n} \binom{-n}{k} \left(a\right)_{-n}$.

where

n = 1,2,.....

a = 1,2,.....

First:

How to compute or what is the mathematical representation of the binomial coefficient $\binom{-n}{k}$ ?

Second:

How to compute or what is the mathematical representation of the Pochhammer’s Symbol $\left(a\right)_{-n}$ ?

It is obvious that the binomial coefficient of $\binom{n}{k}$ is $\frac{n!}{(n-k)!k!}$.

The Pochhammer’s Symbol $\left(a\right)_{n}$ is defined as $ \left(a\right)_{n} = \frac{\Gamma\left(a+n\right)}{\Gamma\left(a\right)}$ .

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If $k$ is positive integer and $n$ is a real number, then you could define $\displaystyle {n \choose k}=\dfrac{n(n-1)\cdots(n-k+1)}{k!}$ together with $\displaystyle {n \choose 0}=1$.

So $\displaystyle {-n \choose k}= (-1)^k {n+k-1 \choose k}$.

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  • $\begingroup$ I have edited the question so that it will be more precise. Thank you. $\endgroup$ – sky-light Jul 6 '15 at 13:41
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(I prefer the notation $a^{\overline n}$ for the rising power $a(a+1)\ldots(a+n-1)$.) In order to preserve the exponential formula

$$a^{\overline{m+n}}=a^{\overline m}(a+m)^{\overline n}\;,$$

we must have (with $m=-n$)

$$a^{\overline{-n}}=\frac1{(a-n)^{\overline n}}=\frac1{(a-1)^{\underline n}};.$$

Since

$$\binom{x}k=\frac{x^{\underline k}}{k!}\;,$$

we have

$$\binom{-n}ka^{\overline{-n}}=\frac{(-n)^{\underline k}}{k!(a-n)^{\overline n}}=\frac{(-1)^k(n+k-1)^{\underline k}}{k!(a-1)^{\underline n}}=\frac{(-1)^k}{(a-1)^{\underline n}}\binom{n+k-1}k\;.$$

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