求证：$\dfrac{1}{\sin^2\dfrac{\pi}{2n+1}}+\dfrac{1}{\sin^2\dfrac{2\pi}{2n+1}}+\cdots +\dfrac{1}{\sin^2\dfrac{2n\pi}{2n+1}}=\dfrac 43n(n+1)$．
证明 设$\theta_k=\dfrac{k\pi}{2n+1}$，$k=1,2,\cdots ,n$，则
\[\begin{split} \cos(2n+1)\theta_k+{\rm i}\sin (2n+1)\theta_k&=\left(\cos\theta_k+{\rm i}\sin\theta_k\right)^{2n+1}\\
&={\rm C}_{2n+1}^0\cos^{2n+1}\theta_k+{\rm C}_{2n+1}^1\cos^{2n}\theta_k\cdot \sin\theta_k\cdot {\rm i}+\cdots +{\rm C}_{2n+1}^{2n+1}\left(\sin\theta_k\cdot{\rm i}\right)^{2n+1}
,\end{split} \]于是可得$${\rm C}_{2n+1}^1\cos^{2n}\theta_k-{\rm C}_{2n+1}^3\cos^{2n-2}\theta_k\sin^2\theta_k+\cdots +(-1)^n{\rm C}_{2n+1}^{2n+1}\sin^{2n}\theta_k=\sin(2n+1)\theta_k=0,$$也即$${\rm C}_{2n+1}^{1}\left(1-\sin^2\theta_k\right)^n-{\rm C}_{2n+1}^3\left(1-\sin ^2\theta_k\right)^{n-1}\sin^2\theta_k+\cdots +(-1)^n{\rm C}_{2n+1}^{2n+1}\sin^{2n}\theta_k=0,$$也即$${\rm C}_{2n+1}^{1}\left(\sin^{-2}\theta_k-1\right)^n-{\rm C}_{2n+1}^3\left(\sin^{-2}\theta_k-1\right)^{n-1}+\cdots +(-1)^n{\rm C}_{2n+1}^{2n+1}=0,$$因此$x=\dfrac{1}{\sin^2\theta_k}$($k=1,2\cdots ,n$)是关于$x$的方程$${\rm C}_{2n+1}^{1}\left(x-1\right)^n-{\rm C}_{2n+1}^3\left(x-1\right)^{n-1}+\cdots +(-1)^n{\rm C}_{2n+1}^{2n+1}=0$$的$n$个实数根，它们的和$$\sum_{k=1}^n\dfrac{1}{\sin^2\theta_k}=\dfrac{{\rm C}_n^1\cdot {\rm C}_{2n+1}^1+{\rm C}_{2n+1}^3}{{\rm C}_{2n+1}^1}=\dfrac{2n(n+1)}3,$$因此$$LHS=2\sum_{k=1}^n\dfrac{1}{\sin^2\theta_k}=\dfrac 43n(n+1).$$