<s>[INST] Showing $\binom{2n}{n} = (-4)^n \binom{-1/2}{n}$ 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.
$$\binom{2n}{n} = (-4)^n \binom{-\frac{1}{2}}{n}$$
 [/INST] $$\begin{align}
 \binom{-1/2}{ n} &= (-1/2)(-3/2)(-5/2)\ldots(-(2n-1)/2) 
/n! \\
&= (-1)^n 1\cdot 3\cdot 5\cdot\ldots\cdot(2n-1)/ (2^n  n!) \\
&= (-1)^n (2n)!/ (2^n \cdot n! \cdot 2 \cdot 4 \cdot 6\cdot \ldots \cdot (2n)) \\
&= (-1)^n (2n)!/ (2^n  n!)^2 \\
&= (-1/4)^n \binom{2n}{n}. 
\end{align}$$
Hence:
$$ (1-4X)^{-1/2} $$
generates (link)
$$ \binom{2n}{n}.$$
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