Question Number 67148 by Cmr 237 last updated on 29/Aug/19
$$\mathrm{explicitez}\:\:\:\mathrm{la}\:\mathrm{suite}\:\mathrm{u}_{\mathrm{n}} \mathrm{definie}\:\mathrm{par}\:\mathrm{la}\:\mathrm{relation}; \\ $$$$\begin{cases}{\mathrm{u}_{\mathrm{0}} =\mathrm{0},\:\mathrm{u}_{\mathrm{1}} =\mathrm{1}}\\{\mathrm{u}_{\mathrm{n}+\mathrm{2}} =\mathrm{u}_{\mathrm{n}+\mathrm{1}} +\mathrm{u}_{\mathrm{n}} \:\:\:\forall\mathrm{n}\in\nmid\boldsymbol{\mathrm{N}}}\end{cases} \\ $$$$\boldsymbol{{u}}_{\boldsymbol{{n}}} =???????? \\ $$$$−\mathrm{calculer}\:\mathrm{la}\:\mathrm{lim}\underset{\mathrm{n}\rightarrow\infty} {\:}\frac{\mathrm{u}_{\mathrm{n}+\mathrm{1}} }{\mathrm{u}_{\mathrm{n}} }=??? \\ $$$$−\mathrm{montre}\:\mathrm{que}\:\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}\mathrm{u}_{\mathrm{k}} =\mathrm{u}_{\mathrm{n}+\mathrm{2}} −\mathrm{1} \\ $$$$\:\:\:\:\:\:\mathrm{voila}^{'} \\ $$
Commented by Tony Lin last updated on 23/Aug/19
![Fibonacci Sequence u_n =(1/( (√5)))[(((1+(√5))/2))^n −(((1−(√5))/2))^n ]](https://www.tinkutara.com/question/Q67150.png)
$${Fibonacci}\:{Sequence} \\ $$$${u}_{{n}} =\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left[\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} −\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} \right] \\ $$
Commented by mathmax by abdo last updated on 23/Aug/19
$${u}_{{n}+\mathrm{2}} ={u}_{{n}+\mathrm{1}} \:+{u}_{{n}} \:\Rightarrow{u}_{{n}+\mathrm{2}} −{u}_{{n}+\mathrm{1}} \:−{u}_{{n}} =\mathrm{0}\: \\ $$$${caracteristic}\:{equation}\:\rightarrow{x}^{\mathrm{2}} −{x}−\mathrm{1}=\mathrm{0} \\ $$$$\Delta\:=\mathrm{1}−\mathrm{4}\left(−\mathrm{1}\right)=\mathrm{5}\:\Rightarrow{x}_{\mathrm{1}} =\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:\:{and}\:{x}_{\mathrm{2}} =\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\Rightarrow\:{u}_{{n}} ={a}\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} \:+{b}\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} \\ $$$${u}_{\mathrm{0}} =\mathrm{0}\:={a}+{b} \\ $$$${u}_{\mathrm{1}} ={a}\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)+{b}\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\:=\mathrm{1}\:\Rightarrow\left(\mathrm{1}+\sqrt{\mathrm{5}}\right){a}+\left(\mathrm{1}−\sqrt{\mathrm{5}}\right){b}\:=\mathrm{2}\:{we}?{get} \\ $$$${the}\:{system}\:\begin{cases}{{a}+{b}=\mathrm{0}}\\{\left(\mathrm{1}+\sqrt{\mathrm{5}}\right){a}+\left(\mathrm{1}−\sqrt{\mathrm{5}}\right){b}\:=\mathrm{2}}\end{cases} \\ $$$$\Delta\:=\begin{vmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\\{\mathrm{1}+\sqrt{\mathrm{5}}\:\:\:\:\:\mathrm{1}−\sqrt{\mathrm{5}}}\end{vmatrix}\:\:=\mathrm{1}−\sqrt{\mathrm{5}}−\mathrm{1}−\sqrt{\mathrm{5}}=−\mathrm{2}\sqrt{\mathrm{5}}\:\Rightarrow \\ $$$${a}\:=\frac{\Delta_{{a}} }{\Delta}\:\:=\:\frac{\begin{vmatrix}{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\\{\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}−\sqrt{\mathrm{5}}}\end{vmatrix}}{−\mathrm{2}\sqrt{\mathrm{5}}}\:=\frac{−\mathrm{2}}{−\mathrm{2}\sqrt{\mathrm{5}}}\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}} \\ $$$${b}\:=\frac{\Delta_{{b}} }{\Delta}\:=\:\frac{\begin{vmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}}\\{\mathrm{1}+\sqrt{\mathrm{5}}\:\:\:\:\:\:\mathrm{2}}\end{vmatrix}\:\:}{−\mathrm{2}\sqrt{\mathrm{5}}}\:=\frac{\mathrm{2}}{−\mathrm{2}\sqrt{\mathrm{5}}}\:=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\:\:\Rightarrow \\ $$$${u}_{{n}} =\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} −\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} \: \\ $$
Commented by mathmax by abdo last updated on 23/Aug/19
$${we}\:{have}\:{u}_{{n}} ={u}_{{n}+\mathrm{2}} −{u}_{{n}+\mathrm{1}} \:\Rightarrow\sum_{{k}=\mathrm{0}} ^{{n}} \:{u}_{{k}} =\sum_{{k}=\mathrm{0}} ^{{n}} \left({u}_{{k}+\mathrm{2}} −{u}_{{k}+\mathrm{1}} \right) \\ $$$$={u}_{\mathrm{2}} −{u}_{\mathrm{1}} +{u}_{\mathrm{3}} −{u}_{\mathrm{2}} +….+{u}_{{n}+\mathrm{2}} −{u}_{{n}+\mathrm{1}} ={u}_{{n}+\mathrm{2}} −{u}_{\mathrm{1}} ={u}_{{n}+\mathrm{2}} −\mathrm{1} \\ $$
Commented by Kunal12588 last updated on 23/Aug/19
$${what}\:{are}\:\Delta_{{a}} \:\&\:\Delta_{{b}} \:{sir},\:{seems}\:{useful}. \\ $$$${can}\:{they}\:{help}\:{to}\:{solve}\:{for}\:\mathrm{3}\:{unknows} \\ $$
Commented by mathmax by abdo last updated on 23/Aug/19
$${yes}\:{its}\:{general}\:{for}\:{unknown}\:\left({x}_{{i}} \right)\:\:{this}\:{is}\:{the}\:{kramer}\:{method}\:{in} \\ $$$${case}\:{if}\:{the}\:{system}\:{have}\:{one}\:{solution}. \\ $$
Commented by mathmax by abdo last updated on 23/Aug/19
$${let}\:{q}=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:{and}\:{q}^{'} =\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\:\Rightarrow\sqrt{\mathrm{5}}{u}_{{n}} ={q}^{{n}} −{q}^{'{n}} ={q}^{{n}} \left\{\mathrm{1}−\left(\frac{{q}^{'} }{{q}}\right)^{{n}} \right\} \\ $$$$\frac{{q}^{'} }{{q}}\:=\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{1}+\sqrt{\mathrm{5}}}\:=\mathrm{1}−\frac{\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{1}+\sqrt{\mathrm{5}}}\:\Rightarrow\mid\frac{{q}^{'} }{{q}}\mid<\mathrm{1}\:\Rightarrow\sqrt{\mathrm{5}}{u}_{{n}} \sim{q}^{{n}} \:{and} \\ $$$$\sqrt{\mathrm{5}}{u}_{{n}+\mathrm{1}} \sim{q}^{{n}+\mathrm{1}} \:\Rightarrow\:\frac{{u}_{{n}+\mathrm{1}} }{{u}_{{n}} }\sim\:{q}\:\Rightarrow{lim}_{{n}\rightarrow+\infty} \frac{{u}_{{n}+\mathrm{1}} }{{u}_{{n}} }\:=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:\:\left({golden}\:{number}\:\varphi\right) \\ $$