y-1-5-2-10-Prove-y-123-55-5-2-

Question Number 222271 by MrGaster last updated on 21/Jun/25

$${y}=\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{10}} ,\mathrm{Prove}:{y}=\frac{\mathrm{123}+\mathrm{55}\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$

Answered by Rasheed.Sindhi last updated on 21/Jun/25

let x=((1+(√5) )/2) ⇒y=x^(10) 2x−1=(√5) 4x^2 −4x+1=5 x^2 −x−1=0 x^2 =x+1 x^4 =x^2 +2x+1=(x+1)+2x+1=3x+2 x^5 =3x^2 +2x=3(x+1)+2x=5x+3 x^(10) =25x^2 +30x+9=25(x+1)+30x+9 =55x+34 =55(((1+(√5) )/2))+34 =((55+55(√5) +68)/2) y=((123+55(√5) )/2)

$${let}\:{x}=\frac{\mathrm{1}+\sqrt{\mathrm{5}}\:}{\mathrm{2}}\:\Rightarrow{y}={x}^{\mathrm{10}} \\ $$$$\:\:\:\:\:\mathrm{2}{x}−\mathrm{1}=\sqrt{\mathrm{5}}\: \\ $$$$\:\:\:\mathrm{4}{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{1}=\mathrm{5} \\ $$$$\:\:{x}^{\mathrm{2}} −{x}−\mathrm{1}=\mathrm{0} \\ $$$$\:\:{x}^{\mathrm{2}} ={x}+\mathrm{1} \\ $$$$\:\:{x}^{\mathrm{4}} ={x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}=\left({x}+\mathrm{1}\right)+\mathrm{2}{x}+\mathrm{1}=\mathrm{3}{x}+\mathrm{2} \\ $$$$\:{x}^{\mathrm{5}} =\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}{x}=\mathrm{3}\left({x}+\mathrm{1}\right)+\mathrm{2}{x}=\mathrm{5}{x}+\mathrm{3} \\ $$$${x}^{\mathrm{10}} =\mathrm{25}{x}^{\mathrm{2}} +\mathrm{30}{x}+\mathrm{9}=\mathrm{25}\left({x}+\mathrm{1}\right)+\mathrm{30}{x}+\mathrm{9} \\ $$$$\:\:\:\:\:\:=\mathrm{55}{x}+\mathrm{34} \\ $$$$\:\:\:\:\:\:\:=\mathrm{55}\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}\:}{\mathrm{2}}\right)+\mathrm{34} \\ $$$$\:\:\:\:\:\:=\frac{\mathrm{55}+\mathrm{55}\sqrt{\mathrm{5}}\:+\mathrm{68}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:{y}=\frac{\mathrm{123}+\mathrm{55}\sqrt{\mathrm{5}}\:}{\mathrm{2}} \\ $$

Answered by MrGaster last updated on 21/Jun/25

φ=((1+(√5))/2) φ^n =F_n φ+F_(n−1) F_n =F_(n−1) +F_(n−2) ,F_0 =0,F_1 =1 F_2 =F_1 +F_0 =1 F_3 =F_2 +F_1 =2 F_4 =F_3 +F_2 =3 F_5 =F_4 +F_3 =5 F_6 =F_5 +F_4 =8 F_7 =F_6 +F_5 =13 F_8 =F_7 +F_6 =21 F_9 =F_8 +F_7 =34 F_(10) =F_9 +F_8 =55 φ^(10) =F_(10) φ+F_9 =55∙((1+(√5))/2)+34 =((55(1+(√5)))/2)+34=((55+55(√5))/2)+((68)/2)=((55+55(√5)+68)/2)=((123+55(√5))/2)

$$\phi=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\phi^{{n}} ={F}_{{n}} \phi+{F}_{{n}−\mathrm{1}} \\ $$$${F}_{{n}} ={F}_{{n}−\mathrm{1}} +{F}_{{n}−\mathrm{2}} ,{F}_{\mathrm{0}} =\mathrm{0},{F}_{\mathrm{1}} =\mathrm{1} \\ $$$${F}_{\mathrm{2}} ={F}_{\mathrm{1}} +{F}_{\mathrm{0}} =\mathrm{1} \\ $$$${F}_{\mathrm{3}} ={F}_{\mathrm{2}} +{F}_{\mathrm{1}} =\mathrm{2} \\ $$$${F}_{\mathrm{4}} ={F}_{\mathrm{3}} +{F}_{\mathrm{2}} =\mathrm{3} \\ $$$${F}_{\mathrm{5}} ={F}_{\mathrm{4}} +{F}_{\mathrm{3}} =\mathrm{5} \\ $$$${F}_{\mathrm{6}} ={F}_{\mathrm{5}} +{F}_{\mathrm{4}} =\mathrm{8} \\ $$$${F}_{\mathrm{7}} ={F}_{\mathrm{6}} +{F}_{\mathrm{5}} =\mathrm{13} \\ $$$${F}_{\mathrm{8}} ={F}_{\mathrm{7}} +{F}_{\mathrm{6}} =\mathrm{21} \\ $$$${F}_{\mathrm{9}} ={F}_{\mathrm{8}} +{F}_{\mathrm{7}} =\mathrm{34} \\ $$$${F}_{\mathrm{10}} ={F}_{\mathrm{9}} +{F}_{\mathrm{8}} =\mathrm{55} \\ $$$$\phi^{\mathrm{10}} ={F}_{\mathrm{10}} \phi+{F}_{\mathrm{9}} =\mathrm{55}\centerdot\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}+\mathrm{34} \\ $$$$=\frac{\mathrm{55}\left(\mathrm{1}+\sqrt{\mathrm{5}}\right)}{\mathrm{2}}+\mathrm{34}=\frac{\mathrm{55}+\mathrm{55}\sqrt{\mathrm{5}}}{\mathrm{2}}+\frac{\mathrm{68}}{\mathrm{2}}=\frac{\mathrm{55}+\mathrm{55}\sqrt{\mathrm{5}}+\mathrm{68}}{\mathrm{2}}=\frac{\mathrm{123}+\mathrm{55}\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$

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