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Question Number 29164 by abdo imad last updated on 04/Feb/18

let put α= 1+i(√3)     simlify  A_n = Σ_(k=0) ^n   α^k    .

$${let}\:{put}\:\alpha=\:\mathrm{1}+{i}\sqrt{\mathrm{3}}\:\:\:\:\:{simlify} \\ $$$${A}_{{n}} =\:\sum_{{k}=\mathrm{0}} ^{{n}} \:\:\alpha^{{k}} \:\:\:. \\ $$

Commented by abdo imad last updated on 06/Feb/18

e have α≠1 so A_n = ((1−α^(n+1) )/(1−α))=((1−(1+i(√3))^(n+1) )/(−i(√3)))    =(i/(√3))(1−(1+i(√3))^(n+1)  but we have  1+i(√3)=2( (1/2) +i((√3)/2))=2 e^(i(π/3))  ⇒(1+i(√3))^n = 2^n  e^(i((nπ)/3))   ⇒  A_n =(i/(√3))( 1−2^n e^(i((nπ)/3)) )=(i/(√3))(1−2^n cos(((nπ)/3))−i2^n sin(((nπ)/3)))  =(2^n /(√3))sin(((nπ)/3)) +(i/(√3))(1−2^n  cos(((nπ)/3)).

$${e}\:{have}\:\alpha\neq\mathrm{1}\:{so}\:{A}_{{n}} =\:\frac{\mathrm{1}−\alpha^{{n}+\mathrm{1}} }{\mathrm{1}−\alpha}=\frac{\mathrm{1}−\left(\mathrm{1}+{i}\sqrt{\mathrm{3}}\right)^{{n}+\mathrm{1}} }{−{i}\sqrt{\mathrm{3}}}\:\: \\ $$$$=\frac{{i}}{\sqrt{\mathrm{3}}}\left(\mathrm{1}−\left(\mathrm{1}+{i}\sqrt{\mathrm{3}}\right)^{{n}+\mathrm{1}} \:{but}\:{we}\:{have}\right. \\ $$$$\mathrm{1}+{i}\sqrt{\mathrm{3}}=\mathrm{2}\left(\:\frac{\mathrm{1}}{\mathrm{2}}\:+{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)=\mathrm{2}\:{e}^{{i}\frac{\pi}{\mathrm{3}}} \:\Rightarrow\left(\mathrm{1}+{i}\sqrt{\mathrm{3}}\right)^{{n}} =\:\mathrm{2}^{{n}} \:{e}^{{i}\frac{{n}\pi}{\mathrm{3}}} \\ $$$$\Rightarrow\:\:{A}_{{n}} =\frac{{i}}{\sqrt{\mathrm{3}}}\left(\:\mathrm{1}−\mathrm{2}^{{n}} {e}^{{i}\frac{{n}\pi}{\mathrm{3}}} \right)=\frac{{i}}{\sqrt{\mathrm{3}}}\left(\mathrm{1}−\mathrm{2}^{{n}} {cos}\left(\frac{{n}\pi}{\mathrm{3}}\right)−{i}\mathrm{2}^{{n}} {sin}\left(\frac{{n}\pi}{\mathrm{3}}\right)\right) \\ $$$$=\frac{\mathrm{2}^{{n}} }{\sqrt{\mathrm{3}}}{sin}\left(\frac{{n}\pi}{\mathrm{3}}\right)\:+\frac{{i}}{\sqrt{\mathrm{3}}}\left(\mathrm{1}−\mathrm{2}^{{n}} \:{cos}\left(\frac{{n}\pi}{\mathrm{3}}\right).\right. \\ $$

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