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Question Number 90749 by abdomathmax last updated on 25/Apr/20

find Σ_(k=0) ^n C_n ^k cos(((kπ)/n))

$${find}\:\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:{cos}\left(\frac{{k}\pi}{{n}}\right) \\ $$

Commented by mathmax by abdo last updated on 26/Apr/20

A_n =Σ_(k=0) ^n C_n ^k cos(((kπ)/n)) ⇒A_n =Re(Σ_(k=0) ^n C_n ^k e^((ikπ)/n) ) we have Σ_(k=0) ^n C_n ^k e^((ikπ)/n) =Σ_(k=0) ^n C_n ^k (e^((iπ)/n) )^k =(1+e^((iπ)/n) )^n =(1+cos((π/n))+isin((π/n)))^n =(2cos^2 ((π/(2n)))+2isin((π/(2n)))cos((π/(2n)))^n =(2cos((π/(2n)))e^((iπ)/(2n)) )^n =2^n cos^n ((π/(2n)))e^(i(π/2)) =2^n cos^n ((π/(2n)))i ⇒A_n =0

$${A}_{{n}} =\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:{cos}\left(\frac{{k}\pi}{{n}}\right)\:\Rightarrow{A}_{{n}} ={Re}\left(\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:{e}^{\frac{{ik}\pi}{{n}}} \right) \\ $$$${we}\:{have}\:\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:{e}^{\frac{{ik}\pi}{{n}}} \:=\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\left({e}^{\frac{{i}\pi}{{n}}} \right)^{{k}} \\ $$$$=\left(\mathrm{1}+{e}^{\frac{{i}\pi}{{n}}} \right)^{{n}} \:=\left(\mathrm{1}+{cos}\left(\frac{\pi}{{n}}\right)+{isin}\left(\frac{\pi}{{n}}\right)\right)^{{n}} \\ $$$$=\left(\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}{n}}\right)+\mathrm{2}{isin}\left(\frac{\pi}{\mathrm{2}{n}}\right){cos}\left(\frac{\pi}{\mathrm{2}{n}}\right)^{{n}} \right. \\ $$$$=\left(\mathrm{2}{cos}\left(\frac{\pi}{\mathrm{2}{n}}\right){e}^{\frac{{i}\pi}{\mathrm{2}{n}}} \right)^{{n}} \:=\mathrm{2}^{{n}} \:{cos}^{{n}} \left(\frac{\pi}{\mathrm{2}{n}}\right){e}^{{i}\frac{\pi}{\mathrm{2}}} \:=\mathrm{2}^{{n}} \:{cos}^{{n}} \left(\frac{\pi}{\mathrm{2}{n}}\right){i}\:\Rightarrow{A}_{{n}} =\mathrm{0} \\ $$

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