求助一个二项式展开的相关问题

lemondian

\[
\begin{aligned}
(-1)^m & =e^{i m \pi} \\
& =\left(e^{i \frac{m \pi}{2 n+1}}\right)^{2 n+1} \\
& =\left(\cos \frac{m \pi}{2 n+1}+i \sin \frac{m \pi}{2 n+1}\right)^{2 n+1} \\
& =\sum_{k=0}^{2 n+1}\binom{2 n+1}{k} \cos ^k \frac{m \pi}{2 n+1}\left(i \sin \frac{m \pi}{2 n+1}\right)^{2 n-k+1}\\
& \text { 两端取虚部并同除 } \cos ^{2 n+1} \frac{m \pi}{2 n+1} \text { 得: } \\
&\color{red}{\Downarrow}\\
&\sum_{k=0}^n\binom{2 n+1}{2k}\left(-\tan ^2 \frac{m \pi}{2 n+1}\right)^{n-k}=0
\end{aligned}
\]
红色箭头处是如何得到的？（两端取虚部并同除那个东东）
算不出最后一行的结果呢？哪出问题了？

lemondian

那位高人帮忙解答一下吧，谢谢了

战巡

拆开不就知道了嘛
\[\left(\cos(\frac{m\pi}{2n+1})+i\sin(\frac{m\pi}{2n+1})\right)^{2n+1}\]
\[\cos^{2n+1}(\frac{m\pi}{2n+1})+C_{2n+1}^1i\cos^{2n}(\frac{m\pi}{2n+1})\sin(\frac{m\pi}{2n+1})+C_{2n+1}^2i^2\cos^{2n-1}(\frac{m\pi}{2n+1})\sin^2(\frac{m\pi}{2n+1})+C_{2n+1}^3i^3\cos^{2n-2}(\frac{m\pi}{2n+1})\sin^3(\frac{m\pi}{2n+1})+...\]
虚部当然是$i^1,i^3,i^5,...$这些奇数次项的总和啊
即
\[Im(\mbox{原式})=\sum_{k=0}^nC_{2n+1}^{2k+1}i^{2k+1}\cos^{2n-2k}(\frac{m\pi}{2n+1})\sin^{2k+1}(\frac{m\pi}{2n+1})\]
\[=i\sum_{k=0}^nC_{2n+1}^{2k+1}(-1)^k\cos^{2n-2k}(\frac{m\pi}{2n+1})\sin^{2k+1}(\frac{m\pi}{2n+1})\]
除以$\cos^{2n+1}(\frac{m\pi}{2n+1})$得到
\[=i\sum_{k=0}^{n}C_{2n+1}^{2k+1}(-1)^k\tan^{2k+1}(\frac{m\pi}{2n+1})\]
重新编码，令$j=n-k$，就得到
\[=i\sum_{j=0}^nC_{2n+1}^{2n+1-2j}(-1)^{n-j}\tan^{2n+1-2j}(\frac{m\pi}{2n+1})\]
\[=i\tan(\frac{m\pi}{2n+1})\sum_{j=0}^nC_{2n+1}^{2j}(-1)^{n-j}\tan^{2n-2j}(\frac{m\pi}{2n+1})\]