The first two terms of the {a_n } series are defind as a_n =a_(n−1) +a_(n−2) for the general term a_1 =5, a_2 =8 and n≥3 . since the L=lim_(n→∞) (a_(n+1) /a


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Question Number 143519 by tugu last updated on 15/Jun/21
$The first two terms of the {a_n } series are defind as a_n =a_(n−1) +a_(n−2) for the general term a_1 =5, a_2 =8 and n≥3 . since the L=lim_(n→∞) (a_(n+1) /a_n ) what is the value of L$
$T h e f i r s t t w o t e r m s o f t h e {a_{n}} s e r i e s a r e d e f i n d a s a_{n} = a_{n - 1} + a_{n - 2} f o r t h e g e n e r a l t e r m$ $a_{1} = 5, a_{2} = 8 a n d n ⩾ 3 .$ $s i n c e t h e L = l i \underset{n \to \infty}{m} \frac{a_{n + 1}}{a_{n}} w h a t i s t h e v a l u e o f L$
	Answered by mathmax by abdo last updated on 15/Jun/21
	$a_n =a_(n−1) +a_(n−2) ⇒a_(n+2) =a_(n+1) +a_n ⇒a_(n+2) −a_(n+1) −a_n =0 (ec)→x^2 −x−1=0 →Δ=1+4=5 ⇒x_1 =((1+(√5))/2) and x_2 =((1−(√5))/2) ⇒a_n =α(((1+(√5))/2))^n +β(((1−(√5))/2))^n a_(1 ) =5 ⇒α(((1+(√5))/2))+β(((1−(√5))/2))=5 a_2 =8 ⇒α(((1+(√5))/2))^2 +β(((1−(√5))/2))^2 =8 we get the system { ((((1+(√5))/2)α +((1−(√5))/2)β=5)),(((((1+(√5))/2))^2 α +(((1−(√5))/2))^2 β=8)) :} Δ_s =(((1+(√5))(1−(√5))^2 )/8)−(((1−(√5))(1+(√5))^2 )/8) =−(1/2)(1−(√5))+(1/2)(1+(√5)) =(1/2)(1+(√5)−1+(√5))=(√5) Δ_α = determinant (((5 ((1−(√5))/2))),((8 (((1−(√5))/2))^2 )))=(5/4)(1−(√5))^2 −4(1−(√5)) =(5/4)(6−2(√5))−4+4(√5)=(5/2)(3−(√5))−4+4(√5)=((15)/2)−4−((5(√5))/2)+4(√5) =(7/2)+(3/2)(√(5 )) ⇒α=((7+3(√5))/(2(√5))) Δ_β = determinant (((((1+(√5))/2) 5)),(((((1+(√5))/2))^2 8)))=4(1+(√5))−(5/4)(1+(√5))^2 =4+4(√5)−(5/4)(6+2(√5)) =4+4(√5)−(5/2)(3+(√5))=4+4(√5)−((15)/2)−((5(√5))/2) =−(7/2) +(3/2)(√5) ⇒β =((−7+3(√5))/(2(√5))) ⇒ a_n =(((7+3(√5))/(2(√5))))(((1+(√5))/2))^n +(((−7+3(√5))/(2(√5))))(((1−(√5))/2))^n ⇒a_n =(((7+3(√5))(1+(√5))^n )/(2^(n+1) ((√5)))) +(((−7+3(√5))(1−(√5))^n )/(2^(n+1) ((√5)))) =(1/(((√5))2^(n+1) )){ (7+3(√5))(1+(√5))^n −(7−3(√5))(1−(√5))^n } rest to calculate (a_(n+1) /a_n ) ....be continued...$
	$a_{n} = a_{n - 1} + a_{n - 2} \Rightarrow a_{n + 2} = a_{n + 1} + a_{n} \Rightarrow a_{n + 2} - a_{n + 1} - a_{n} = 0$ $(ec) \to x^{2} - x - 1 = 0 \to Δ = 1 + 4 = 5 \Rightarrow x_{1} = \frac{1 + \sqrt{5}}{2} and x_{2} = \frac{1 - \sqrt{5}}{2}$ $\Rightarrow a_{n} = α {(\frac{1 + \sqrt{5}}{2})}^{n} + β {(\frac{1 - \sqrt{5}}{2})}^{n}$ $a_{1} = 5 \Rightarrow α (\frac{1 + \sqrt{5}}{2}) + β (\frac{1 - \sqrt{5}}{2}) = 5$ $a_{2} = 8 \Rightarrow α {(\frac{1 + \sqrt{5}}{2})}^{2} + β {(\frac{1 - \sqrt{5}}{2})}^{2} = 8 we get the system$ ${\begin{cases} \frac{1 + \sqrt{5}}{2} α + \frac{1 - \sqrt{5}}{2} β = 5 \\ {(\frac{1 + \sqrt{5}}{2})}^{2} α + {(\frac{1 - \sqrt{5}}{2})}^{2} β = 8 \end{cases}$ $Δ_{s} = \frac{(1 + \sqrt{5}) {(1 - \sqrt{5})}^{2}}{8} - \frac{(1 - \sqrt{5}) {(1 + \sqrt{5})}^{2}}{8}$ $= - \frac{1}{2} (1 - \sqrt{5}) + \frac{1}{2} (1 + \sqrt{5}) = \frac{1}{2} (1 + \sqrt{5} - 1 + \sqrt{5}) = \sqrt{5}$ $Δ_{α} = \| \begin{matrix} 5 \frac{1 - \sqrt{5}}{2} \\ 8 {(\frac{1 - \sqrt{5}}{2})}^{2} \end{matrix} \| = \frac{5}{4} {(1 - \sqrt{5})}^{2} - 4 (1 - \sqrt{5})$ $= \frac{5}{4} (6 - 2 \sqrt{5}) - 4 + 4 \sqrt{5} = \frac{5}{2} (3 - \sqrt{5}) - 4 + 4 \sqrt{5} = \frac{15}{2} - 4 - \frac{5 \sqrt{5}}{2} + 4 \sqrt{5}$ $= \frac{7}{2} + \frac{3}{2} \sqrt{5} \Rightarrow α = \frac{7 + 3 \sqrt{5}}{2 \sqrt{5}}$ $Δ_{β} = \| \begin{matrix} \frac{1 + \sqrt{5}}{2} 5 \\ {(\frac{1 + \sqrt{5}}{2})}^{2} 8 \end{matrix} \| = 4 (1 + \sqrt{5}) - \frac{5}{4} {(1 + \sqrt{5})}^{2}$ $= 4 + 4 \sqrt{5} - \frac{5}{4} (6 + 2 \sqrt{5}) = 4 + 4 \sqrt{5} - \frac{5}{2} (3 + \sqrt{5}) = 4 + 4 \sqrt{5} - \frac{15}{2} - \frac{5 \sqrt{5}}{2}$ $= - \frac{7}{2} + \frac{3}{2} \sqrt{5} \Rightarrow β = \frac{- 7 + 3 \sqrt{5}}{2 \sqrt{5}} \Rightarrow$ $a_{n} = (\frac{7 + 3 \sqrt{5}}{2 \sqrt{5}}) {(\frac{1 + \sqrt{5}}{2})}^{n} + (\frac{- 7 + 3 \sqrt{5}}{2 \sqrt{5}}) {(\frac{1 - \sqrt{5}}{2})}^{n}$ $\Rightarrow a_{n} = \frac{(7 + 3 \sqrt{5}) {(1 + \sqrt{5})}^{n}}{2^{n + 1} (\sqrt{5})} + \frac{(- 7 + 3 \sqrt{5}) {(1 - \sqrt{5})}^{n}}{2^{n + 1} (\sqrt{5})}$ $= \frac{1}{(\sqrt{5}) 2^{n + 1}} {(7 + 3 \sqrt{5}) {(1 + \sqrt{5})}^{n} - (7 - 3 \sqrt{5}) {(1 - \sqrt{5})}^{n}}$ $rest to calculate \frac{a_{n + 1}}{a_{n}} . . . . be continued . . .$
	Answered by mr W last updated on 15/Jun/21

