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We've been asked to prove an identity using binomial coefficients, but there's a negative fraction and im not sure how to solve it. I saw a similar post that helped and I wanted to know what you would do if instead of $-1/2$ you had $-3/2$.

So using this method: $$\binom{n}{k} = \frac{n!}{k! \, (n-k)!} = \prod_{m=1}^k \frac{n-m+1}{m}.$$ So for $n = -1/2$, we have $$\begin{align*} \binom{-{\textstyle \frac{1}{2}}}{k} &= \prod_{m=1}^k \frac{\frac{1}{2} - m}{m} = \prod_{m=1}^k \frac{1 - 2m}{2m} = \prod_{m=1}^k \frac{(1 - 2m)(2m)}{4m^2} \\ &= 4^{-k} \frac{(2k)!}{k! \, k!} = 4^{-k} \binom{2k}{k}. \end{align*}$$

But I want to use $n = -3/2$ and am having trouble getting to a similar result.

Thank you.

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  • $\begingroup$ Your $4^{-k}\binom{2k}k$ is missing a sign factor $(-1)^k$, e.g. the value of $\binom{-1/2}1$ is $-1/2$, not $1/2$. $\endgroup$ – bof Feb 25 '14 at 23:28
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$$\begin{align*} \binom{-3/2}{k} &= \prod_{m=1}^k \frac{-1/2 - m}{m} = (-1)^k \prod_{m=1}^k \frac{1 + 2m}{2m} = (-1)^k \prod_{m=1}^k \frac{(1 + 2m)(2m)}{4m^2} \\ &= (-4)^{-k} \frac{(2k+1)!}{k! \, k!} = \frac{2k+1}{(-4)^k} \binom{2k}{k} = \frac{k+1}{(-4)^k} \binom{2k+1}{k}. \end{align*}$$

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  • $\begingroup$ @bof fixed, thanks $\endgroup$ – gt6989b Feb 25 '14 at 23:31
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You can formally replace factorials with gamma function:

$\frac{n!}{k!(n-k)!}=\frac{\Gamma(n+1)}{\Gamma(k+1)\Gamma(n-k+1)}$

The latter formula is valid for any real or complex $n,k$, except for negative integers. In particular, for $n=-1/2$ you have $\Gamma(n+1)=\Gamma(1/2)=\sqrt\pi$.

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$$ \binom n k = \underbrace{\frac{n!}{k!(n-k)!}}_{\text{FIRST}} = \underbrace{\frac{n(n-1)(n-2)(n-3)\cdots(n-k+1)}{k!}}_{\text{SECOND}} $$

The expression above the FIRST $\underbrace{\text{underbrace}}$ is valid when $n$ is a nonnegative integer. The expression above the SECOND $\underbrace{\text{underbrace}}$ works just as well if $n$ is any complex number at all, including $-3/2$. For example, $$ \binom{-3/2}{4} = \frac{(-3/2)(-5/2)(-7/2)(-9/2)}{4\cdot3\cdot2\cdot1}. $$

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