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Question-189464




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
$$\mathrm{4}{a}_{{n}+\mathrm{1}} ^{\mathrm{2}} −\mathrm{4}{a}_{{n}} {a}_{{n}+\mathrm{1}} +{a}_{{n}} ^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$${n}=\mathrm{1}: \\ $$$$\mathrm{4}{a}_{\mathrm{2}} ^{\mathrm{2}} −\mathrm{4}{a}_{\mathrm{1}} {a}_{\mathrm{2}} +{a}_{\mathrm{1}} ^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$${a}_{\mathrm{2}} =\frac{\mathrm{4}{a}_{\mathrm{1}} \pm\sqrt{\mathrm{16}−\mathrm{4}\left(\mathrm{4}\right)\left({a}_{\mathrm{1}} ^{\mathrm{2}} −\mathrm{1}\right)}\:}{\mathrm{2}\left(\mathrm{4}\right)} \\ $$$$\:\:\:\:\:=\frac{\mathrm{4}{a}_{\mathrm{1}} \pm\mathrm{4}\sqrt{\mathrm{2}−{a}_{\mathrm{1}} ^{\mathrm{2}} }\:}{\mathrm{2}\left(\mathrm{4}\right)} \\ $$$$\:\:\:\:\:=\frac{{a}_{\mathrm{1}} \pm\sqrt{\mathrm{2}−{a}_{\mathrm{1}} ^{\mathrm{2}} }\:}{\mathrm{2}}\in\mathbb{Z}\Rightarrow{a}_{\mathrm{1}} =\pm\mathrm{1} \\ $$
Commented by Rupesh123 last updated on 18/Mar/23
Excellent!

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