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The-first-two-terms-of-the-a-n-series-are-defind-as-a-n-a-n-1-a-n-2-for-the-general-term-a-1-5-a-2-8-and-n-3-since-the-L-lim-n-a-n-1-a-n-what-is-the-value-of-L-

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Question Number 143519 by tugu last updated on 15/Jun/21
The first two terms of the {a_n } series are defind as a_n =a_(n−1) +a_(n−2) for the general term a_1 =5, a_2 =8 and n≥3 . since the L=lim_(n→∞) (a_(n+1) /a_n ) what is the value of L
$$ \\ $$$${The}\:{first}\:{two}\:{terms}\:{of}\:{the}\:\left\{{a}_{{n}} \right\}\:{series}\:\:\:{are}\:{defind}\:{as}\:{a}_{{n}} ={a}_{{n}−\mathrm{1}} +{a}_{{n}−\mathrm{2}} \:\:{for}\:{the}\:{general}\:{term} \\ $$$$\:{a}_{\mathrm{1}} =\mathrm{5},\:{a}_{\mathrm{2}} =\mathrm{8}\:{and}\:{n}\geqslant\mathrm{3}\:. \\ $$$${since}\:{the}\:{L}={li}\underset{{n}\rightarrow\infty} {{m}}\frac{{a}_{{n}+\mathrm{1}} }{{a}_{{n}} }\:\:{what}\:{is}\:{the}\:{value}\:{of}\:{L} \\ $$$$ \\ $$
Answered by mathmax by abdo last updated on 15/Jun/21
a_n =a_(n−1) +a_(n−2) ⇒a_(n+2) =a_(n+1) +a_n ⇒a_(n+2) −a_(n+1) −a_n =0 (ec)→x^2 −x−1=0 →Δ=1+4=5 ⇒x_1 =((1+(√5))/2) and x_2 =((1−(√5))/2) ⇒a_n =α(((1+(√5))/2))^n +β(((1−(√5))/2))^n a_(1 ) =5 ⇒α(((1+(√5))/2))+β(((1−(√5))/2))=5 a_2 =8 ⇒α(((1+(√5))/2))^2 +β(((1−(√5))/2))^2 =8 we get the system { ((((1+(√5))/2)α +((1−(√5))/2)β=5)),(((((1+(√5))/2))^2 α +(((1−(√5))/2))^2 β=8)) :} Δ_s =(((1+(√5))(1−(√5))^2 )/8)−(((1−(√5))(1+(√5))^2 )/8) =−(1/2)(1−(√5))+(1/2)(1+(√5)) =(1/2)(1+(√5)−1+(√5))=(√5) Δ_α = determinant (((5 ((1−(√5))/2))),((8 (((1−(√5))/2))^2 )))=(5/4)(1−(√5))^2 −4(1−(√5)) =(5/4)(6−2(√5))−4+4(√5)=(5/2)(3−(√5))−4+4(√5)=((15)/2)−4−((5(√5))/2)+4(√5) =(7/2)+(3/2)(√(5 )) ⇒α=((7+3(√5))/(2(√5))) Δ_β = determinant (((((1+(√5))/2) 5)),(((((1+(√5))/2))^2 8)))=4(1+(√5))−(5/4)(1+(√5))^2 =4+4(√5)−(5/4)(6+2(√5)) =4+4(√5)−(5/2)(3+(√5))=4+4(√5)−((15)/2)−((5(√5))/2) =−(7/2) +(3/2)(√5) ⇒β =((−7+3(√5))/(2(√5))) ⇒ a_n =(((7+3(√5))/(2(√5))))(((1+(√5))/2))^n +(((−7+3(√5))/(2(√5))))(((1−(√5))/2))^n ⇒a_n =(((7+3(√5))(1+(√5))^n )/(2^(n+1) ((√5)))) +(((−7+3(√5))(1−(√5))^n )/(2^(n+1) ((√5)))) =(1/(((√5))2^(n+1) )){ (7+3(√5))(1+(√5))^n −(7−3(√5))(1−(√5))^n } rest to calculate (a_(n+1) /a_n ) ....be continued...
