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Question Number 121832 by bemath last updated on 12/Nov/20

Let { ((u_1 =1)),((u_2 =1)) :} and u_(n+2) = u_(n+1) + u_n find u_n .

$$\:\:{Let}\:\begin{cases}{{u}_{\mathrm{1}} =\mathrm{1}}\\{{u}_{\mathrm{2}} =\mathrm{1}}\end{cases}\:{and}\:{u}_{{n}+\mathrm{2}} \:=\:{u}_{{n}+\mathrm{1}} \:+\:{u}_{{n}} \\ $$$${find}\:{u}_{{n}} . \\ $$

Commented by bemath last updated on 12/Nov/20

thank you both

$${thank}\:{you}\:{both} \\ $$

Answered by mr W last updated on 12/Nov/20

q^2 −q−1=0 q=((1±(√5))/2) u_n =A(((1+(√5))/2))^n +B(((1−(√5))/2))^n u_0 =A+B=u_2 −u_1 =1−1=0 ⇒A=−B u_1 =A(((1+(√5))/2))+B(((1−(√5))/2))=1 A[((1+(√5))/2)−((1−(√5))/2)]=1 ⇒A=(1/( (√5))) ⇒B=−(1/( (√5))) ⇒u_n =(1/( (√5)))[(((1+(√5))/2))^n −(((1−(√5))/2))^n ]

$${q}^{\mathrm{2}} −{q}−\mathrm{1}=\mathrm{0} \\ $$$${q}=\frac{\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$${u}_{{n}} ={A}\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} +{B}\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} \\ $$$${u}_{\mathrm{0}} ={A}+{B}={u}_{\mathrm{2}} −{u}_{\mathrm{1}} =\mathrm{1}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{A}=−{B} \\ $$$${u}_{\mathrm{1}} ={A}\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)+{B}\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)=\mathrm{1} \\ $$$${A}\left[\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}−\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right]=\mathrm{1} \\ $$$$\Rightarrow{A}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}} \\ $$$$\Rightarrow{B}=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}} \\ $$$$\Rightarrow{u}_{{n}} =\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left[\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} −\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} \right] \\ $$

Answered by Dwaipayan Shikari last updated on 12/Nov/20

u_(n+2) =u_(n+1) +u_n r^(n+2) =r^(n+1) +r^n r^2 −r−1=0⇒r=((1±(√5))/2) u_n =Λ(((1+(√5))/2))^n +Γ(((1−(√5))/2))^n u_1 =((Λ+Γ)/2)+(√5)((Λ−Γ)/2)=1 u_2 =3((Λ+Γ)/2)+(√5)((Λ−Γ)/2)=1 ⇒Λ+Γ=0⇒Λ=−Γ 1=(((√5)(−2Γ))/2)⇒Γ=−(1/( (√5))) and Λ=(1/( (√5))) u_n =(1/( (√5)))(((1+(√5))/2))^n −(1/( (√5)))(((1−(√5))/2))^n Fibonocci sequence

$${u}_{{n}+\mathrm{2}} ={u}_{{n}+\mathrm{1}} +{u}_{{n}} \\ $$$${r}^{{n}+\mathrm{2}} ={r}^{{n}+\mathrm{1}} +{r}^{{n}} \\ $$$${r}^{\mathrm{2}} −{r}−\mathrm{1}=\mathrm{0}\Rightarrow{r}=\frac{\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$${u}_{{n}} =\Lambda\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} +\Gamma\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} \\ $$$${u}_{\mathrm{1}} =\frac{\Lambda+\Gamma}{\mathrm{2}}+\sqrt{\mathrm{5}}\frac{\Lambda−\Gamma}{\mathrm{2}}=\mathrm{1} \\ $$$${u}_{\mathrm{2}} =\mathrm{3}\frac{\Lambda+\Gamma}{\mathrm{2}}+\sqrt{\mathrm{5}}\frac{\Lambda−\Gamma}{\mathrm{2}}=\mathrm{1}\:\:\:\:\:\Rightarrow\Lambda+\Gamma=\mathrm{0}\Rightarrow\Lambda=−\Gamma \\ $$$$\mathrm{1}=\frac{\sqrt{\mathrm{5}}\left(−\mathrm{2}\Gamma\right)}{\mathrm{2}}\Rightarrow\Gamma=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\:\:\:\:{and}\:\:\Lambda=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}} \\ $$$${u}_{{n}} =\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} −\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} \\ $$$$\boldsymbol{\mathrm{Fibonocci}}\:\boldsymbol{\mathrm{sequence}} \\ $$

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