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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^′

$$\mathrm{explicitez}\:\:\:\mathrm{la}\:\mathrm{suite}\:\mathrm{u}_{\mathrm{n}} \mathrm{definie}\:\mathrm{par}\:\mathrm{la}\:\mathrm{relation}; \\ $$$$\begin{cases}{\mathrm{u}_{\mathrm{0}} =\mathrm{0},\:\mathrm{u}_{\mathrm{1}} =\mathrm{1}}\\{\mathrm{u}_{\mathrm{n}+\mathrm{2}} =\mathrm{u}_{\mathrm{n}+\mathrm{1}} +\mathrm{u}_{\mathrm{n}} \:\:\:\forall\mathrm{n}\in\nmid\boldsymbol{\mathrm{N}}}\end{cases} \\ $$$$\boldsymbol{{u}}_{\boldsymbol{{n}}} =???????? \\ $$$$−\mathrm{calculer}\:\mathrm{la}\:\mathrm{lim}\underset{\mathrm{n}\rightarrow\infty} {\:}\frac{\mathrm{u}_{\mathrm{n}+\mathrm{1}} }{\mathrm{u}_{\mathrm{n}} }=??? \\ $$$$−\mathrm{montre}\:\mathrm{que}\:\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}\mathrm{u}_{\mathrm{k}} =\mathrm{u}_{\mathrm{n}+\mathrm{2}} −\mathrm{1} \\ $$$$\:\:\:\:\:\:\mathrm{voila}^{'} \\ $$

Commented by Tony Lin last updated on 23/Aug/19

Fibonacci Sequence u_n =(1/(√5))[(((1+(√5))/2))^n −(((1−(√5))/2))^n ]

$${Fibonacci}\:{Sequence} \\ $$$${u}_{{n}} =\frac{\mathrm{1}}{\sqrt{\mathrm{5}}}\left[\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} −\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} \right] \\ $$

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

$${u}_{{n}+\mathrm{2}} ={u}_{{n}+\mathrm{1}} \:+{u}_{{n}} \:\Rightarrow{u}_{{n}+\mathrm{2}} −{u}_{{n}+\mathrm{1}} \:−{u}_{{n}} =\mathrm{0}\: \\ $$$${caracteristic}\:{equation}\:\rightarrow{x}^{\mathrm{2}} −{x}−\mathrm{1}=\mathrm{0} \\ $$$$\Delta\:=\mathrm{1}−\mathrm{4}\left(−\mathrm{1}\right)=\mathrm{5}\:\Rightarrow{x}_{\mathrm{1}} =\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:\:{and}\:{x}_{\mathrm{2}} =\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\Rightarrow\:{u}_{{n}} ={a}\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} \:+{b}\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} \\ $$$${u}_{\mathrm{0}} =\mathrm{0}\:={a}+{b} \\ $$$${u}_{\mathrm{1}} ={a}\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)+{b}\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\:=\mathrm{1}\:\Rightarrow\left(\mathrm{1}+\sqrt{\mathrm{5}}\right){a}+\left(\mathrm{1}−\sqrt{\mathrm{5}}\right){b}\:=\mathrm{2}\:{we}?{get} \\ $$$${the}\:{system}\:\begin{cases}{{a}+{b}=\mathrm{0}}\\{\left(\mathrm{1}+\sqrt{\mathrm{5}}\right){a}+\left(\mathrm{1}−\sqrt{\mathrm{5}}\right){b}\:=\mathrm{2}}\end{cases} \\ $$$$\Delta\:=\begin{vmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\\{\mathrm{1}+\sqrt{\mathrm{5}}\:\:\:\:\:\mathrm{1}−\sqrt{\mathrm{5}}}\end{vmatrix}\:\:=\mathrm{1}−\sqrt{\mathrm{5}}−\mathrm{1}−\sqrt{\mathrm{5}}=−\mathrm{2}\sqrt{\mathrm{5}}\:\Rightarrow \\ $$$${a}\:=\frac{\Delta_{{a}} }{\Delta}\:\:=\:\frac{\begin{vmatrix}{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\\{\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}−\sqrt{\mathrm{5}}}\end{vmatrix}}{−\mathrm{2}\sqrt{\mathrm{5}}}\:=\frac{−\mathrm{2}}{−\mathrm{2}\sqrt{\mathrm{5}}}\:=\frac{\mathrm{1}}{\sqrt{\mathrm{5}}} \\ $$$${b}\:=\frac{\Delta_{{b}} }{\Delta}\:=\:\frac{\begin{vmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}}\\{\mathrm{1}+\sqrt{\mathrm{5}}\:\:\:\:\:\:\mathrm{2}}\end{vmatrix}\:\:}{−\mathrm{2}\sqrt{\mathrm{5}}}\:=\frac{\mathrm{2}}{−\mathrm{2}\sqrt{\mathrm{5}}}\:=−\frac{\mathrm{1}}{\sqrt{\mathrm{5}}}\:\:\Rightarrow \\ $$$${u}_{{n}} =\frac{\mathrm{1}}{\sqrt{\mathrm{5}}}\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} −\frac{\mathrm{1}}{\sqrt{\mathrm{5}}}\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} \: \\ $$

