常用或不常用数学结论
\[\sum_{k = 1}^{2m} \cos^{2m + 1}{(\frac{2k - 1}{4m + 1}\pi)} = \frac{1}{2}
\]
\(step 1.\)
拆分 \(\cos^{2m + 1}{\theta}\) .
\[\cos^{2m + 1}{\theta} = (\frac{e^{i\theta}+e^{-i\theta}}{2})^{2m + 1} = \frac{1}{2^{2m + 1}}[\sum_{k = 0}^{2m + 1} C_{2m + 1}^{k}(e^{(2k - 2m - 1)i\theta} + e^{(2k - 2m - 1)-i\theta})]
\]
\[= \frac{1}{2^{2m}}(\sum_{k = 0}^{m} C_{2m + 1}^{k}\cos{[(2m + 1 - 2k)\theta]})
\]
\[\equiv \frac{1}{2^{2m}}(\sum_{i = 0}^{m} C_{2m + 1}^{i}\cos{[(2m + 1 - 2i)\theta]})
\]
\[\equiv \frac{1}{2^{2m}}(\sum_{i = 0}^{m} C_{2m + 1}^{m - i}\cos{[(2i + 1)\theta]})
\]
\[\therefore 原式 = \frac{1}{2^{2m}}\sum_{k = 1}^{2m}(\sum_{i = 0}^{m} C_{2m + 1}^{m - i}\cos{[\frac{(2i + 1)(2k - 1)}{4m + 1}\pi]})
\]
\[= \frac{1}{2 ^ {2m}} \sum_{i = 0}^{m}C_{2m + 1}^{m - i}(\sum_{k = 1}^{2m}\cos{[\frac{2i + 1}{4m + 1}(2k - 1)\pi]})
\]
先求 \(\sum_{k = 1}^{2m}\cos{[\frac{2i + 1}{4m + 1}(2k - 1)\pi]}\)，设 \(sin\alpha\)，
\[\sum_{k = 1}^{2m}\cos{[\frac{2i + 1}{4m + 1}(2k - 1)\pi]} = \frac{\sin{\alpha}\sum_{k = 1}^{2m}\cos{[\frac{2i + 1}{4m + 1}(2k - 1)\pi]}}{\sin{\alpha}}
\]
\[= \frac{\sum_{k = 1}^{2m}\sin{\alpha}\cos{[\frac{2i + 1}{4m + 1}(2k - 1)\pi]}}{\sin{\alpha}}
\]
\[分子 \times 2 = \\\sin{(\frac{2i + 1}{4m + 1}1\pi + \alpha)} - \sin{(\frac{2i + 1}{4m + 1}1\pi - \alpha)}+
\\
\sin{(\frac{2i + 1}{4m + 1}3\pi + \alpha)} - \sin{(\frac{2i + 1}{4m + 1}3\pi - \alpha)}
+
\\
\dots
\\
\sin{(\frac{2i + 1}{4m + 1}(4m - 1)\pi + \alpha)} - \sin{(\frac{2i + 1}{4m + 1}(4m - 1)\pi - \alpha)}
\]
则令 \(\alpha = \frac{2i + 1}{4m + 1}\pi\)，发现分子中两两可相消，有：
\[分子 = \frac{\sin{(\frac{2i + 1}{4m + 1}4m\pi)}}{2} = \frac{\sin{[(2i + 1)\pi - \frac{2i + 1}{4m + 1}\pi]}}{2} = \frac{\sin{(\frac{2i + 1}{4m + 1}\pi)}}{2} = \frac{\sin\alpha}{2}
\]
\[\therefore 上式 = \frac{\frac{\sin{\alpha}}{2}}{\sin{\alpha}} = \frac{1}{2}
\]
则原式 \(=\)
\[\frac{1}{2^{2m + 1}} \sum_{i = 0}^{m} C_{2m + 1}^{m - i}
\]
\[\equiv \frac{1}{2^{2m + 1}} \sum_{i = 0}^{m} C_{2m + 1}^{i} = \frac{1}{2^{2m + 2}} (\sum_{i = 0}^{m} C_{2m + 1}^{i} + \sum_{i = 0}^{m} C_{2m + 1}^{i})
\]
\[= \frac{1}{2^{2m + 2}} (\sum_{i = 0}^{m} C_{2m + 1}^{i} + \sum_{i = m + 1}^{2m + 1} C_{2m + 1}^{i}) = \frac{\sum_{i = 0}^{2m + 1} C_{2m + 1}^{i}}{2^{2m + 2}}
\]
\[ = \frac{1}{2}
\]
\(QED.\)