$$\mathrm{a}_{\mathrm{n}} =\mathrm{a}_{\mathrm{n}−\mathrm{1}} +\mathrm{a}_{\mathrm{n}−\mathrm{2}} \:\Rightarrow\mathrm{a}_{\mathrm{n}+\mathrm{2}} =\mathrm{a}_{\mathrm{n}+\mathrm{1}} +\mathrm{a}_{\mathrm{n}} \:\Rightarrow\mathrm{a}_{\mathrm{n}+\mathrm{2}} −\mathrm{a}_{\mathrm{n}+\mathrm{1}} −\mathrm{a}_{\mathrm{n}} =\mathrm{0} \\ $$$$\left(\mathrm{ec}\right)\rightarrow\mathrm{x}^{\mathrm{2}} −\mathrm{x}−\mathrm{1}=\mathrm{0}\:\rightarrow\Delta=\mathrm{1}+\mathrm{4}=\mathrm{5}\:\Rightarrow\mathrm{x}_{\mathrm{1}} =\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:\mathrm{and}\:\mathrm{x}_{\mathrm{2}} =\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{a}_{\mathrm{n}} =\alpha\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{n}} \:+\beta\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{n}} \\ $$$$\mathrm{a}_{\mathrm{1}\:} =\mathrm{5}\:\Rightarrow\alpha\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)+\beta\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)=\mathrm{5} \\ $$$$\mathrm{a}_{\mathrm{2}} =\mathrm{8}\:\Rightarrow\alpha\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\beta\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{2}} \:=\mathrm{8}\:\:\mathrm{we}\:\mathrm{get}\:\mathrm{the}\:\mathrm{system} \\ $$$$\begin{cases}{\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\alpha\:+\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\beta=\mathrm{5}}\\{\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{2}} \alpha\:+\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{2}} \beta=\mathrm{8}}\end{cases} \\ $$$$\Delta_{\mathrm{s}} =\frac{\left(\mathrm{1}+\sqrt{\mathrm{5}}\right)\left(\mathrm{1}−\sqrt{\mathrm{5}}\right)^{\mathrm{2}} }{\mathrm{8}}−\frac{\left(\mathrm{1}−\sqrt{\mathrm{5}}\right)\left(\mathrm{1}+\sqrt{\mathrm{5}}\right)^{\mathrm{2}} }{\mathrm{8}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\sqrt{\mathrm{5}}\right)+\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\sqrt{\mathrm{5}}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\sqrt{\mathrm{5}}−\mathrm{1}+\sqrt{\mathrm{5}}\right)=\sqrt{\mathrm{5}} \\ $$$$\Delta_{\alpha} =\begin{vmatrix}{\mathrm{5}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}}\\{\mathrm{8}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{2}} }\end{vmatrix}=\frac{\mathrm{5}}{\mathrm{4}}\left(\mathrm{1}−\sqrt{\mathrm{5}}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{1}−\sqrt{\mathrm{5}}\right) \\ $$$$=\frac{\mathrm{5}}{\mathrm{4}}\left(\mathrm{6}−\mathrm{2}\sqrt{\mathrm{5}}\right)−\mathrm{4}+\mathrm{4}\sqrt{\mathrm{5}}=\frac{\mathrm{5}}{\mathrm{2}}\left(\mathrm{3}−\sqrt{\mathrm{5}}\right)−\mathrm{4}+\mathrm{4}\sqrt{\mathrm{5}}=\frac{\mathrm{15}}{\mathrm{2}}−\mathrm{4}−\frac{\mathrm{5}\sqrt{\mathrm{5}}}{\mathrm{2}}+\mathrm{4}\sqrt{\mathrm{5}} \\ $$$$=\frac{\mathrm{7}}{\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{2}}\sqrt{\mathrm{5}\:}\:\Rightarrow\alpha=\frac{\mathrm{7}+\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{2}\sqrt{\mathrm{5}}} \\ $$$$\Delta_{\beta} =\begin{vmatrix}{\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{5}}\\{\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{2}} \:\:\:\:\:\:\:\:\mathrm{8}}\end{vmatrix}=\mathrm{4}\left(\mathrm{1}+\sqrt{\mathrm{5}}\right)−\frac{\mathrm{5}}{\mathrm{4}}\left(\mathrm{1}+\sqrt{\mathrm{5}}\right)^{\mathrm{2}} \\ $$$$=\mathrm{4}+\mathrm{4}\sqrt{\mathrm{5}}−\frac{\mathrm{5}}{\mathrm{4}}\left(\mathrm{6}+\mathrm{2}\sqrt{\mathrm{5}}\right)\:=\mathrm{4}+\mathrm{4}\sqrt{\mathrm{5}}−\frac{\mathrm{5}}{\mathrm{2}}\left(\mathrm{3}+\sqrt{\mathrm{5}}\right)=\mathrm{4}+\mathrm{4}\sqrt{\mathrm{5}}−\frac{\mathrm{15}}{\mathrm{2}}−\frac{\mathrm{5}\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$=−\frac{\mathrm{7}}{\mathrm{2}}\:+\frac{\mathrm{3}}{\mathrm{2}}\sqrt{\mathrm{5}}\:\Rightarrow\beta\:=\frac{−\mathrm{7}+\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{2}\sqrt{\mathrm{5}}}\:\Rightarrow \\ $$$$\mathrm{a}_{\mathrm{n}} =\left(\frac{\mathrm{7}+\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{2}\sqrt{\mathrm{5}}}\right)\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{n}} \:+\left(\frac{−\mathrm{7}+\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{2}\sqrt{\mathrm{5}}}\right)\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{n}} \\ $$$$\Rightarrow\mathrm{a}_{\mathrm{n}} =\frac{\left(\mathrm{7}+\mathrm{3}\sqrt{\mathrm{5}}\right)\left(\mathrm{1}+\sqrt{\mathrm{5}}\right)^{\mathrm{n}} }{\mathrm{2}^{\mathrm{n}+\mathrm{1}} \left(\sqrt{\mathrm{5}}\right)}\:+\frac{\left(−\mathrm{7}+\mathrm{3}\sqrt{\mathrm{5}}\right)\left(\mathrm{1}−\sqrt{\mathrm{5}}\right)^{\mathrm{n}} }{\mathrm{2}^{\mathrm{n}+\mathrm{1}} \left(\sqrt{\mathrm{5}}\right)} \\ $$$$=\frac{\mathrm{1}}{\left(\sqrt{\mathrm{5}}\right)\mathrm{2}^{\mathrm{n}+\mathrm{1}} }\left\{\:\left(\mathrm{7}+\mathrm{3}\sqrt{\mathrm{5}}\right)\left(\mathrm{1}+\sqrt{\mathrm{5}}\right)^{\mathrm{n}} −\left(\mathrm{7}−\mathrm{3}\sqrt{\mathrm{5}}\right)\left(\mathrm{1}−\sqrt{\mathrm{5}}\right)^{\mathrm{n}} \right\} \\ $$$$\mathrm{rest}\:\mathrm{to}\:\mathrm{calculate}\:\frac{\mathrm{a}_{\mathrm{n}+\mathrm{1}} }{\mathrm{a}_{\mathrm{n}} }\:\:….\mathrm{be}\:\mathrm{continued}… \\ $$
Answered by mr W last updated on 15/Jun/21
x^2 −x−1=0 x=((1±(√5))/2) a_n =A(((1+(√5))/2))^n +B(((1−(√5))/2))^n a_0 =a_2 −a_1 =8−5=3=A+B ...(i) a_1 =5=A(((1+(√5))/2))+B(((1−(√5))/2)) ...(ii) ⇒A=((15+7(√5))/(10)) ⇒B=((15−7(√5))/(10)) a_n =(((15+7(√5))/(10)))(((1+(√5))/2))^n +(((15−7(√5))/(10)))(((1−(√5))/2))^n a_n =(((1+(√5))/2))^n [(((15+7(√5))/(10)))+(((15−7(√5))/(10)))(((1−(√5))/(1+(√5))))^n ] (a_(n+1) /a_n )=(((1+(√5))/2))×(((((15+7(√5))/(10)))+(((15−7(√5))/(10)))(((1−(√5))/(1+(√5))))^(n+1) )/((((15+7(√5))/(10)))+(((15−7(√5))/(10)))(((1−(√5))/(1+(√5))))^n )) lim_(n→∞) (a_(n+1) /a_n )=(((1+(√5))/2))×(((((15+7(√5))/(10)))+(((15−7(√5))/(10)))×0)/((((15+7(√5))/(10)))+(((15−7(√5))/(10)))×0))=((1+(√5))/2)=ϕ
