Question-111213

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

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

let S_n =Σ_(k=1) ^(2n) cos^2 (((kπ)/(2n+1))) ⇒S_n =(1/2)Σ_(k=1) ^(2n) (1+cos(((2kπ)/(2n+1)))) =((2n)/2) +(1/2)Σ_(k=1) ^(2n) cos(((2kπ)/(2n+1))) =n +(1/2)Σ_(k=1) ^(2n) cos(((2kπ)/(2n+1))) snd Σ_(k=1) ^(2n) cos(((2kπ)/(2n+1))) =Re(Σ_(k=0) ^(2n) e^((2ikπ)/(2n+1)) −1) Σ_(k=0) ^(2n) (e^((((2iπ)/(2n+1)))) )^k =((1−(e^((2iπ)/(2n+1)) )^(2n+1) )/(1−e^((2iπ)/(2n+1)) )) =0 ⇒Re(...) =−1 ⇒ S_n =n+(1/2)(−1) =n−(1/2)

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

Commented by mnjuly1970 last updated on 03/Sep/20

$${thank}\:{you}\:{very}\:{much}\:{sir}… \\ $$

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

Σ_(k=1) ^(2n) cos^2 (((kπ)/(2n+1))) =Σ_(k=1) ^(2n) (1−sin^2 (((kπ)/(2n+1)))) =2n−Σ_(k=1) ^(2n) sin^2 (((kπ)/(2n+1))) =2n−(n−(1/2)) =n+(1/2)

$$\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{2n}} \:\mathrm{cos}^{\mathrm{2}} \left(\frac{\mathrm{k}\pi}{\mathrm{2n}+\mathrm{1}}\right)\:=\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{2n}} \left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \left(\frac{\mathrm{k}\pi}{\mathrm{2n}+\mathrm{1}}\right)\right) \\ $$$$=\mathrm{2n}−\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{2n}} \:\mathrm{sin}^{\mathrm{2}} \left(\frac{\mathrm{k}\pi}{\mathrm{2n}+\mathrm{1}}\right)\:=\mathrm{2n}−\left(\mathrm{n}−\frac{\mathrm{1}}{\mathrm{2}}\right)\:=\mathrm{n}+\frac{\mathrm{1}}{\mathrm{2}} \\ $$

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