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 will have a product of odd values,nowmultiply both numerator and discriminator bythe product of even values, and rewrite using the expression
$$
(2n)!!=2^n n!
$$
