Question Number 73487 by abdomathmax last updated on 13/Nov/19
$${let}\:\:\:\:\alpha\:{and}\:\beta\:{roots}\:{of}\:\:{the}\:{equation}\:\:{x}^{\mathrm{2}} −{x}+\mathrm{2}=\mathrm{0} \\ $$$${simplify}\:\:\:{A}_{{p}} =\:\alpha^{{p}} \:+\beta^{{p}} \:{and}\:{calculate} \\ $$$$\sum_{{p}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{A}_{{p}} \:\:{and}\:\sum_{{p}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:{A}_{{p}} ^{\mathrm{2}} \\ $$
Commented by mathmax by abdo last updated on 15/Nov/19
$${letsolve}\:{x}^{\mathrm{2}} −{x}\:+\mathrm{2}\:=\mathrm{0}\:\rightarrow\Delta=\mathrm{1}−\mathrm{8}=−\mathrm{7}\:\:\Rightarrow \\ $$$$\alpha=\frac{\mathrm{1}+{i}\sqrt{\mathrm{7}}}{\mathrm{2}}\:\:{and}\:\beta=\frac{\mathrm{1}−{i}\sqrt{\mathrm{7}}}{\mathrm{2}} \\ $$$$\mid\alpha\mid=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{1}+\mathrm{7}}=\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{2}}\:=\sqrt{\mathrm{2}}\:\Rightarrow\alpha=\sqrt{\mathrm{2}}{e}^{{iarctan}\left(\sqrt{\mathrm{7}}\right)} \:\:{and}\:\beta\:=\sqrt{\mathrm{2}}{e}^{−{iarctan}\left(\sqrt{\mathrm{7}}\right)} \\ $$$${A}_{{p}} =\alpha^{{p}} \:+\beta^{{p}} \:=\left(\sqrt{\mathrm{2}}\right)^{{p}} \:{e}^{{ip}\:{arctan}\left(\sqrt{\mathrm{7}}\right)} \:+\left(\sqrt{\mathrm{2}}\right)^{{p}} \:{e}^{−{ip}\:{arctan}\left(\sqrt{\mathrm{7}}\right)} \\ $$$$=\mathrm{2}^{\frac{{p}}{\mathrm{2}}} ×\mathrm{2}\:{Re}\left({e}^{{ip}\:{arctan}\left(\sqrt{\mathrm{7}}\right)} \right)\:=\mathrm{2}^{\mathrm{1}+\frac{{p}}{\mathrm{2}}} \:\:{cos}\left({p}\:{arctan}\left(\sqrt{\mathrm{7}}\right)\right) \\ $$$$\sum_{{p}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{A}_{{p}} =\sum_{{p}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\mathrm{2}^{\mathrm{1}+\frac{{p}}{\mathrm{2}}} {cos}\left({parctan}\left(\sqrt{\mathrm{7}}\right)\right) \\ $$$$={Re}\left(\mathrm{2}\sum_{{p}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\left(\sqrt{\mathrm{2}}\right)^{{p}} \:{e}^{{ip}\:{arctan}\left(\sqrt{\mathrm{7}}\right)} \right)\:\:{we}\:{have} \\ $$$$\left.\sum_{{p}=\mathrm{0}} ^{{n}−\mathrm{1}} \left(\sqrt{\mathrm{2}}\right)^{{p}} \:{e}^{{ip}\:{arctan}\left(\sqrt{\mathrm{7}}\right)} \right)\:=\sum_{{p}=\mathrm{0}} ^{{n}−\mathrm{1}} \left(\sqrt{\mathrm{2}}{e}^{{i}\:{arctan}\left(\sqrt{\mathrm{7}}\right)} \right)^{{p}} \\ $$$$=\frac{\mathrm{1}−\left(\sqrt{\mathrm{2}}{e}^{{i}\:{arctan}\left(\sqrt{\mathrm{7}}\right)} \right)^{{n}} }{\mathrm{1}−\sqrt{\mathrm{2}}{e}^{{i}\:{arctan}\left(\sqrt{\mathrm{7}}\right)} }\:=\frac{\mathrm{1}−\left(\sqrt{\mathrm{2}}\right)^{{n}} \:{e}^{{in}\:{arctan}\left(\sqrt{\mathrm{7}}\right)} }{\mathrm{1}−\sqrt{\mathrm{2}}{e}^{{isrctan}\left(\sqrt{\mathrm{7}}\right)} } \\ $$$$=\frac{\left(\mathrm{1}−\left(\sqrt{\mathrm{2}}\right)^{{n}} \left(\:{cos}\left({narctan}\left(\sqrt{\mathrm{7}}\right)\:+{isin}\left({narctan}\left(\sqrt{\mathrm{7}}\right)\right)\right.\right.\right.}{\mathrm{1}−\sqrt{\mathrm{2}}\left({cos}\left({arctan}\left(\sqrt{\mathrm{7}}\right)+{isin}\left({arctan}\left(\sqrt{\mathrm{7}}\right)\right)\right.\right.} \\ $$$${rest}\:{to}\:{extract}\:\:{Re}\:\left({of}\:{this}\:{quantity}…\right) \\ $$