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

$$letput\alpha =1+i\sqrt{3}simlify$$$$A_n=\sum _{k=0}^n{\alpha }^k.$$

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

$$ehave\alpha \ne 1so{A}_n=\frac{1-{\alpha }^{n+1}}{1-\alpha }=\frac{1-{(1+i\sqrt{3})}^{n+1}}{-i\sqrt{3}}$$$$=\frac{i}{\sqrt{3}}(1-{(1+i\sqrt{3})}^{n+1}butwehave$$$$1+i\sqrt{3}=2(\frac{1}{2}+i\frac{\sqrt{3}}{2})=2e^{i\frac{\pi }{3}}\Rightarrow {(1+i\sqrt{3})}^n=2^n{e}^{i\frac{n\pi }{3}}$$$$\Rightarrow A_n=\frac{i}{\sqrt{3}}(1-2^n{e}^{i\frac{n\pi }{3}})=\frac{i}{\sqrt{3}}(1-2^{n}cos\left(\frac{n\pi }{3}\right)-i{2}^{n}sin\left(\frac{n\pi }{3}\right))$$$$=\frac{2^n}{\sqrt{3}}sin\left(\frac{n\pi }{3}\right)+\frac{i}{\sqrt{3}}(1-2^{n}cos\left(\frac{n\pi }{3}\right).$$

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