Question Number 30739 by abdo imad last updated on 25/Feb/18
$${let}\:\left({u}_{{n}} \right)\:/\:{u}_{\mathrm{1}} =\mathrm{1}−{i}\:{and}\:\:\forall{p}\in\left\{\mathrm{2},\mathrm{3},...{n}\right\}\:{u}_{{p}} ={u}_{{p}−\mathrm{1}} {j}\:{with} \\ $$$${j}={e}^{{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \\ $$$$\left.\mathrm{1}\right){verify}\:{that}\:{u}_{\mathrm{1}} \:+{u}_{\mathrm{2}} \:+{u}_{\mathrm{3}} =\mathrm{0} \\ $$$$\left.\mathrm{2}\right){prove}\:{that}\:\forall{p}\in\:\left\{\mathrm{4},\mathrm{5},...,{n}\right\}\:\:{u}_{{p}} ={u}_{{p}−\mathrm{3}} \\ $$$$\left.\mathrm{3}\right){find}\:{the}\:{value}\:{of}\:{S}_{{n}} \:=\sum_{{i}=\mathrm{1}} ^{{n}} \:{u}_{{i}} \\ $$$$\left.\mathrm{4}\right){calculate}\:\:\alpha_{{n}} =\:\sum_{{p}=\mathrm{0}} ^{{n}−\mathrm{1}} {cos}\left(−\frac{\pi}{\mathrm{4}}\:+\frac{\mathrm{2}{p}\pi}{\mathrm{3}}\right)\:{and} \\ $$$$\beta_{{n}} =\:\sum_{{p}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{sin}\left(−\frac{\pi}{\mathrm{4}}\:+\frac{\mathrm{2}{p}\pi}{\mathrm{3}}\right). \\ $$