观察下列等式： \[\begin{split}\left(\sin\dfrac{\mathrm \pi} {3}\right)^{-2}+\left(\sin\dfrac{2{\mathrm \pi} }{3}\right)^{-2}&=\dfrac{4}{3}\times 1\times 2,\\ \left(\sin\dfrac{\mathrm \pi} {5}\right)^{-2}+\left(\sin\dfrac{2{\mathrm \pi} }{5}\right)^{-2}+\left(\sin\dfrac{3{\mathrm \pi} }{5}\right)^{-2}+\left(\sin\dfrac{4{\mathrm \pi} }{5}\right)^{-2}&=\dfrac{4}{3}\times 2\times 3,\\ \left(\sin\dfrac{\mathrm \pi} {7}\right)^{-2}+\left(\sin\dfrac{2{\mathrm \pi} }{7}\right)^{-2}+\left(\sin\dfrac{3{\mathrm \pi} }{7}\right)^{-2}+\cdots+\left(\sin\dfrac{6{\mathrm \pi} }{7}\right)^{-2}&=\dfrac{4}{3}\times 3\times 4,\\ \left(\sin\dfrac{\mathrm \pi} {9}\right)^{-2}+\left(\sin\dfrac{2{\mathrm \pi} }{9}\right)^{-2}+\left(\sin\dfrac{3{\mathrm \pi} }{9}\right)^{-2}+\cdots+\left(\sin\dfrac{8{\mathrm \pi} }{9}\right)^{-2}&=\dfrac{4}{3}\times 4\times 5,\\ &\cdots\cdots\end{split}\] 照此规律， $\left(\sin\dfrac{\mathrm \pi} {2n+1}\right)^{-2}+\left(\sin\dfrac{2{\mathrm \pi} }{2n+1}\right)^{-2}+\left(\sin\dfrac{3{\mathrm \pi} }{2n+1}\right)^{-2}+\cdots+\left(\sin\dfrac{2n{\mathrm \pi} }{2n+1}\right)^{-2}=$ _______．
答案 $\dfrac{4}{3}n\left(n+1\right)$．
解析 归纳，猜想\[\left(\sin\dfrac{\mathrm \pi} {2n+1}\right)^{-2}+\left(\sin\dfrac{2{\mathrm \pi} }{2n+1}\right)^{-2}+\left(\sin\dfrac{3{\mathrm \pi} }{2n+1}\right)^{-2}+\cdots+\left(\sin\dfrac{2n{\mathrm \pi} }{2n+1}\right)^{-2}=\dfrac 43n(n+1).\] 记题中代数式为 $m$，$\theta_k=\dfrac{2k\pi}{2n+1}$，其中 $k=0,1,2,\cdots,2n$，则\[m=\sum_{k=1}^{2n}\dfrac{2}{1-\cos\theta_k},\]由 $\left(\cos\theta_k+{\rm i}\sin\theta_k\right)^{2n+1}=1$ 可得$$\mathop{\rm C}\nolimits_{2n+1}^0\cos^{2n+1}\theta_k-\mathop{\rm C}\nolimits_{2n+1}^2\cos^{2n-1}\theta_k\sin^2\theta_k+\cdots +(-1)^n\mathop{\rm C}\nolimits_{2n+1}^{2n}\cos\theta_k\sin^{2n}\theta_k=1,$$记 $\dfrac{2}{1-\cos\theta_k}=x_k$，则\[\cos\theta_k=1-\dfrac{2}{x_k},\]因此$$\mathop{\rm C}\nolimits_{2n+1}^0\left(1-\dfrac{2}{x_k}\right)^{2n+1}-\mathop{\rm C}\nolimits_{2n+1}^2\left(1-\dfrac{2}{x_k}\right)^{2n-1}\left(\dfrac{4}{x_k}-\dfrac{4}{x_k^2}\right)+\cdots +(-1)^n\mathop{\rm C}\nolimits_{2n+1}^{2n}\left(1-\dfrac{2}{x_k}\right)\left(\dfrac{4}{x_k}-\dfrac{4}{x_k^2}\right)^n=1,$$也即\[\mathop{\rm C}\nolimits_{2n+1}^0\left(x_k-2\right)^{2n+1}-\mathop{\rm C}\nolimits_{2n+1}^2\left(x_k-2\right)^{2n-1}\left(4x_k-4\right)+\cdots +(-1)^n\mathop{\rm C}\nolimits_{2n+1}^{2n}\left(x_k-2\right)\left(4x_k-4\right)^n=x_k^{2n+1},\]这是一个关于 $x_k$ 的 $2n$ 次方程，$x_1,x_2,\cdots,x_{2n}$ 是该方程的 $2n$ 个根，其最高项系数为\[\mathop{\rm C}\nolimits_{2n+1}^0\mathop{\rm C}\nolimits_{2n+1}^{2n}\cdot (-2)-\mathop{\rm C}\nolimits_{2n+1}^2\mathop{\rm C}\nolimits_{2n-1}^{2n-1}\cdot 4=-2(2n+1)^2,\]而次高项系数为\[\mathop{\rm C}\nolimits_{2n+1}^0\mathop{\rm C}\nolimits_{2n+1}^{2n-1}(-2)^2-\mathop{\rm C}\nolimits_{2n+1}^2\mathop{\rm C}\nolimits_{2n-1}^{2n-1}(-4)-\mathop{\rm C}\nolimits_{2n+1}^2\mathop{\rm C}\nolimits_{2n-1}^{2n-2}\cdot (-2)\cdot 4+\mathop{\rm C}\nolimits_{2n+1}^4\mathop{\rm C}\nolimits_{2n-3}^{2n-3}\cdot 4^2=\dfrac 83n(n+1)(2n+1)^2,\]因此根据韦达定理，有\[m=\sum_{k=1}^{2n}x_k=\dfrac{4n(n+1)}3.\]