Proof a binomial identity for Newton's binomial coefficients

Question

I am supposed to prove this identity. I am unable to write fractions for some reason, so I'm providing an image. I have absolutely no idea how to go about this.

Thank you very much!

$$\binom{\frac{-1}{2}}{n} = \frac{(-1)^n}{4^n}\binom{2n}{n}$$

Answer 1

First it may be good to recall what $\binom{x}{n}$ is.

It is the polynomial of degree $n:$ $\frac{x(x-1)\dots(x-n+1)}{n!}$

If we put in $x=\frac{1}{2}$ we get $\frac{(-1/2)(-3/2)\dots(-(2n-1)/2)}{n!}$

we can rewrite this as $\frac{(1\cdot3\dots (2n-1))}{(-2)^n n!}$

So now all we have to do is realize that $1\cdot3\dots\cdot(2n-1)= \frac{(2n)!}{n!2^n}$.

To see that just write the right hand side as $\frac{1\cdot2\dots2n}{2\cdot 4\dots 2n}$ and see that all of the even terms in the numerator cancel out with the denominator.

Answer 2

Hint: $$\binom{\frac{-1}{2}}{n} = \frac{\frac{-1}2 \times \frac{-3}2 \times \frac{-5}2 \times \cdots \times \frac{-(2n-1)}{2}}{n \times (n-1) \times (n-2) \times \cdots \times 1} $$

$$\binom{2n}{n} = \frac{2n \times (2n-1) \times (2n-2) \times \cdots \times 2 \times 1}{(n \times (n-1) \times \cdots \times 1)^2} $$

and note $\frac{2n}{n}=\frac{2n-2}{n-1}=\cdots=\frac21 =2$