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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.

David
  • 796

2 Answers2

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you will have a product of odd values,nowmultiply both numerator and discriminator bythe product of even values, and rewrite using the expression

$$ (2n)!!=2^n n! $$

Alex
  • 19,262
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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 $$(-1)^n4^n{-1/2\choose n}=4^n\frac{\left(\frac{1}{2}\right)^n \times 1 \times 3 \times 5 \times 7 \ldots\times (2n-1)}{ n!} \\ = 2^n \frac{ 1 \times 3 \times 5 \times 7 \ldots(2n-1)}{ n!} \times \frac{n!}{n!} \\ = \frac{ 1 \times 3 \times 5 \times 7 \ldots(2n-1)}{ n! } \times \frac{2 \times 4 \times 6 \times \cdots \times 2n }{n!} \\ = \frac{(2n)!}{n!\,n!} \\= {2n \choose n}$$

Henry
  • 157,058