explicitez la suite u_n definie par la relation; { ((u_0 =0, u_1 =1)),((u_(n+2) =u_(n+1) +u_n ∀n∈∤N)) :} u_n =???????? −calculer la lim _(n→∞) (u_(n+1) /u_n )=??? −montre que Σ_(k=0) ^n u_k =u


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Question Number 67148 by Cmr 237 last updated on 29/Aug/19
$explicitez la suite u_n definie par la relation; { ((u_0 =0, u_1 =1)),((u_(n+2) =u_(n+1) +u_n ∀n∈∤N)) :} u_n =???????? −calculer la lim _(n→∞) (u_(n+1) /u_n )=??? −montre que Σ_(k=0) ^n u_k =u_(n+2) −1 voila^′$
$explicitez la suite u_{n} definie par la relation;$ ${\begin{cases} u_{0} = 0, u_{1} = 1 \\ u_{n + 2} = u_{n + 1} + u_{n} \forall n \in∤ N \end{cases}$ $u_{n} = ? ? ? ? ? ? ? ?$ $- calculer la \lim \underset{n \to \infty}{} \frac{u_{n + 1}}{u_{n}} = ? ? ?$ $- montre que \underset{k = 0}{\sum^{n}} u_{k} = u_{n + 2} - 1$ ${voila}^{^{'}}$
Commented by Tony Lin last updated on 23/Aug/19

$F i b o n a c c i S e q u e n c e$ $u_{n} = \frac{1}{\sqrt{5}} [{(\frac{1 + \sqrt{5}}{2})}^{n} - {(\frac{1 - \sqrt{5}}{2})}^{n}]$
Commented by mathmax by abdo last updated on 23/Aug/19
$u_(n+2) =u_(n+1) +u_n ⇒u_(n+2) −u_(n+1) −u_n =0 caracteristic equation →x^2 −x−1=0 Δ =1−4(−1)=5 ⇒x_1 =((1+(√5))/2) and x_2 =((1−(√5))/2) ⇒ u_n =a(((1+(√5))/2))^n +b(((1−(√5))/2))^n u_0 =0 =a+b u_1 =a(((1+(√5))/2))+b(((1−(√5))/2)) =1 ⇒(1+(√5))a+(1−(√5))b =2 we?get the system { ((a+b=0)),(((1+(√5))a+(1−(√5))b =2)) :} Δ = determinant (((1 1)),((1+(√5) 1−(√5)))) =1−(√5)−1−(√5)=−2(√5) ⇒ a =(Δ_a /Δ) = ( determinant (((0 1)),((2 1−(√5))))/(−2(√5))) =((−2)/(−2(√5))) =(1/(√5)) b =(Δ_b /Δ) = (( determinant (((1 0)),((1+(√5) 2))) )/(−2(√5))) =(2/(−2(√5))) =−(1/(√5)) ⇒ u_n =(1/(√5))(((1+(√5))/2))^n −(1/(√5))(((1−(√5))/2))^n$
$u_{n + 2} = u_{n + 1} + u_{n} \Rightarrow u_{n + 2} - u_{n + 1} - u_{n} = 0$ $c a r a c t e r i s t i c e q u a t i o n \to x^{2} - x - 1 = 0$ $Δ = 1 - 4 (- 1) = 5 \Rightarrow x_{1} = \frac{1 + \sqrt{5}}{2} a n d x_{2} = \frac{1 - \sqrt{5}}{2}$ $\Rightarrow u_{n} = a {(\frac{1 + \sqrt{5}}{2})}^{n} + b {(\frac{1 - \sqrt{5}}{2})}^{n}$ $u_{0} = 0 = a + b$ $u_{1} = a (\frac{1 + \sqrt{5}}{2}) + b (\frac{1 - \sqrt{5}}{2}) = 1 \Rightarrow (1 + \sqrt{5}) a + (1 - \sqrt{5}) b = 2 w e ? g e t$ $t h e s y s t e m {\begin{cases} a + b = 0 \\ (1 + \sqrt{5}) a + (1 - \sqrt{5}) b = 2 \end{cases}$ $Δ = \| \begin{matrix} 1 1 \\ 1 + \sqrt{5} 1 - \sqrt{5} \end{matrix} \| = 1 - \sqrt{5} - 1 - \sqrt{5} = - 2 \sqrt{5} \Rightarrow$ $a = \frac{Δ_{a}}{Δ} = \frac{\| \begin{matrix} 0 1 \\ 2 1 - \sqrt{5} \end{matrix} \|}{- 2 \sqrt{5}} = \frac{- 2}{- 2 \sqrt{5}} = \frac{1}{\sqrt{5}}$ $b = \frac{Δ_{b}}{Δ} = \frac{\| \begin{matrix} 1 0 \\ 1 + \sqrt{5} 2 \end{matrix} \|}{- 2 \sqrt{5}} = \frac{2}{- 2 \sqrt{5}} = - \frac{1}{\sqrt{5}} \Rightarrow$ $u_{n} = \frac{1}{\sqrt{5}} {(\frac{1 + \sqrt{5}}{2})}^{n} - \frac{1}{\sqrt{5}} {(\frac{1 - \sqrt{5}}{2})}^{n}$
Commented by mathmax by abdo last updated on 23/Aug/19

