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If-the-roots-of-the-equation-x-2-x-1-0-are-and-provided-that-x-n-n-n-Find-x-16-

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Question Number 107812 by I want to learn more last updated on 12/Aug/20
If the roots of the equation x^2 − x − 1 = 0 are α and β, provided that x_n = α^n + β^n . Find x_(16) .
$$\mathrm{If}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation}\:\:\:\mathrm{x}^{\mathrm{2}} \:\:−\:\:\mathrm{x}\:\:−\:\:\mathrm{1}\:\:\:=\:\:\mathrm{0}\:\:\:\mathrm{are}\:\:\:\alpha\:\:\mathrm{and}\:\:\beta, \\ $$$$\mathrm{provided}\:\mathrm{that}\:\:\:\:\:\:\mathrm{x}_{\mathrm{n}} \:\:=\:\:\alpha^{\mathrm{n}} \:\:+\:\:\beta^{\mathrm{n}} \:\:.\:\:\:\mathrm{Find}\:\:\:\:\mathrm{x}_{\mathrm{16}} . \\ $$
Answered by mr W last updated on 12/Aug/20
α+β=1 αβ=−1 α^2 +β^2 =(α+β)^2 −2αβ=1−2(−1)=3 α^4 +β^4 =(α^2 +β^2 )^2 −2(αβ)^2 =3^2 −2=7 α^8 +β^8 =(α^4 +β^4 )^2 −2(αβ)^4 =7^2 −2=47 α^(16) +β^(16) =(α^8 +β^8 )^2 −2(αβ)^8 =47^2 −2=2207
$$\alpha+\beta=\mathrm{1} \\ $$$$\alpha\beta=−\mathrm{1} \\ $$$$\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} =\left(\alpha+\beta\right)^{\mathrm{2}} −\mathrm{2}\alpha\beta=\mathrm{1}−\mathrm{2}\left(−\mathrm{1}\right)=\mathrm{3} \\ $$$$\alpha^{\mathrm{4}} +\beta^{\mathrm{4}} =\left(\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{2}\left(\alpha\beta\right)^{\mathrm{2}} =\mathrm{3}^{\mathrm{2}} −\mathrm{2}=\mathrm{7} \\ $$$$\alpha^{\mathrm{8}} +\beta^{\mathrm{8}} =\left(\alpha^{\mathrm{4}} +\beta^{\mathrm{4}} \right)^{\mathrm{2}} −\mathrm{2}\left(\alpha\beta\right)^{\mathrm{4}} =\mathrm{7}^{\mathrm{2}} −\mathrm{2}=\mathrm{47} \\ $$$$\alpha^{\mathrm{16}} +\beta^{\mathrm{16}} =\left(\alpha^{\mathrm{8}} +\beta^{\mathrm{8}} \right)^{\mathrm{2}} −\mathrm{2}\left(\alpha\beta\right)^{\mathrm{8}} =\mathrm{47}^{\mathrm{2}} −\mathrm{2}=\mathrm{2207} \\ $$
Commented by I want to learn more last updated on 12/Aug/20
I really appreciate sir. Sir is there a way to solve higher power like x_(91) ??
$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{sir}. \\ $$$$\mathrm{Sir}\:\mathrm{is}\:\mathrm{there}\:\mathrm{a}\:\mathrm{way}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{higher}\:\mathrm{power}\:\mathrm{like}\:\:\:\:\mathrm{x}_{\mathrm{91}} \:\:?? \\ $$
Commented by mr W last updated on 12/Aug/20
α,β=((1±(√5))/2) α^n +β^n =(((1+(√5))/2))^n +(((1−(√5))/2))^n
$$\alpha,\beta=\frac{\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\alpha^{{n}} +\beta^{{n}} =\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} +\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} \\ $$
Commented by I want to learn more last updated on 13/Aug/20
Thanks sir
$$\mathrm{Thanks}\:\mathrm{sir} \\ $$
Answered by hgrocks last updated on 12/Aug/20
x^(2 ) = x + 1 x^(n+1) = x^n + x^(n−1) α^(n +1) = α^n + α^(n−1) (1) β^( n+1) = β^( n) + β^( n−1) (2) (1)+(2)→ x_(n+1) = x_n + x_(n−1) x_0 = 2 x_1 = α + β = 1 x_2 = 2+1 = 3 x_3 = 1+3 = 4 Recursive Sequence : 2,1,3,4,7,11............
$$\: \\ $$$$\mathrm{x}^{\mathrm{2}\:} \:=\:\mathrm{x}\:+\:\mathrm{1} \\ $$$$\mathrm{x}^{\mathrm{n}+\mathrm{1}} \:=\:\mathrm{x}^{\mathrm{n}} \:+\:\mathrm{x}^{\mathrm{n}−\mathrm{1}} \\ $$$$\alpha^{\mathrm{n}\:+\mathrm{1}} \:=\:\alpha^{\mathrm{n}} \:+\:\alpha^{\mathrm{n}−\mathrm{1}} \:\:\:\:\:\:\:\left(\mathrm{1}\right) \\ $$$$\beta^{\:\mathrm{n}+\mathrm{1}} \:=\:\beta^{\:\mathrm{n}} \:+\:\beta^{\:\mathrm{n}−\mathrm{1}} \:\:\:\:\:\:\:\left(\mathrm{2}\right) \\ $$$$ \\ $$$$\left(\mathrm{1}\right)+\left(\mathrm{2}\right)\rightarrow \\ $$$$\mathrm{x}_{\mathrm{n}+\mathrm{1}} \:=\:\mathrm{x}_{\mathrm{n}} \:+\:\mathrm{x}_{\mathrm{n}−\mathrm{1}} \\ $$$$\mathrm{x}_{\mathrm{0}} \:=\:\mathrm{2} \\ $$$$\mathrm{x}_{\mathrm{1}} \:=\:\alpha\:+\:\beta\:=\:\mathrm{1} \\ $$$$\mathrm{x}_{\mathrm{2}} \:=\:\mathrm{2}+\mathrm{1}\:=\:\mathrm{3}\: \\ $$$$\mathrm{x}_{\mathrm{3}} \:=\:\mathrm{1}+\mathrm{3}\:=\:\mathrm{4} \\ $$$$\mathrm{Recursive}\:\mathrm{Sequence}\:: \\ $$$$\mathrm{2},\mathrm{1},\mathrm{3},\mathrm{4},\mathrm{7},\mathrm{11}………… \\ $$
Commented by I want to learn more last updated on 13/Aug/20
Thanks sir.
$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$

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