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

Commented by TITA last updated on 24/Dec/20

$pleaae help$
	Answered by physicstutes last updated on 24/Dec/20
	$Sir please type your questions next time they are really bad for the eyes. Solution Fibonacci sequence = { ((F_(n+1) = F_n + F_(n−1) )),((F_1 = F_2 = 1,)) :} ∀ n ∈ N , n≥ 2 (a) F_3 = F_2 + F_1 = 1 + 1 = 2 F_4 = F_3 + F_2 = 2 + 1 = 3 F_5 = F_4 + F_3 = 3 + 2 = 5 now gcd (2,3,5) = 1 ⇒ F_3 ,F_4 and F_5 are relatively prime. (b) we have proven for n = 3 and n = 4 ⇒ n = k gcd (F_k , F_(k+1) ) = 1 prove for n=k+1 ⇒ gcd (F_(k+1) , F_(k+2) ) = 1 since gcd(F_(k+1) ,F_(n+1) +F_1 ) = 1.$
	$Sir please type your questions next time$ $they are really bad for the eyes .$ $Solution$ $Fibonacci sequence = {\begin{cases} F_{n + 1} = F_{n} + F_{n - 1} \\ F_{1} = F_{2} = 1, \end{cases} \forall n \in N, n ⩾ 2$ $(a) F_{3} = F_{2} + F_{1} = 1 + 1 = 2$ $F_{4} = F_{3} + F_{2} = 2 + 1 = 3$ $F_{5} = F_{4} + F_{3} = 3 + 2 = 5$ $now \gcd (2, 3, 5) = 1 \Rightarrow F_{3}, F_{4} and F_{5} are relatively prime .$ $(b) we have proven for n = 3 and n = 4$ $\Rightarrow n = k$ $\gcd (F_{k}, F_{k + 1}) = 1$ $prove for n = k + 1 \Rightarrow \gcd (F_{k + 1}, F_{k + 2}) = 1$ $since \gcd (F_{k + 1}, F_{n + 1} + F_{1}) = 1 .$
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