Prove gamma (n+1/2) = (2npi^1/2)/(n4^n) by induction

[Pami]

I tried solving this question this way:
Gamma(n+1/2)
=(n+1/2-1)gamma(n+1/2-1)
=(n-1/2)gamma(n-1/2)
=(2n-1)/2 gamma (2 n-1)/2
Don't know what to do next

 

[Svein]

To begin with $\Gamma (\frac{1}{2})=\sqrt{\pi}$. From there: $\Gamma (1+\frac{1}{2})= \frac{1}{2}\Gamma(\frac{1}{2})=\frac{1}{2}\sqrt{\pi}$. Checks against the formula.
Assume that the formula is correct for n. Then $\Gamma(n+1+\frac{1}{2})=(n+\frac{1}{2})\Gamma(n+\frac{1}{2})$...