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Question Number 49952 by maxmathsup by imad last updated on 12/Dec/18

$$1)simplify{A}_n=\frac{1}{{(2+i\sqrt{3})}^n}+\frac{1}{{(2-i\sqrt{3})}^n}$$$$2)smplify{B}_n=\frac{1}{{(2+\sqrt{3})}^n}+\frac{1}{{(2-\sqrt{3})}^n}$$$$nintegrnatural.$$

Commented by Abdo msup. last updated on 14/Dec/18

$$1)wehave{A}_n=\frac{{(2+i\sqrt{3})}^n+{(2-i\sqrt{3})}^n}{7^n}$$$$=\frac{2Re\left({(2+i\sqrt{3})}^n\right)}{7^n}but2+i\sqrt{3}=\sqrt{7}(\frac{2}{\sqrt{7}}+\frac{i\sqrt{3}}{\sqrt{7}})=r{e}^{i\theta }\Rightarrow$$$$r=\sqrt{7}andcos\theta =\frac{2}{\sqrt{7}},sin\theta =\frac{\sqrt{3}}{\sqrt{7}}\Rightarrow tan\theta =\frac{\sqrt{3}}{2}\Rightarrow \theta =arctan\left(\frac{\sqrt{3}}{2}\right)$$$$\Rightarrow {(2+i\sqrt{3})}^n={\left(\sqrt{7}\right)}^n{e}^{inarctan\left(\frac{\sqrt{3}}{2}\right)}\Rightarrow \frac{2{\left(\sqrt{7}\right)}^{n}cos\left(narctan\left(\frac{\sqrt{3}}{2}\right)\right)}{7^n}$$$$=\frac{2}{{\left(\sqrt{7}\right)}^n}cos\left(narctan\left(\frac{\sqrt{3}}{2}\right)\right).$$$$$$

Commented by Abdo msup. last updated on 14/Dec/18

$$2)wehsve{B}_n=\frac{{(2+\sqrt{3})}^n+{(2-\sqrt{3})}^n}{{(2^2-{\left(\sqrt{3}\right)}^2)}^n}$$$$=\sum _{k=0}^n{C}_n^k{2}^{n-k}{\left(\sqrt{3}\right)}^k+\sum _{k=0}^n{C}_n^k{2}^{n-k}{(-\sqrt{3})}^k$$$$=\sum _{k=0}^n{C}_n^k\{{\left(\sqrt{3}\right)}^k+{(-\sqrt{3})}^k)2^{n-k}$$$$=\sum _{p=0}^{\left[\frac{n}{2}\right]}{C}_n^{2p}\{3^p+3^p)2^{n-2p}$$$$=2^{n+1}\sum _{p=0}^{\left[\frac{n}{2}\right]}{\left(\frac{3}{4}\right)}^p{C}_n^{2p}$$

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