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Question Number 189464 by Rupesh123 last updated on 17/Mar/23

Answered by Rasheed.Sindhi last updated on 17/Mar/23

4a_(n+1) ^2 −4a_n a_(n+1) +a_n ^2 −1=0 n=1: 4a_2 ^2 −4a_1 a_2 +a_1 ^2 −1=0 a_2 =((4a_1 ±(√(16−4(4)(a_1 ^2 −1))) )/(2(4))) =((4a_1 ±4(√(2−a_1 ^2 )) )/(2(4))) =((a_1 ±(√(2−a_1 ^2 )) )/2)∈Z⇒a_1 =±1

4an+12−4anan+1+an2−1=0n=1:4a22−4a1a2+a12−1=0a2=4a1±16−4(4)(a12−1)2(4)=4a1±42−a122(4)=a1±2−a122∈Z⇒a1=±1

Commented by Rupesh123 last updated on 18/Mar/23

Excellent!

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