Commented by mathmax by abdo last updated on 23/Aug/19

we have u_n =u_(n+2) −u_(n+1) ⇒Σ_(k=0) ^n u_k =Σ_(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

$${we}\:{have}\:{u}_{{n}} ={u}_{{n}+\mathrm{2}} −{u}_{{n}+\mathrm{1}} \:\Rightarrow\sum_{{k}=\mathrm{0}} ^{{n}} \:{u}_{{k}} =\sum_{{k}=\mathrm{0}} ^{{n}} \left({u}_{{k}+\mathrm{2}} −{u}_{{k}+\mathrm{1}} \right) \\ $$$$={u}_{\mathrm{2}} −{u}_{\mathrm{1}} +{u}_{\mathrm{3}} −{u}_{\mathrm{2}} +....+{u}_{{n}+\mathrm{2}} −{u}_{{n}+\mathrm{1}} ={u}_{{n}+\mathrm{2}} −{u}_{\mathrm{1}} ={u}_{{n}+\mathrm{2}} −\mathrm{1} \\ $$

Commented by Kunal12588 last updated on 23/Aug/19

what are Δ_a & Δ_b sir, seems useful. can they help to solve for 3 unknows

$${what}\:{are}\:\Delta_{{a}} \:\&\:\Delta_{{b}} \:{sir},\:{seems}\:{useful}. \\ $$$${can}\:{they}\:{help}\:{to}\:{solve}\:{for}\:\mathrm{3}\:{unknows} \\ $$

Commented by mathmax by abdo last updated on 23/Aug/19

yes its general for unknown (x_i ) this is the kramer method in case if the system have one solution.

$${yes}\:{its}\:{general}\:{for}\:{unknown}\:\left({x}_{{i}} \right)\:\:{this}\:{is}\:{the}\:{kramer}\:{method}\:{in} \\ $$$${case}\:{if}\:{the}\:{system}\:{have}\:{one}\:{solution}. \\ $$

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)

$${let}\:{q}=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:{and}\:{q}^{'} =\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\:\Rightarrow\sqrt{\mathrm{5}}{u}_{{n}} ={q}^{{n}} −{q}^{'{n}} ={q}^{{n}} \left\{\mathrm{1}−\left(\frac{{q}^{'} }{{q}}\right)^{{n}} \right\} \\ $$$$\frac{{q}^{'} }{{q}}\:=\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{1}+\sqrt{\mathrm{5}}}\:=\mathrm{1}−\frac{\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{1}+\sqrt{\mathrm{5}}}\:\Rightarrow\mid\frac{{q}^{'} }{{q}}\mid<\mathrm{1}\:\Rightarrow\sqrt{\mathrm{5}}{u}_{{n}} \sim{q}^{{n}} \:{and} \\ $$$$\sqrt{\mathrm{5}}{u}_{{n}+\mathrm{1}} \sim{q}^{{n}+\mathrm{1}} \:\Rightarrow\:\frac{{u}_{{n}+\mathrm{1}} }{{u}_{{n}} }\sim\:{q}\:\Rightarrow{lim}_{{n}\rightarrow+\infty} \frac{{u}_{{n}+\mathrm{1}} }{{u}_{{n}} }\:=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:\:\left({golden}\:{number}\:\varphi\right) \\ $$

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