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Question Number 126780 by TITA last updated on 24/Dec/20

Commented by TITA last updated on 24/Dec/20

pleaae help

$$\mathrm{pleaae}\:\mathrm{help} \\ $$

Answered by physicstutes last updated on 24/Dec/20

Sir please type your questions next time they are really bad for the eyes. Solution Fibonacci sequence = { ((F_(n+1) = F_n + F_(n−1) )),((F_1 = F_2 = 1,)) :} ∀ n ∈ N , n≥ 2 (a) F_3 = F_2 + F_1 = 1 + 1 = 2 F_4 = F_3 + F_2 = 2 + 1 = 3 F_5 = F_4 + F_3 = 3 + 2 = 5 now gcd (2,3,5) = 1 ⇒ F_3 ,F_4 and F_5 are relatively prime. (b) we have proven for n = 3 and n = 4 ⇒ n = k gcd (F_k , F_(k+1) ) = 1 prove for n=k+1 ⇒ gcd (F_(k+1) , F_(k+2) ) = 1 since gcd(F_(k+1) ,F_(n+1) +F_1 ) = 1.

$$\mathrm{Sir}\:\mathrm{please}\:\mathrm{type}\:\mathrm{your}\:\mathrm{questions}\:\mathrm{next}\:\mathrm{time} \\ $$$$\mathrm{they}\:\mathrm{are}\:\mathrm{really}\:\mathrm{bad}\:\mathrm{for}\:\mathrm{the}\:\mathrm{eyes}. \\ $$$$\:\:\boldsymbol{\mathrm{Solution}} \\ $$$$\:\mathrm{Fibonacci}\:\mathrm{sequence}\:=\:\begin{cases}{{F}_{{n}+\mathrm{1}} \:=\:{F}_{{n}} \:+\:{F}_{{n}−\mathrm{1}} }\\{{F}_{\mathrm{1}} \:=\:{F}_{\mathrm{2}} \:=\:\mathrm{1},}\end{cases}\:\forall\:{n}\:\in\:\mathbb{N}\:,\:{n}\geqslant\:\mathrm{2} \\ $$$$\left(\mathrm{a}\right)\:{F}_{\mathrm{3}} \:=\:{F}_{\mathrm{2}} +\:{F}_{\mathrm{1}} \:=\:\mathrm{1}\:+\:\mathrm{1}\:=\:\mathrm{2} \\ $$$$\:\:\:\:\:\:\:{F}_{\mathrm{4}} \:=\:{F}_{\mathrm{3}} \:+\:{F}_{\mathrm{2}} \:=\:\mathrm{2}\:+\:\mathrm{1}\:=\:\mathrm{3} \\ $$$$\:\:\:\:\:\:\:{F}_{\mathrm{5}} \:=\:{F}_{\mathrm{4}} \:+\:{F}_{\mathrm{3}} \:=\:\mathrm{3}\:+\:\mathrm{2}\:=\:\mathrm{5} \\ $$$$\mathrm{now}\:\mathrm{gcd}\:\left(\mathrm{2},\mathrm{3},\mathrm{5}\right)\:=\:\mathrm{1}\:\Rightarrow\:\:{F}_{\mathrm{3}} ,{F}_{\mathrm{4}} \mathrm{and}\:{F}_{\mathrm{5}} \:\mathrm{are}\:\mathrm{relatively}\:\mathrm{prime}. \\ $$$$\:\left(\mathrm{b}\right)\:\mathrm{we}\:\mathrm{have}\:\mathrm{proven}\:\mathrm{for}\:{n}\:=\:\mathrm{3}\:\mathrm{and}\:{n}\:=\:\mathrm{4} \\ $$$$\Rightarrow\:{n}\:=\:{k} \\ $$$$\:\:\mathrm{gcd}\:\left({F}_{{k}} ,\:{F}_{{k}+\mathrm{1}} \right)\:=\:\mathrm{1} \\ $$$$\mathrm{prove}\:\mathrm{for}\:{n}={k}+\mathrm{1}\:\Rightarrow\:\mathrm{gcd}\:\left({F}_{{k}+\mathrm{1}} ,\:{F}_{{k}+\mathrm{2}} \right)\:=\:\mathrm{1} \\ $$$$\mathrm{since}\:\mathrm{gcd}\left({F}_{{k}+\mathrm{1}} ,{F}_{{n}+\mathrm{1}} +{F}_{\mathrm{1}} \right)\:=\:\mathrm{1}. \\ $$

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