求 $1+\dfrac{1}{1+\tan\dfrac{\pi}{2n+1}}+\dfrac{1}{1+\tan\dfrac{2\pi}{2n+1}}+\cdots +\dfrac{1}{1+\tan\dfrac{2n\pi}{2n+1}}$ 的值．
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【答案】
$1+\sum_\limits{k=1}^{2n}\dfrac{1}{1+\tan\dfrac{k\pi}{2n+1}}=\begin{cases} 2n+1,& 2 \mid n,\\ 0,&2\nmid n.\end{cases}$
【解析】
记 $\theta_k=\dfrac{k}{2n+1}\cdot 2\pi$，其中 $k=0,1,2,\cdots, 2n$，则由 $\left(\cos\theta_k+{\rm i}\sin\theta_k\right)^{2n+1}=1$ 可得$${\rm C}_{2n+1}^0\cos^{2n+1}\theta_k-{\rm C}_{2n+1}^2\cos^{2n-1}\theta_k\sin^2\theta_k+\cdots +(-1)^n{\rm C}_{2n+1}^{2n}\cos\theta_k\sin^{2n}\theta_k=1,$$令 $x_k=\dfrac{1}{\cos\theta_k}$，移项后两边同除以 $\cos^{2n+1}\theta_k$ 整理得到$$x_k^{2n+1}+(-1)^{n+1}{\rm C}_{2n+1}^{2n}\left(x_k^2-1\right)^n+\cdots +{\rm C}_{2n+1}^2\left(x_k^2-1\right)-{\rm C}_{2n+1}^0=0,$$这个 $2n+1$ 次方程的 $2n+1$ 个根恰为 $x_k,k=0,1,\cdots,2n$，于是$$\sum_{k=0}^{2n}x_k=(-1)^n{\rm C}_{2n+1}^{2n}=(-1)^n\cdot \left(2n+1\right),$$又$$\cos\left(\dfrac{k}{2n+1}\cdot {2\pi}\right)=\cos\left(\dfrac{2n+1-k}{2n+1}\cdot 2\pi\right),$$于是$$\sum_{k=1}^nx_k=\dfrac 12\left(\sum_{k=1}^{2n}{x_k}\right)=\dfrac{(-1)^n(2n+1)-1}2=\begin{cases} n,& 2\mid n,\\ -(n+1),&2\nmid n.\end{cases}$$而原式等于\[\begin{split} 1+\sum_{k=1}^{2n}\dfrac{1}{1+\tan\dfrac{k\pi}{2n+1}}&=1+\sum_{k=1}^{2n}\dfrac{\cos\dfrac {k\pi}{2n+1}\left(\cos\dfrac {k\pi}{2n+1}-\sin\dfrac{k\pi}{2n+1}\right)}{\cos\left(\dfrac{k}{2n+1}\cdot 2\pi\right)}\\
&=1+\dfrac 12\sum_{k=1}^n{\left(1-\tan\dfrac{2k\pi}{2n+1}+\dfrac 1{\cos{\dfrac {2k\pi}{2n+1}}}\right)}\\
&=(n+1)+\sum_{k=1}^{n}x_k\\
&=\begin{cases} 2n+1,& 2 \mid n,\\ 0,&2\nmid n.\end{cases} \end{split}\]
&=1+\dfrac 12\sum_{k=1}^n{\left(1-\tan\dfrac{2k\pi}{2n+1}+\dfrac 1{\cos{\dfrac {2k\pi}{2n+1}}}\right)}\\
&=(n+1)+\sum_{k=1}^{n}x_k\\
&=\begin{cases} 2n+1,& 2 \mid n,\\ 0,&2\nmid n.\end{cases} \end{split}\]
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