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Question Number 202328 by MATHEMATICSAM last updated on 24/Dec/23

If n ≥ 2 and U_n = (3 + (√5))^n + (3 − (√5))^n then prove that U_(n + 1) = 6U_n − 4U_(n − 1) .

Ifn⩾2andUn=(3+5)n+(3−5)nthenprovethatUn+1=6Un−4Un−1.

Commented by aleks041103 last updated on 24/Dec/23

It is true even for n=1

Itistrueevenforn=1

Answered by aleks041103 last updated on 24/Dec/23

U_(n+1) +4U_(n−1) −6U_n = =(3+(√5))^(n−1) ((3+(√5))^2 +4−6(3+(√5)))+ (3−(√5))^(n−1) ((3−(√5))^2 +4−6(3−(√5))) (3+(√5))^2 =14+6(√5) ⇒(14+6(√5))+4−18−6(√5)=0 (3−(√5))^2 =14−6(√5) (14−6(√5))+4−18+6(√5)=0 ⇒U_(n+1) −6U_n +4U_(n−1) = =(3+(√5))^(n−1) (0)+(3−(√5))^(n−1) (0)=0 ⇒U_(n+1) =6U_n −4U_(n−1)

Un+1+4Un−1−6Un==(3+5)n−1((3+5)2+4−6(3+5))+(3−5)n−1((3−5)2+4−6(3−5))(3+5)2=14+65⇒(14+65)+4−18−65=0(3−5)2=14−65(14−65)+4−18+65=0⇒Un+1−6Un+4Un−1==(3+5)n−1(0)+(3−5)n−1(0)=0⇒Un+1=6Un−4Un−1

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