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Question Number 214186 by hardmath last updated on 30/Nov/24

If x + (1/x) = 1 find x^(61) + (1/x^(61) )+ 4 = ?

$$\mathrm{If}\:\:\:\mathrm{x}\:+\:\frac{\mathrm{1}}{\mathrm{x}}\:=\:\mathrm{1}\:\:\:\mathrm{find}\:\:\:\mathrm{x}^{\mathrm{61}} \:+\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{61}} }+\:\mathrm{4}\:\:=\:? \\ $$

Answered by Rasheed.Sindhi last updated on 30/Nov/24

x+(1/x)=1; x^(61) +(1/x^(61) )+4=? x+(1/x)=1⇒x^2 =x−1 x^3 =x^2 −x=(x−1)−x=−1 (x^3 )^(20) =(−1)^(20) =1⇒x^(60) =1 x^(61) +(1/x^(61) )+4=x^(60) .x+(1/(x^(60) .x))+4 =x+(1/x)+4=1+4=5

$${x}+\frac{\mathrm{1}}{{x}}=\mathrm{1};\:{x}^{\mathrm{61}} +\frac{\mathrm{1}}{{x}^{\mathrm{61}} }+\mathrm{4}=? \\ $$$${x}+\frac{\mathrm{1}}{{x}}=\mathrm{1}\Rightarrow{x}^{\mathrm{2}} ={x}−\mathrm{1} \\ $$$${x}^{\mathrm{3}} ={x}^{\mathrm{2}} −{x}=\left({x}−\mathrm{1}\right)−{x}=−\mathrm{1} \\ $$$$\left({x}^{\mathrm{3}} \right)^{\mathrm{20}} =\left(−\mathrm{1}\right)^{\mathrm{20}} =\mathrm{1}\Rightarrow{x}^{\mathrm{60}} =\mathrm{1} \\ $$$$\: \\ $$$${x}^{\mathrm{61}} +\frac{\mathrm{1}}{{x}^{\mathrm{61}} }+\mathrm{4}={x}^{\mathrm{60}} .{x}+\frac{\mathrm{1}}{{x}^{\mathrm{60}} .{x}}+\mathrm{4} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={x}+\frac{\mathrm{1}}{{x}}+\mathrm{4}=\mathrm{1}+\mathrm{4}=\mathrm{5} \\ $$

Answered by BaliramKumar last updated on 01/Dec/24

x + (1/x) = 2cos((π/3)) x^(61) + (1/x^(61) ) = 2cos(((61π)/3)) x^n + (1/x^n ) = 2cos(nθ) x^(61) + (1/x^(61) ) = 2cos(((60π + π)/3)) ⇒ 2cos(20π + (π/3)) x^(61) + (1/x^(61) ) = 2cos((π/3)) x^(61) + (1/x^(61) ) = 1 x^(61) + (1/x^(61) ) + 4 = 1 + 4 = 5

$${x}\:+\:\:\frac{\mathrm{1}}{{x}}\:=\:\mathrm{2}{cos}\left(\frac{\pi}{\mathrm{3}}\right) \\ $$$${x}^{\mathrm{61}} \:+\:\:\frac{\mathrm{1}}{{x}^{\mathrm{61}} }\:=\:\mathrm{2}{cos}\left(\frac{\mathrm{61}\pi}{\mathrm{3}}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}^{{n}} \:+\:\frac{\mathrm{1}}{{x}^{{n}} }\:=\:\mathrm{2}{cos}\left({n}\theta\right)\:\:\:\:\:\:\:\:\:\: \\ $$$${x}^{\mathrm{61}} \:+\:\:\frac{\mathrm{1}}{{x}^{\mathrm{61}} }\:=\:\mathrm{2}{cos}\left(\frac{\mathrm{60}\pi\:+\:\pi}{\mathrm{3}}\right)\:\:\:\Rightarrow\:\mathrm{2}{cos}\left(\mathrm{20}\pi\:+\:\frac{\pi}{\mathrm{3}}\right)\:\:\:\:\:\:\:\:\:\: \\ $$$${x}^{\mathrm{61}} \:+\:\:\frac{\mathrm{1}}{{x}^{\mathrm{61}} }\:=\:\mathrm{2}{cos}\left(\frac{\pi}{\mathrm{3}}\right) \\ $$$${x}^{\mathrm{61}} \:+\:\:\frac{\mathrm{1}}{{x}^{\mathrm{61}} }\:=\:\mathrm{1} \\ $$$${x}^{\mathrm{61}} \:+\:\:\frac{\mathrm{1}}{{x}^{\mathrm{61}} }\:+\:\mathrm{4}\:=\:\mathrm{1}\:+\:\mathrm{4}\:=\:\mathrm{5} \\ $$$$ \\ $$

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