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Question Number 194709 by MM42 last updated on 13/Jul/23
Show that  in fibonacci sequence              f_(3n) =f_n ^3 +f_(n+1) ^3 −f_(n−1) ^3
Showthatinfibonaccisequencef3n=fn3+fn+13fn13
Answered by TheHoneyCat last updated on 14/Jul/23
Let ϕ:= ((1+(√5))/2) and ψ:=((1−(√5))/2)  You might know that f_n =((ϕ^n −ψ^n )/( (√5)))  from that we can compute:  X=f_n ^(  3) +f_(n+1) ^(          3) −f_(n−1) ^(          3)   =(1/( (√5)))×(1/5)((ϕ^n −ψ^n )^3 +(ϕ^(n+1) −ψ^(n+1) )^3 −(ϕ^(n−1) −ψ^(n−1) )^3 )  =(1/( (√5)))×(1/5)(ϕ^(3n) −3ϕ^(2n) ψ^n +3ϕ^n ψ^(2n) −ψ^(3n)   +ϕ^(3n+3) −3ϕ^(2n+2) ψ^(n+1) +3ϕ^(n+1) ψ^(2n+2) −ψ^(3n+3)   −ϕ^(3n−3) +3ϕ^(2n−2) ψ^(n−1) −3ϕ^(n−1) ψ^(2n−2) +ψ^(3n−3) )  =(1/5)(f_(3n) +f_(3n+3) −f_(3n−3) )+(3/(5×(√5)))(−ϕ^(2n) ψ^n +ϕ^n ψ^(2n)   −ϕ^(2n+2) ψ^(n+1) +ϕ^(n+1) ψ^(2n+2) +ϕ^(2n−2) ψ^(n−1) −ϕ^(n−1) ψ^(2n−2) )  =(1/5)(f_(3n) +f_(3n+1) +f_(3n+2) −f_(3n−3) )  +((3ϕ^(n−1) ψ^(n−1) )/(5×(√5)))(−ϕ^(n+1) ψ+ϕψ^(n+1) −ϕ^(n+3) ψ^2   +ϕ^2 ψ^(n+3) +ϕ^(n−1) −ψ^(n−1) )  =(1/5)(f_(3n) +f_(3n) +f_(3n−1) +f_(3n+1) +f_(3n) −f_(3n−3) )  +((3ϕ^(n−1) ψ^(n−1) )/(5×(√5)))(−ϕ^(n+1) ψ+ϕψ^(n+1) −ϕ^(n+3) ψ^2   +ϕ^2 ψ^(n+3) +ϕ^(n−1) −ψ^(n−1) )    Now this is the moment where you use the fact  that ψ=((−1)/ϕ)  X=(1/5)(3f_(3n) +f_(3n−1) +f_(3n) +f_(3n−1) −f_(3n−3) )  +((3(−1)^(n+1) )/(5(√5)))(ϕ^n −ψ^n −ϕ^(n+1) +ψ^(n−1) +ϕ^(n−1) −ψ^(n−1) )  =(1/5)(4f_(3n) +f_(3n−1) +f_(3n−2) +f_(3n−3) −f_(3n−3) )  +((3(−1)^(n+1) )/5)(f_(n−1) +f_n −f_(n+1) )  =(1/5)(4f_(3n) +f_(3n−1) +f_(3n−2) )+0  =(1/5)(4f_(3n) +f_(3n) )  =f_(3n)   _□     There you go.
Letφ:=1+52andψ:=152Youmightknowthatfn=φnψn5fromthatwecancompute:X=fn3+fn+13fn13=15×15((φnψn)3+(φn+1ψn+1)3(φn1ψn1)3)=15×15(φ3n3φ2nψn+3φnψ2nψ3n+φ3n+33φ2n+2ψn+1+3φn+1ψ2n+2ψ3n+3φ3n3+3φ2n2ψn13φn1ψ2n2+ψ3n3)=15(f3n+f3n+3f3n3)+35×5(φ2nψn+φnψ2nφ2n+2ψn+1+φn+1ψ2n+2+φ2n2ψn1φn1ψ2n2)=15(f3n+f3n+1+f3n+2f3n3)+3φn1ψn15×5(φn+1ψ+φψn+1φn+3ψ2+φ2ψn+3+φn1ψn1)=15(f3n+f3n+f3n1+f3n+1+f3nf3n3)+3φn1ψn15×5(φn+1ψ+φψn+1φn+3ψ2+φ2ψn+3+φn1ψn1)Nowthisisthemomentwhereyouusethefactthatψ=1φX=15(3f3n+f3n1+f3n+f3n1f3n3)+3(1)n+155(φnψnφn+1+ψn1+φn1ψn1)=15(4f3n+f3n1+f3n2+f3n3f3n3)+3(1)n+15(fn1+fnfn+1)=15(4f3n+f3n1+f3n2)+0=15(4f3n+f3n)=f3n◻Thereyougo.

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