Given $\Gamma\left(\frac1{2}\right)=\sqrt{\pi}$, $$\sqrt{\pi}=\Gamma\left(\frac1{2}\right)=\Gamma\left(-\frac1{2}+1\right)=-\frac{1}{2}\Gamma\left(-\frac1{2}\right)$$ and so $\Gamma\left(-\frac1{2}\right)=-2\sqrt{\pi}$ $$-2\sqrt{\pi}=\Gamma\left(-\frac1{2}\right)=\Gamma\left(-\frac{3}{2}+1\right)=-\frac{3}{2}\Gamma\left(-\frac{3}{2}\right)$$ and so $\Gamma\left(-\frac{3}{2}\right)=\frac{4\sqrt{\pi}}{3}$
And so I'm trying to determine a closed form working this way and prove it using induction. The formula that I seem to be deducing is
$$\Gamma\left(\frac1{2}-n\right)=\frac{(-2)^n\sqrt{\pi}}{(2n-1)!!}$$
After showing base cases and assuming above is true for $n=k$, I work like this:
$$\Gamma\left(\frac1{2}-k\right)=\Gamma\left(\frac{3}{2}-(k+1)\right)=\Gamma\left(\frac{1}{2}-(k+1)+1\right)=\frac{2k+1}{-2}\Gamma\left(\frac{1}{2}-(k+1)\right)$$
Now I am in a position to solve for $\Gamma\left(\frac{1}{2}-(k+1)\right)$. However, I feel like this is an invalid inductive argument. I am used to doing induction problems where I plug in $k+1$ for $n$ as opposed to extracting it from the assumption step. Yet this is how I was able to conjecture the closed form I have, from extraacting successive terms from the previous. So my two question are these:
1) Is the closed form correct? I feel it is because given what I extracted, I can solve for $\Gamma\left(\frac{1}{2}-(k+1)\right)$ and get the formula for $k+1$.
2) Is this a valid proof if it is correct (and I suppose even if it isn't, it this extraction method valid during induction proofs)?