Question Number 49244 by maxmathsup by imad last updated on 04/Dec/18
$${let}\:{w}\:{from}\:{C}\:{and}\:{w}^{{n}} \:=\mathrm{1}\:{find}\:{the}\:{value}\:{of}\: \\ $$$${S}\:=\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{C}_{{n}} ^{{k}} \:{w}^{{k}} \:. \\ $$
Answered by Smail last updated on 05/Dec/18
$${S}=\left(\mathrm{1}+{w}\right)^{{n}−\mathrm{1}} \\ $$$${w}={e}^{\frac{\mathrm{2}{ik}\pi}{{n}}} \\ $$$${S}=\left(\mathrm{1}+{e}^{{i}\theta} \right)^{{n}−\mathrm{1}} \:\:{with}\:\theta=\frac{\mathrm{2}{k}\pi}{{n}} \\ $$$${S}=\left(\mathrm{1}+{cos}\theta+{isin}\theta\right)^{{n}−\mathrm{1}} \\ $$$$=\left(\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{\theta}{\mathrm{2}}\right)+\mathrm{2}{isin}\left(\frac{\theta}{\mathrm{2}}\right){cos}\left(\frac{\theta}{\mathrm{2}}\right)\right)^{{n}−\mathrm{1}} \\ $$$$=\left(\mathrm{2}{cos}\left(\frac{\theta}{\mathrm{2}}\right)\right)^{{n}−\mathrm{1}} \left({cos}\left(\frac{\theta}{\mathrm{2}}\right)+{isin}\left(\frac{\theta}{\mathrm{2}}\right)\right)^{{n}−\mathrm{1}} \\ $$$$=\mathrm{2}^{{n}−\mathrm{1}} {cos}^{{n}−\mathrm{1}} \left(\frac{\theta}{\mathrm{2}}\right)×{e}^{{i}\frac{\theta}{\mathrm{2}}\left({n}−\mathrm{1}\right)} \\ $$$$=\mathrm{2}^{{n}−\mathrm{1}} {cos}^{{n}−\mathrm{1}} \left(\frac{{k}\pi}{{n}}\right)×{e}^{{i}\left(\frac{{n}−\mathrm{1}}{{n}}\right){k}\pi} \\ $$$$=\mathrm{2}^{{n}−\mathrm{1}} {cos}^{{n}−\mathrm{1}} \left(\frac{{k}\pi}{{n}}\right){e}^{{i}\left(\mathrm{1}−\frac{\mathrm{1}}{{n}}\right){k}\pi} \\ $$$$=\mathrm{2}^{{n}−\mathrm{1}} {cos}^{{n}−\mathrm{1}} \left(\frac{{k}\pi}{{n}}\right)\left(−\mathrm{1}\right)^{{k}} \left({cos}\left(\frac{{k}\pi}{{n}}\right)−{isin}\left(\frac{{k}\pi}{{n}}\right)\right) \\ $$$${S}=\mathrm{2}^{{n}−\mathrm{1}} \left(−\mathrm{1}\right)^{{k}} {cos}^{{n}} \left(\frac{{k}\pi}{{n}}\right)\left(\mathrm{1}−{icot}\left(\frac{{k}\pi}{{n}}\right)\right) \\ $$