	$x^{2} - x - 1 = 0$ $x = \frac{1 \pm \sqrt{5}}{2}$ $a_{n} = A {(\frac{1 + \sqrt{5}}{2})}^{n} + B {(\frac{1 - \sqrt{5}}{2})}^{n}$ $a_{0} = a_{2} - a_{1} = 8 - 5 = 3 = A + B . . . (i)$ $a_{1} = 5 = A (\frac{1 + \sqrt{5}}{2}) + B (\frac{1 - \sqrt{5}}{2}) . . . (i i)$ $\Rightarrow A = \frac{15 + 7 \sqrt{5}}{10}$ $\Rightarrow B = \frac{15 - 7 \sqrt{5}}{10}$ $a_{n} = (\frac{15 + 7 \sqrt{5}}{10}) {(\frac{1 + \sqrt{5}}{2})}^{n} + (\frac{15 - 7 \sqrt{5}}{10}) {(\frac{1 - \sqrt{5}}{2})}^{n}$ $a_{n} = {(\frac{1 + \sqrt{5}}{2})}^{n} [(\frac{15 + 7 \sqrt{5}}{10}) + (\frac{15 - 7 \sqrt{5}}{10}) {(\frac{1 - \sqrt{5}}{1 + \sqrt{5}})}^{n}]$ $\frac{a_{n + 1}}{a_{n}} = (\frac{1 + \sqrt{5}}{2}) \times \frac{(\frac{15 + 7 \sqrt{5}}{10}) + (\frac{15 - 7 \sqrt{5}}{10}) {(\frac{1 - \sqrt{5}}{1 + \sqrt{5}})}^{n + 1}}{(\frac{15 + 7 \sqrt{5}}{10}) + (\frac{15 - 7 \sqrt{5}}{10}) {(\frac{1 - \sqrt{5}}{1 + \sqrt{5}})}^{n}}$ $\lim_{n \to \infty} \frac{a_{n + 1}}{a_{n}} = (\frac{1 + \sqrt{5}}{2}) \times \frac{(\frac{15 + 7 \sqrt{5}}{10}) + (\frac{15 - 7 \sqrt{5}}{10}) \times 0}{(\frac{15 + 7 \sqrt{5}}{10}) + (\frac{15 - 7 \sqrt{5}}{10}) \times 0} = \frac{1 + \sqrt{5}}{2} = φ$
	Answered by mr W last updated on 15/Jun/21

	$a_{n + 1} = a_{n} + a_{n - 1}$ $\frac{a_{n + 1}}{a_{n}} = 1 + \frac{1}{\frac{a_{n}}{a_{n - 1}}}$ $\lim_{n \to \infty} \frac{a_{n + 1}}{a_{n}} = 1 + \frac{1}{\lim_{n \to \infty} \frac{a_{n}}{a_{n - 1}}}$ $L = 1 + \frac{1}{L} > 1$ $L^{2} - L - 1 = 0$ $\Rightarrow L = \frac{1 + \sqrt{5}}{2}$
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