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Question Number 111213 by mnjuly1970 last updated on 02/Sep/20

Answered by mathmax by abdo last updated on 02/Sep/20

$$\mathrm{let}{\mathrm{S}}_{\mathrm{n}}=\sum _{\mathrm{k}=1}^{2\mathrm{n}}{\cos }^2\left(\frac{\mathrm{k}\pi }{2\mathrm{n}+1}\right)\Rightarrow {\mathrm{S}}_{\mathrm{n}}=\frac{1}{2}\sum _{\mathrm{k}=1}^{2\mathrm{n}}(1+\cos \left(\frac{2\mathrm{k}\pi }{2\mathrm{n}+1}\right))$$$$=\frac{2\mathrm{n}}{2}+\frac{1}{2}\sum _{\mathrm{k}=1}^{2\mathrm{n}}\cos \left(\frac{2\mathrm{k}\pi }{2\mathrm{n}+1}\right)=\mathrm{n}+\frac{1}{2}\sum _{\mathrm{k}=1}^{2\mathrm{n}}\cos \left(\frac{2\mathrm{k}\pi }{2\mathrm{n}+1}\right)\mathrm{snd}$$$$\sum _{\mathrm{k}=1}^{2\mathrm{n}}\cos \left(\frac{2\mathrm{k}\pi }{2\mathrm{n}+1}\right)=\mathrm{Re}(\sum _{\mathrm{k}=0}^{2\mathrm{n}}{\mathrm{e}}^{\frac{2\mathrm{ik}\pi }{2\mathrm{n}+1}}-1)$$$$\sum _{\mathrm{k}=0}^{2\mathrm{n}}{\left({\mathrm{e}}^{\left(\frac{2\mathrm{i}\pi }{2\mathrm{n}+1}\right)}\right)}^{\mathrm{k}}=\frac{1-{\left({\mathrm{e}}^{\frac{2\mathrm{i}\pi }{2\mathrm{n}+1}}\right)}^{2\mathrm{n}+1}}{1-{\mathrm{e}}^{\frac{2\mathrm{i}\pi }{2\mathrm{n}+1}}}=0\Rightarrow \mathrm{Re}(...)=-1\Rightarrow$$$${\mathrm{S}}_{\mathrm{n}}=\mathrm{n}+\frac{1}{2}(-1)=\mathrm{n}-\frac{1}{2}$$$$$$

Commented by mnjuly1970 last updated on 03/Sep/20

$$thankyouverymuchsir...$$

Answered by mathmax by abdo last updated on 02/Sep/20

$$\sum _{\mathrm{k}=1}^{2\mathrm{n}}{\cos }^2\left(\frac{\mathrm{k}\pi }{2\mathrm{n}+1}\right)=\sum _{\mathrm{k}=1}^{2\mathrm{n}}(1-{\sin }^2\left(\frac{\mathrm{k}\pi }{2\mathrm{n}+1}\right))$$$$=2\mathrm{n}-\sum _{\mathrm{k}=1}^{2\mathrm{n}}{\sin }^2\left(\frac{\mathrm{k}\pi }{2\mathrm{n}+1}\right)=2\mathrm{n}-(\mathrm{n}-\frac{1}{2})=\mathrm{n}+\frac{1}{2}$$

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