$w e h a v e u_{n} = u_{n + 2} - u_{n + 1} \Rightarrow \sum_{k = 0}^{n} u_{k} = \sum_{k = 0}^{n} (u_{k + 2} - u_{k + 1})$ $= u_{2} - u_{1} + u_{3} - u_{2} + . . . . + u_{n + 2} - u_{n + 1} = u_{n + 2} - u_{1} = u_{n + 2} - 1$
Commented by Kunal12588 last updated on 23/Aug/19

$w h a t a r e Δ_{a} & Δ_{b} s i r, s e e m s u s e f u l .$ $c a n t h e y h e l p t o s o l v e f o r 3 u n k n o w s$
Commented by mathmax by abdo last updated on 23/Aug/19

$y e s i t s g e n e r a l f o r u n k n o w n (x_{i}) t h i s i s t h e k r a m e r m e t h o d i n$ $c a s e i f t h e s y s t e m h a v e o n e s o l u t i o n .$
Commented by mathmax by abdo last updated on 23/Aug/19
$let q=((1+(√5))/2) and q^′ =((1−(√5))/2) ⇒(√5)u_n =q^n −q^(′n) =q^n {1−((q^′ /q))^n } (q^′ /q) =((1−(√5))/(1+(√5))) =1−((2(√5))/(1+(√5))) ⇒∣(q^′ /q)∣<1 ⇒(√5)u_n ∼q^n and (√5)u_(n+1) ∼q^(n+1) ⇒ (u_(n+1) /u_n )∼ q ⇒lim_(n→+∞) (u_(n+1) /u_n ) =((1+(√5))/2) (golden number ϕ)$
$l e t q = \frac{1 + \sqrt{5}}{2} a n d q^{^{'}} = \frac{1 - \sqrt{5}}{2} \Rightarrow \sqrt{5} u_{n} = q^{n} - q^{^{'} n} = q^{n} {1 - {(\frac{q^{^{'}}}{q})}^{n}}$ $\frac{q^{^{'}}}{q} = \frac{1 - \sqrt{5}}{1 + \sqrt{5}} = 1 - \frac{2 \sqrt{5}}{1 + \sqrt{5}} \Rightarrow∣ \frac{q^{^{'}}}{q} ∣< 1 \Rightarrow \sqrt{5} u_{n} \sim q^{n} a n d$ $\sqrt{5} u_{n + 1} \sim q^{n + 1} \Rightarrow \frac{u_{n + 1}}{u_{n}} \sim q \Rightarrow {l i m}_{n \to + \infty} \frac{u_{n + 1}}{u_{n}} = \frac{1 + \sqrt{5}}{2} (g o l d e n n u m b e r φ)$
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