Calculate $(-1)^k \left(\begin{matrix} -1/2 \\ k\end{matrix}\right)$. Is it equal to $\frac {(2k)!}{2^{2k}(k!)^2}$?

Question

I woul like to know what is the value of the binomial number$$(-1)^k\left(\begin{matrix} -1/2 \\ k\end{matrix}\right)$$ It can possibly be equal to the number $$\frac {(2k)!}{2^{2k}(k!)^2}$$ I need to know if this equation holds for a much bigger problem.

Thanks in advance for trying to solve this.

Answer 1

For proving this, we use the definition of the extended binomial coefficient! $$\binom{-n}{r} = \frac{(-n)(-n-1)(-n-2)…(-n-r+1)}{r!} = \frac{(-1)^rn(n+1)(n+2)…(n+r-1)}{r!} = (-1)^r \binom{n+r-1}{r}$$

Now, substituting $n=\frac12$ and $r=k$, should give you the desired result.