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Question Number 191527 by MATHEMATICSAM last updated on 25/Apr/23

$$x+y=1\mathrm{and}{x}^2+y^2=2.\mathrm{Find}\mathrm{the}\mathrm{value}$$$$\mathrm{of}{x}^{11}+y^{11}.$$

Answered by Rasheed.Sindhi last updated on 25/Apr/23

$${(x+y)}^2=1\Rightarrow x^2+y^2+2xy=1$$$$\Rightarrow 2+2xy=1\Rightarrow xy=-\frac{1}{2}$$$${(x+y)}^3=1\Rightarrow x^3+y^3+3xy(x+y)=1$$$$x^3+y^3+3(-\frac{1}{2})\left(1\right)=1$$$$\bullet x^3+y^3=1+\frac{3}{2}=\frac{5}{2}$$$$(x^2+y^2)(x^3+y^3)=\left(2\right)\left(\frac{5}{2}\right)$$$$\Rightarrow x^5+y^5+x^2{y}^2(x+y)=5$$$$\Rightarrow x^5+y^5+{(-\frac{1}{2})}^2\left(1\right)=5$$$$x^5+y^5=5-\frac{1}{4}=\frac{19}{4}$$$$\overline{)\begin{array}{c}{x}^5+y^5=\frac{19}{4}\end{array}}.......\left(i\right)$$$${(x^3+y^3)}^2=\frac{25}{4}\Rightarrow x^6+y^6+2x^3{y}^3=\frac{25}{4}$$$$\bullet x^6+y^6+2{(-\frac{1}{2})}^3=\frac{25}{4}\Rightarrow x^6+y^6=\frac{13}{2}$$$$\overline{)\begin{array}{c}{x}^6+y^6=\frac{13}{2}\end{array}}........\left(ii\right)$$$$\left(i\right)\times \left(ii\right):$$$$x^{11}+y^{11}+x^5{y}^5(x+y)=\left(\frac{19}{4}\right)\left(\frac{13}{2}\right)=\frac{247}{8}$$$$x^{11}+y^{11}+{(-\frac{1}{2})}^5\left(1\right)=\left(\frac{19}{4}\right)\left(\frac{13}{2}\right)=\frac{247}{8}$$$$x^{11}+y^{11}=\frac{247}{8}+\frac{1}{32}=\frac{989}{32}$$$$\overline{)\begin{array}{c}{x}^{11}+y^{11}=\frac{989}{32}\end{array}}$$

Answered by mr W last updated on 25/Apr/23

$$\mathit{M}\mathit{e}\mathit{t}\mathit{h}\mathit{o}\mathit{d}\mathit{I}\mathit{I}$$$$p=e_1=1$$$$p_2=e_1{p}_1-2e_2\Rightarrow 2=1-2e_2\Rightarrow e_2=-\frac{1}{2}$$$$p_n=e_1{p}_{n-1}-e_2{p}_{n-1}$$$$\Rightarrow p_n-p_{n-1}-\frac{1}{2}{p}_{n-1}=0$$$$r^2-r-\frac{1}{2}=0$$$$r=\frac{1\pm \sqrt{3}}{2}$$$$\Rightarrow p_n={\left(\frac{1+\sqrt{3}}{2}\right)}^n+{\left(\frac{1-\sqrt{3}}{2}\right)}^n$$$$examples:$$$$p_5={\left(\frac{1+\sqrt{3}}{2}\right)}^5+{\left(\frac{1-\sqrt{3}}{2}\right)}^5=\frac{19}{4}$$$$p_6={\left(\frac{1+\sqrt{3}}{2}\right)}^6+{\left(\frac{1-\sqrt{3}}{2}\right)}^6=\frac{13}{2}$$$$p_{11}={\left(\frac{1+\sqrt{3}}{2}\right)}^{11}+{\left(\frac{1-\sqrt{3}}{2}\right)}^{11}=\frac{989}{32}✓$$$$p_{15}={\left(\frac{1+\sqrt{3}}{2}\right)}^{15}+{\left(\frac{1-\sqrt{3}}{2}\right)}^{15}=\frac{13775}{128}$$$$p_{20}={\left(\frac{1+\sqrt{3}}{2}\right)}^{20}+{\left(\frac{1-\sqrt{3}}{2}\right)}^{20}=\frac{261575}{512}$$

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