How to show that ${2n \choose n} p^nq^n = (-1)^n {-1/2 \choose n} (4pq)^n$?

Question

(Feller Vol.1, P.273) The probability of a return at the $2n$th trial is given by $$u_{2n} = {2n \choose n} p^nq^n = (-1)^n {-1/2 \choose n} (4pq)^n.$$

I can't figure out why the second equality holds. If I expand the right hand side, I get $$(-4)^n \frac{(-1/2)(-1/2 -1)(-1/2 -2)\cdot\cdot\cdot (-1/2 -n +1)}{n!}p^nq^n = \frac{(2)(2+4)(2+8) \cdot\cdot\cdot(2+4n-4)}{n!}p^nq^n.$$

This implies that $(2)(2+4)(2+8) \cdot\cdot\cdot(2+4n-4) = (2n)(2n-1)\cdot\cdot\cdot (n+1)$ (the numerator of ${2n \choose n}$), but I don't see that this is true. I would appreciate if you give some help.

Answer 1

You have a coefficient $$\frac{2\times6\times10\times\cdots\times(4n-2)}{n!}.$$ This equals $$2^n\frac{1\times3\times5\times\cdots\times(2n-1)}{n!} =2^n\frac{(2n)!}{n!(2\times 4\times\cdots\times(2n))}=\frac{(2n)!}{n!^2} =\binom{2n}n.$$

Answer 2

$$(-1)^n {-1/2 \choose n} (4pq)^n = (-1)^{n}(-1)^{n} \frac{(\frac{1}{2})(\frac{1}{2}+1)\cdots(\frac{1}{2}+n-1)}{n!}2^{n}2^{n}(pq)^{n}$$

$$ = \frac{1 \cdot 3 \cdot 5 \cdots}{n!}2^{n}(pq)^{n} = \frac{(2n-1)!!}{n!}2^{n}(pq)^{n}= \frac{(2n)!!}{(2n)!!}\frac{(2n-1)!!}{n!}2^{n}(pq)^{n} = \frac{(2n)!}{2^{n}(n!)^{2}} \cdot 2^{n}(pq)^{n} = \binom{2n}{n}(pq)^{n}$$

Where $(2n-1)!!$ denotes the Double factorial