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Question Number 182973 by HeferH last updated on 17/Dec/22

Answered by Rasheed.Sindhi last updated on 18/Dec/22

$$x^2+x+1=0;\underset{n=1}{\stackrel{27}{\mathrm{\Sigma }}}{(x^n+\frac{1}{x^n})}^2=?$$$${(x^n+\frac{1}{x^n})}^2=x^{2n}+\frac{1}{x^{2n}}+2$$$$S_{27}=\underset{n=1}{\sum ^{27}}{(x^n+\frac{1}{x^n})}^2=\underset{n=1}{\stackrel{27}{\mathrm{\Sigma }}}{x}^{2n}+\underset{n=1}{\stackrel{27}{\mathrm{\Sigma }}}\frac{1}{x^{2n}}+\underset{n=1}{\stackrel{27}{\mathrm{\Sigma }}}2$$$$=27\left(2\right)+(x^2+x^4+x^6+...x^{54})+(x^{-2}+x^{-4}+x^{-6}+...x^{-54})$$$$=54+\frac{x^2(1-{\left(x^2\right)}^{27})}{1-x^2}+\frac{x^{-2}(1-{\left(x^{-2}\right)}^{27})}{1-x^{-2}}$$$$=54+\frac{x^2-x^{56}}{1-x^2}+\frac{x^{-2}-x^{-56}}{1-x^{-2}}★$$$$$$$$x^2+x+1=0\Rightarrow x^2=-x-1$$$$\Rightarrow x^3=-x^2-x=-(-x-1)-x=1$$$$\Rightarrow x^3=1$$$$\bullet x^{56}={\left(x^3\right)}^{18}\cdot x^2=1^{18}(-x-1)=-x-1$$$$\bullet x^{-56}={\left(x^3\right)}^{-19}\cdot x={\left(1\right)}^{-19}\cdot x=x$$$$\bullet x^{-2}={\left(x^3\right)}^{-1}\cdot x={\left(1\right)}^{-1}\cdot x=x$$$$$$$$S_{27}=54+\frac{x^2-x^{56}}{1-x^2}+\frac{x^{-2}-x^{-56}}{1-x^{-2}}$$$$=54+\frac{(-x-1)-(-x-1)}{1-(-x-1)}+\frac{\left(x\right)-\left(x\right)}{1-\left(x\right)}$$$$=54$$

Commented by Rasheed.Sindhi last updated on 18/Dec/22

$$\mathrm{Or}\mathrm{in}\mathrm{somewhat}\mathrm{shorter}\mathrm{way}$$$$★S_{27}=54+\frac{x^2-x^{56}}{1-x^2}+\frac{x^{-2}-x^{-56}}{1-x^{-2}}$$$$$$$$x^2+x+1=0\Rightarrow x^2=-x-1$$$$\Rightarrow x^3=-x^2-x=-(-x-1)-x=1$$$$$$$$S_{27}=54+\frac{x^2-{\left(x^3\right)}^{18}\cdot x^2}{1-x^2}+\frac{x^{-2}-{\left(x^3\right)}^{-18}\cdot x^{-2}}{1-x^{-2}}$$$$S_{27}=54+\frac{x^2-{\left(1\right)}^{18}\cdot x^2}{1-x^2}+\frac{x^{-2}-{\left(1\right)}^{-18}\cdot x^{-2}}{1-x^{-2}}$$$$S_{27}=54+\frac{\cancel{x^2}-\cancel{x^2}}{1-x^2}+\frac{\cancel{x^{-2}}-\cancel{x^{-2}}}{1-x^{-2}}$$$$=54$$

Answered by TANMAY PANACEA last updated on 18/Dec/22

Commented by mr W last updated on 18/Dec/22

$$welcomebacksir!$$$$youseemtohavetakenalongbreak.$$

Commented by Rasheed.Sindhi last updated on 18/Dec/22

$$\mathcal{T}anmaysir,happytoseeyouagain!$$

Answered by TheSupreme last updated on 18/Dec/22

$$x+1+\frac{1}{x}=0\to \frac{1}{x}=-x-1$$$$\underset{n=1}{\sum ^{27}}{x}^{2n}+\frac{1}{x^{2n}}+2=$$$$=\frac{1-x^{2\ast 27-1}}{1-x}-1+\frac{1-\frac{1}{x^{2\ast 27-1}}}{1-\frac{1}{x}}-1+2\cdot 27=$$$$=\frac{1-x^{53}}{1-x}-\frac{1-x^{53}}{x^{52}(1-x)}+52$$$$=\frac{1-x^{53}}{1-x}[1-\frac{1}{x^{52}}]+52=$$$$=$$$$x^2+x+1=0\to x_{1,2}=\frac{-1\pm \sqrt{-3}}{2}=e^{\pm \frac{\pi }{3}i}$$$$\mathrm{\Sigma }{(e^{k\frac{\pi }{3}i}+e^{-k\frac{\pi }{3}i})}^2=\mathrm{\Sigma }4{cos}^2\left(k\frac{\pi }{3}\right)=$$$$=\underset{n=1}{\sum ^9}4{cos}^2\left(0\right)+4{cos}^2\left(\frac{\pi }{3}\right)+4{cos}^2\frac{2\pi }{3}=$$$$=4\underset{n=1}{\sum ^9}1+\frac{1}{4}+\frac{1}{4}=6\cdot 9=54$$$$$$

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