Question Number 194579 by MM42 last updated on 10/Jul/23
![if u_n =(1/( (√5)))[(((1+(√5))/2))^n −(((1−(√5))/2))^n ] then u_(n+1) =u_n +u_(n−1) ? ; n=0,1,2,..](https://www.tinkutara.com/question/Q194579.png)
$${if}\:\:\:\:{u}_{{n}} =\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left[\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} −\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} \right] \\ $$$$\:{then}\:\:\:{u}_{{n}+\mathrm{1}} ={u}_{{n}} +{u}_{{n}−\mathrm{1}} \:\:\:?\:\:\:\:\:;\:\:\:{n}=\mathrm{0},\mathrm{1},\mathrm{2},.. \\ $$
Answered by Frix last updated on 10/Jul/23
$${n}=\left\{\mathrm{0},\:\mathrm{1},\:\mathrm{2},\:\mathrm{3},\:\mathrm{4},\:\mathrm{5},\:\mathrm{6},\:\mathrm{7},\:…\right\} \\ $$$${u}_{{n}} =\left\{\mathrm{0},\:\mathrm{1},\:\mathrm{1},\:\mathrm{2},\:\mathrm{3},\:\mathrm{5},\:\mathrm{8},\:\mathrm{13},\:…\right\} \\ $$
Commented by MM42 last updated on 10/Jul/23
$${fibonacci}\:\:{sequence} \\ $$$${please}\:\:{prove}… \\ $$
Answered by MM42 last updated on 13/Jul/23
![prove u_n +u_(n−1) =(1/( (√5)))[(((1+(√5))/2))^n −(((1−(√5))/2))^n +(((1+(√5))/2))^(n−1) −(((1−(√5))/2))^(n−1) ] =(1/( (√5)))[(((1+(√5))/2))^(n−1) (((1+(√5))/2)+1)−(((1−(√5))/2))^(n−1) (((1−(√5))/2)+1)] =(1/( (√5)))[(((1+(√5))/2))^(n−1) (((3+(√5))/2))−(((1−(√5))/2))^(n−1) (((3−(√5))/2))] =(1/( (√5)))[(((1+(√5))/2))^n −(((1−(√5))/2))^n ]=u_(n+1) metod 2 let a=((1+(√5))/2) & b=((1−(√5))/2) ⇒u_n =(1/( (√5)))(a^n −b^n ) a+b=1 & ab=−1 ⇒ x^2 +x−1=0 ; a,b , are the roots a+b=1⇒a^(n+1) +a^n b=a^n ⇒a^(n+1) −a^(n−1) =a^n (i) b^(n+1) +ab^n =b^n ⇒b^(n+1) −b^(n−1) =b^n (ii) (i)+(ii)⇒a^(n+1) −b^(n+1) =(a^n −b^n )+(a^(n−1) −b^(n−1) ) ⇒u_(n+1) =u_m +u_(n−1)](https://www.tinkutara.com/question/Q194664.png)
$${prove} \\ $$$${u}_{{n}} +{u}_{{n}−\mathrm{1}} =\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left[\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} −\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} +\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}−\mathrm{1}} −\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}−\mathrm{1}} \right] \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left[\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}−\mathrm{1}} \left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}+\mathrm{1}\right)−\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}−\mathrm{1}} \left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}+\mathrm{1}\right)\right] \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left[\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}−\mathrm{1}} \left(\frac{\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)−\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}−\mathrm{1}} \left(\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\right] \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left[\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} −\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} \right]={u}_{{n}+\mathrm{1}} \\ $$$${metod}\:\mathrm{2} \\ $$$${let}\:\:{a}=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:\:\&\:\:{b}=\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\:\:\Rightarrow{u}_{{n}} =\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left({a}^{{n}} −{b}^{{n}} \right) \\ $$$${a}+{b}=\mathrm{1}\:\:\&\:\:{ab}=−\mathrm{1}\:\:\:\Rightarrow\:{x}^{\mathrm{2}} +{x}−\mathrm{1}=\mathrm{0}\:\:;\:{a},{b}\:,\:{are}\:{the}\:{roots}\: \\ $$$${a}+{b}=\mathrm{1}\Rightarrow{a}^{{n}+\mathrm{1}} +{a}^{{n}} {b}={a}^{{n}} \:\Rightarrow{a}^{{n}+\mathrm{1}} −{a}^{{n}−\mathrm{1}} ={a}^{{n}} \:\:\left({i}\right) \\ $$$$\:{b}^{{n}+\mathrm{1}} +{ab}^{{n}} ={b}^{{n}} \:\Rightarrow{b}^{{n}+\mathrm{1}} −{b}^{{n}−\mathrm{1}} ={b}^{{n}} \:\:\left({ii}\right) \\ $$$$\left({i}\right)+\left({ii}\right)\Rightarrow{a}^{{n}+\mathrm{1}} −{b}^{{n}+\mathrm{1}} =\left({a}^{{n}} −{b}^{{n}} \right)+\left({a}^{{n}−\mathrm{1}} −{b}^{{n}−\mathrm{1}} \right) \\ $$$$\Rightarrow{u}_{{n}+\mathrm{1}} ={u}_{{m}} +{u}_{{n}−\mathrm{1}} \\ $$$$ \\ $$