$$B_{\frac{1}{2}}(n,n+1)=\frac{2^{-2 n-1} \left(\Gamma \left(n+\frac{1}{2}\right)+\sqrt{\pi } n \Gamma (n)\right)}{n \Gamma \left(n+\frac{1}{2}\right)}$$ for $n>0$ where $B$ is the incomplete beta function and I have proof of that here. I do not know how to prove it for $n\leq 0$.
Mathematica 9.0 produced this nice graph which is evidence that the equation in question is true for all complex n.
For $\Re(n)>0$ we have: $$\begin{eqnarray*}B\left(\frac{1}{2},n,n+1\right)&=&\int_{0}^{1/2}t^{n-1}(1-t)^{n}\,dt=\int_{0}^{1/2}(t-t^2)^n\frac{dt}{t}\\&=&\int_{0}^{1/2}\left(\frac{1}{4}-u^2\right)^n\frac{du}{1/2-u}\\&=&\frac{1}{4^n}\int_{0}^{1}\left(1-v^2\right)^{n}\frac{1}{1-v}\,dv\\&=&\frac{1}{4^n}\left(\int_{0}^{1}\left(1-v^2\right)^{n-1}\,dv+\int_{0}^{1}v\left(1-v^2\right)^{n-1}\,dv\right)\\&=&\frac{1}{4^n}\left(\int_{0}^{1}\left(1-v^2\right)^{n-1}\,dv+\frac{1}{2n}\right)\\&=&\frac{1}{2^{2n+1}}\left(B\left(\frac{1}{2},n\right)+\frac{1}{n}\right)=\frac{1}{2^{2n+1}}\left(\frac{\Gamma\left(\frac{1}{2}\right)\Gamma(n)}{\Gamma\left(n+\frac{1}{2}\right)}+\frac{1}{n}\right)\end{eqnarray*} $$ as stated. Like for the $\Gamma$ function, analytic continuation is ensured by integration by parts.