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Is there a proof for the following identity that only uses the definition of the (generalized) binomial coefficient and basic transformations?

Let $n$ be a non-negative integer. $$\binom{2n}{n} = (-4)^n \binom{-\frac{1}{2}}{n}$$

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    $\begingroup$ Try to show $1\cdot 3 \cdot 5 \cdots (2n-1)$, the product of the first $n$ odd numbers, is $(2n)!/(2^nn!)$ by inserting factors $2, 4, 6, \dots, 2n$ to make the product $(2n)!$ and then dividing by $2 \cdot 4 \cdot 6 \cdots (2n)$ and then pulling a factor of $2$ from each of those even numbers in the denominator. $\endgroup$ – KCd Jan 13 '15 at 2:17
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You have not specified what is meant by the definition of the generalized binomial coefficient. We take one interpretation, which may not be the intended one.

The term $\binom{-1/2}{n}$ is equal to $$\frac{1}{n!}\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\cdots \left(-\frac{2n-1}{2}\right).$$ Multiply the top by $(2)(4)\cdots (2n)$ and the bottom by the same thing, in the form $2^n n!$. We get $$(-1)^n\frac{1}{n!}\cdot \frac{1}{2^n\cdot 2^n} \frac{2n!}{n!}.$$ The rest of the calculation is straightforward.

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$$\begin{align*} (-4)^n\binom{-1/2}n&=(-4)^n\frac{(-1/2)^{\underline n}}{n!}\\\\ &=\frac{(-4)^n}{n!}\left(-\frac12\right)\left(-\frac32\right)\ldots\left(-\frac{2n-1}2\right)\\\\ &=\frac{(-4)^n}{n!}\cdot\frac{(2n-1)!!}{(-2)^n}\\\\ &=\frac{2^nn!(2n-1)!!}{(n!)^2}\\\\ &=\frac{(2n)!!(2n-1)!!}{(n!)^2}\\\\ &=\frac{(2n)!}{(n!)^2}\\\\ &=\binom{2n}n \end{align*}$$

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$$\begin{align} \binom{-1/2}{ n} &= (-1/2)(-3/2)(-5/2)\ldots(-(2n-1)/2) /n! \\ &= (-1)^n 1\cdot 3\cdot 5\cdot\ldots\cdot(2n-1)/ (2^n n!) \\ &= (-1)^n (2n)!/ (2^n \cdot n! \cdot 2 \cdot 4 \cdot 6\cdot \ldots \cdot (2n)) \\ &= (-1)^n (2n)!/ (2^n n!)^2 \\ &= (-1/4)^n \binom{2n}{n}. \end{align}$$

Hence: $$ (1-4X)^{-1/2} $$ generates (link) $$ \binom{2n}{n}.$$

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I'll assume the generalized binomial coefficient is defined via the gamma function, with $\binom{x}{y}=\frac{\Gamma(x+1)}{\Gamma(y+1)\Gamma(x-y+1)}$ This will match the normal definition for integers, and will follow most of the familiar properties. When dealing with the gamma function, the most important properties to remember are that $\Gamma(x+1)=x\Gamma(x)$ and that the gamma function is defined for all real numbers, except for the negative integers and zero. In my view, this is by far the most natural extension for the binomial coefficients.

\begin{align} (-4)^n\binom{-\frac{1}{2}}{n}&=(-4)^n\frac{\Gamma(\frac{1}{2})}{\Gamma(n+1)\Gamma(\frac{1}{2}-n)}\\&=(-4)^n\frac{(\prod\limits_{k=1}^n\frac{1}{2}-k)\Gamma(\frac{1}{2}-n)}{\Gamma(n+1)\Gamma(\frac{1}{2}-n)}\\&=(-4)^n\frac{(\prod\limits_{k=1}^n\frac{1}{2}-k)}{\Gamma(n+1)}\\&=2^n\frac{\prod\limits_{k=1}^n2k-1}{n!}\\&=2^n\frac{\frac{\prod\limits_{k=1}^n(2k-1)(2k)}{\prod\limits_{k-1}^n2k}}{n!}\\&=2^n\frac{\frac{\prod\limits_{k=1}^{2n}k}{2^n\prod\limits_{k=1}^nk}}{n!}=2^n\frac{\frac{(2n)!}{2^nn!}}{n!}\\&=\frac{(2n)!}{(n!)^2}\\&=\frac{(2n)!}{n!(2n-n)!}\\&=\binom{2n}{n} \end{align}

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