Let $\theta=\frac{\pi}{2n+1}$, and multiply the expression by $2\sin\theta$ gives: $$\begin{align}&2\sin\theta\cos\theta + 2\sin\theta\cos3\theta+\cdots + 2\sin\theta\cos (2n-1)\theta\\=&\sin 2\theta + (\sin 4\theta - \sin 2\theta) +(\sin 6\theta -\sin 4\theta)+\cdots +(\sin 2n\theta - \sin (n-2)\theta)\\=&\sin 2n\theta\\=&\sin\frac{2n\pi}{2n+1}\\=&\sin\frac{\pi}{2n+1}\\=&\sin\theta\end{align}$$

Hence, the original expression equals $\frac{1}{2}$.