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Question Number 72020 by mathmax by abdo last updated on 23/Oct/19

calculate Σ_(k=0) ^n C_n ^k cos(((kπ)/(2n)))

$${calculate}\:\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:{cos}\left(\frac{{k}\pi}{\mathrm{2}{n}}\right) \\ $$

Commented by mathmax by abdo last updated on 24/Oct/19

let A_n =Σ_(k=0) ^n C_n ^k cos(((kπ)/(2n))) ⇒A_n =Re(Σ_(k=0) ^n C_n ^k e^(i((kπ)/(2n))) ) but Σ_(k=0) ^n C_n ^k e^(i((kπ)/(2n))) =Σ_(k=0) ^n C_n ^k (e^((iπ)/(2n)) )^k =(1+e^((iπ)/(2n)) )^n =(1+cos((π/(2n)))+isin((π/(2n))))^n ={2cos^2 ((π/(4n)))+2isin((π/(4n)))cos((π/(4n)))}^n =2^n cos^n ((π/(4n))){e^((iπ)/(4n)) }^n =2^n e^((iπ)/4) cos^n ((π/(4n))) =2^n cos^n ((π/(4n))){((√2)/2) +i((√2)/2)} ⇒A_n =(√2)2^(n−1) cos^n ((π/(4n))) ⇒ A_n =2^(n−(1/2)) cos^n ((π/(4n)))

$${let}\:{A}_{{n}} =\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:{cos}\left(\frac{{k}\pi}{\mathrm{2}{n}}\right)\:\Rightarrow{A}_{{n}} ={Re}\left(\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:{e}^{{i}\frac{{k}\pi}{\mathrm{2}{n}}} \right)\:{but} \\ $$$$\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:{e}^{{i}\frac{{k}\pi}{\mathrm{2}{n}}} =\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\left({e}^{\frac{{i}\pi}{\mathrm{2}{n}}} \right)^{{k}} \:=\left(\mathrm{1}+{e}^{\frac{{i}\pi}{\mathrm{2}{n}}} \right)^{{n}} \\ $$$$=\left(\mathrm{1}+{cos}\left(\frac{\pi}{\mathrm{2}{n}}\right)+{isin}\left(\frac{\pi}{\mathrm{2}{n}}\right)\right)^{{n}} \:=\left\{\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{4}{n}}\right)+\mathrm{2}{isin}\left(\frac{\pi}{\mathrm{4}{n}}\right){cos}\left(\frac{\pi}{\mathrm{4}{n}}\right)\right\}^{{n}} \\ $$$$=\mathrm{2}^{{n}} \:{cos}^{{n}} \left(\frac{\pi}{\mathrm{4}{n}}\right)\left\{{e}^{\frac{{i}\pi}{\mathrm{4}{n}}} \right\}^{{n}} =\mathrm{2}^{{n}} \:{e}^{\frac{{i}\pi}{\mathrm{4}}} \:{cos}^{{n}} \left(\frac{\pi}{\mathrm{4}{n}}\right) \\ $$$$=\mathrm{2}^{{n}} \:{cos}^{{n}} \left(\frac{\pi}{\mathrm{4}{n}}\right)\left\{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:+{i}\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right\}\:\Rightarrow{A}_{{n}} =\sqrt{\mathrm{2}}\mathrm{2}^{{n}−\mathrm{1}} \:{cos}^{{n}} \left(\frac{\pi}{\mathrm{4}{n}}\right)\:\Rightarrow \\ $$$${A}_{{n}} =\mathrm{2}^{{n}−\frac{\mathrm{1}}{\mathrm{2}}} \:{cos}^{{n}} \left(\frac{\pi}{\mathrm{4}{n}}\right) \\ $$

Answered by mind is power last updated on 23/Oct/19

=ReΣ_(k=0) ^n C_n ^k e^((iπk)/(2n)) =Re(1+e^((iπ)/(2n)) )^n 1+e^((iπ)/(2n)) =e^((iπ)/(4n)) (e^((iπ)/(4n)) +e^((−iπ)/(4n)) )=2cos((π/(4n)))e^(i(π/(4n))) (1+e^((iπ)/(2n)) )^n =2^n cos^n ((π/(4n)))e^(i(π/4)) =ReΣ_(k=0) ^n C_n ^k e^((iπk)/(2n)) =Re(1+e^((iπ)/(2n)) )^n =2^n cos((π/(4n))).((√2)/2)

$$=\mathrm{Re}\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}} \mathrm{C}_{\mathrm{n}} ^{\mathrm{k}} \mathrm{e}^{\frac{\mathrm{i}\pi\mathrm{k}}{\mathrm{2n}}} =\mathrm{Re}\left(\mathrm{1}+\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{2n}}} \right)^{\mathrm{n}} \\ $$$$\mathrm{1}+\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{2n}}} =\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4n}}} \left(\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4n}}} +\mathrm{e}^{\frac{−\mathrm{i}\pi}{\mathrm{4n}}} \right)=\mathrm{2cos}\left(\frac{\pi}{\mathrm{4n}}\right)\mathrm{e}^{\mathrm{i}\frac{\pi}{\mathrm{4n}}} \\ $$$$\left(\mathrm{1}+\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{2n}}} \right)^{\mathrm{n}} =\mathrm{2}^{\mathrm{n}} \mathrm{cos}^{\mathrm{n}} \left(\frac{\pi}{\mathrm{4n}}\right)\mathrm{e}^{\mathrm{i}\frac{\pi}{\mathrm{4}}} \\ $$$$=\mathrm{Re}\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}} \mathrm{C}_{\mathrm{n}} ^{\mathrm{k}} \mathrm{e}^{\frac{\mathrm{i}\pi\mathrm{k}}{\mathrm{2n}}} =\mathrm{Re}\left(\mathrm{1}+\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{2n}}} \right)^{\mathrm{n}} =\mathrm{2}^{\mathrm{n}} \mathrm{cos}\left(\frac{\pi}{\mathrm{4n}}\right).\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$ \\ $$$$ \\ $$

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