How to evaluate $-1/2 \choose 2$?

Question

How to evaluate $-1/2 \choose 2$? I came across this here:

But I don't see how they get the second line of the equation.

Answer 1

Two different ways.

1) An analytic function has a unique Taylor series at the origin. By setting $f(x)=\frac{1}{\sqrt{1-x}}$ and by computing $f(0),f'(0),f''(0)$ we get $$ f(x) = 1 + \frac{x}{2} + \frac{3x^2}{8} + O(x^3).$$ Assuming that the identity $$ (1-x)^n = \sum_{k\geq 0}\binom{n}{k}(-x)^k $$ holds for non-natural values of $n$ too, we have $\binom{-1/2}{1}=-\frac{1}{2}$ and $\binom{-1/2}{2}=\frac{3}{8}$.
In general, $\binom{-1/2}{k}=\frac{(-1)^k}{4^k}\binom{2k}{k}$.

2) This makes actually sense since $$ \binom{n}{k}=\frac{n!}{k!(n-k)!} = \frac{\Gamma(n+1)}{\Gamma(k+1)\Gamma(n-k+1)}$$ allows to extend the definition of $\binom{n}{k}$ in the $n=-\frac{1}{2}$ case, too: $$ \binom{-1/2}{k} = \frac{\Gamma(1/2)}{k!\Gamma(1/2-k)}=(-1)^k \frac{(2k-1)!!}{(2k)!!}$$ and the binomial theorem is compatible with this extension.

Answer 2

As shown in this answer, $$ \begin{align} \binom{-1/2}{n} &=\frac{\left(-\frac12\right)\left(-\frac32\right)\cdots\left(-\frac{2n-1}2\right)}{1\cdot2\cdots n}\\[6pt] &=(-1)^n\frac{(2n-1)!!}{2^nn!}\\[6pt] &=(-1)^n\frac{\frac{(2n)!}{2^nn!}}{2^nn!}\\[9pt] &=\left(-\frac14\right)^n\binom{2n}{n} \end{align} $$

Answer 3

$$\binom{n}{k}=\frac{n!}{k!(n-k)!}=\frac{n(n-1)\cdots (n-k+1)}{k!}=\frac{n^{\underline{k}}}{k!}.$$ So just put $\frac{1}{2}$ there.