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Question Number 146924 by Willson last updated on 16/Jul/21

$$\mathbf{Montrer}\mathbf{que}$$$$\underset{\mathit{k}=0}{\sum ^{2\mathit{n}-1}}{\mathit{c}\mathit{o}\mathit{s}}^{2\mathit{n}}(\mathit{\theta }+\frac{\mathit{k}\mathit{\pi }}{2\mathit{n}})=\frac{{\mathit{n}\mathit{C}}_{2\mathit{n}}^{\mathit{n}}}{2^{2\mathit{n}-1}}$$

Answered by mindispower last updated on 17/Jul/21

$${cos}^{2n}\left(x\right)=\frac{1}{2^{2n-1}}.\left(\frac{\underset{k=0}{\sum ^{2n}}{C}_{2n}^k{e}^{ikx}.e^{-i(2n-k)x}}{2}\right)$$$$=\frac{1}{2^{2n-1}}\left(\frac{\underset{k=0}{\sum ^{n-1}}{C}_{2n}^k{e}^{i(k-2n+k)x}+C_{2n}^n+\underset{k=n+1}{\sum ^{2n}}{C}_{2n}^k{e}^{ikx-i(2n-k)x}}{2}\right)$$$$=\frac{1}{2^{2n-1}}(\frac{\underset{k=0}{\sum ^{n-1}}{C}_{2n}^k{e}^{i(2k-2n)x}+\underset{k=0}{\sum ^{n-1}}{C}_{2n}^{2n-k}{e}^{i(2n-k)-ikx}}{2}+\frac{C_n^{2n}}{2})$$$$=\frac{1}{2^{2n-1}}(\frac{\underset{k=0}{\sum ^{n-1}}{C}_{2n}^k(e^{-(2n-2k)x}+e^{i(2n-2k)x})}{2}+\frac{C_n^{2n}}{2}$$$$=\frac{\underset{k=0}{\sum ^{n-1}}{C}_{2n}^{k}cos\left((2n-2k)x\right)}{2^{2n-1}}+\frac{C_n^{2n}}{2}$$$$S=\frac{1}{2^{2n-1}}\underset{k=0}{\sum ^{2n-1}}Re(\underset{m=0}{\sum ^{n-1}}{C}_{2n}^m{e}^{i(2n-2m)(\theta +\frac{k\pi }{2n})}+\frac{C_n^{2n}}{2})$$$$=\frac{{nC}_n^{2b}}{2^{2n-1}}$$$$=\frac{{nC}_n^{2n}}{2}+Re(\underset{m=0}{\sum ^{n-1}}{C}_{2n}^m.e^{i(2n-2m)\theta }\underset{k=0}{\sum ^{2n-1}}{e}^{i(n-m).\frac{k\pi }{n}})$$$$=\frac{{nC}_n^{2n}}{2^{2n-1}}+Re\underset{m=0}{\sum ^{n-1}}{C}_{2n}^m{e}^{i(2n-2m)\theta }.\frac{1-{\left(e^{i(n-m).\frac{\pi }{n}}\right)}^{2n}}{1-e^{i(n-m).\frac{\pi }{n}}}$$$$=\frac{{nC}_n^{2n}}{2^{2n-1}}+Re(\underset{m=0}{\sum ^{n-1}}{C}_{2n}^m{e}^{i(2n-2m)\theta }.\frac{1-e^{i2(n-m)\pi }}{1-e^{i(n-m).\frac{\pi }{n}}})$$$$=\frac{{nC}_n^{2n}}{2^{2n-1}}+0$$$$\Rightarrow \underset{k=0}{\sum ^{2n-1}}{cos}^{2n}(\theta +\frac{k\pi }{2n})=\frac{{nC}_n^{2n}}{2^{2n-1}}$$

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