Binomial identity involving negative values I want to check this identity that appears in Feller's first book on probability:
$${2n\choose n}=(-1)^n4^n{-1/2\choose n}$$
So I apply this definition
$${x\choose r}=\frac{x(x-1)(x-2)\ldots(x-r+1)}{r!}$$
obtaining
$$\frac{1/2(1/2-1)(1/2-2)\ldots(1/2-n+1)}{n!}=\frac{(-1)(-1)^{n-1}(1/2)(1/2+1)(1/2+2)\ldots(1/2+n-1)}{n!}\\=\frac{(-1)^n\left(\frac{1}{2}\right)^n(1+2)(1+4)(1+6)\ldots(1+2n-2)}{n!}$$
But I can't see how to advance from here. I also tried adding the terms (1/2+s) and then extracting the 2 in the denominator but the numerator is a factorial of odd terms and I'm stucked again. Thanks.

You got as far as $${-1/2\choose n}=\frac{(-1)^n\left(\frac{1}{2}\right)^n(1+0)(1+2)(1+4)(1+6)\ldots(1+2n-2)}{n!}$$ though you still need to multiply by $(-1)^n 4^n$. Writing this as
$$(-1)^n4^n{-1/2\choose n}=4^n\frac{\left(\frac{1}{2}\right)^n \times 1 \times 3 \times 5 \times 7 \ldots\times (2n-1)}{ n!} \\ 
= 2^n \frac{ 1 \times 3 \times 5 \times 7 \ldots(2n-1)}{ n!} \times \frac{n!}{n!} \\ 
= \frac{ 1 \times 3 \times 5 \times 7 \ldots(2n-1)}{ n! } \times \frac{2 \times 4 \times 6 \times \cdots \times 2n }{n!} \\ = \frac{(2n)!}{n!\,n!} \\= {2n \choose n}$$
