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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-




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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