$${x}^{\mathrm{2}} −{x}−\mathrm{1}=\mathrm{0} \\ $$$${x}=\frac{\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$${a}_{{n}} ={A}\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} +{B}\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} \\ $$$${a}_{\mathrm{0}} ={a}_{\mathrm{2}} −{a}_{\mathrm{1}} =\mathrm{8}−\mathrm{5}=\mathrm{3}={A}+{B}\:\:\:…\left({i}\right) \\ $$$${a}_{\mathrm{1}} =\mathrm{5}={A}\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)+{B}\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\:\:\:…\left({ii}\right) \\ $$$$\Rightarrow{A}=\frac{\mathrm{15}+\mathrm{7}\sqrt{\mathrm{5}}}{\mathrm{10}} \\ $$$$\Rightarrow{B}=\frac{\mathrm{15}−\mathrm{7}\sqrt{\mathrm{5}}}{\mathrm{10}} \\ $$$${a}_{{n}} =\left(\frac{\mathrm{15}+\mathrm{7}\sqrt{\mathrm{5}}}{\mathrm{10}}\right)\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} +\left(\frac{\mathrm{15}−\mathrm{7}\sqrt{\mathrm{5}}}{\mathrm{10}}\right)\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} \\ $$$${a}_{{n}} =\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} \left[\left(\frac{\mathrm{15}+\mathrm{7}\sqrt{\mathrm{5}}}{\mathrm{10}}\right)+\left(\frac{\mathrm{15}−\mathrm{7}\sqrt{\mathrm{5}}}{\mathrm{10}}\right)\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{1}+\sqrt{\mathrm{5}}}\right)^{{n}} \right] \\ $$$$\frac{{a}_{{n}+\mathrm{1}} }{{a}_{{n}} }=\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)×\frac{\left(\frac{\mathrm{15}+\mathrm{7}\sqrt{\mathrm{5}}}{\mathrm{10}}\right)+\left(\frac{\mathrm{15}−\mathrm{7}\sqrt{\mathrm{5}}}{\mathrm{10}}\right)\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{1}+\sqrt{\mathrm{5}}}\right)^{{n}+\mathrm{1}} }{\left(\frac{\mathrm{15}+\mathrm{7}\sqrt{\mathrm{5}}}{\mathrm{10}}\right)+\left(\frac{\mathrm{15}−\mathrm{7}\sqrt{\mathrm{5}}}{\mathrm{10}}\right)\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{1}+\sqrt{\mathrm{5}}}\right)^{{n}} } \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{a}_{{n}+\mathrm{1}} }{{a}_{{n}} }=\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)×\frac{\left(\frac{\mathrm{15}+\mathrm{7}\sqrt{\mathrm{5}}}{\mathrm{10}}\right)+\left(\frac{\mathrm{15}−\mathrm{7}\sqrt{\mathrm{5}}}{\mathrm{10}}\right)×\mathrm{0}}{\left(\frac{\mathrm{15}+\mathrm{7}\sqrt{\mathrm{5}}}{\mathrm{10}}\right)+\left(\frac{\mathrm{15}−\mathrm{7}\sqrt{\mathrm{5}}}{\mathrm{10}}\right)×\mathrm{0}}=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}=\varphi \\ $$
Answered by mr W last updated on 15/Jun/21
a_(n+1) =a_n +a_(n−1) (a_(n+1) /a_n )=1+(1/(a_n /a_(n−1) )) lim_(n→∞) (a_(n+1) /a_n )=1+(1/(lim_(n→∞) (a_n /a_(n−1) ))) L=1+(1/L) >1 L^2 −L−1=0 ⇒L=((1+(√5))/2)
$${a}_{{n}+\mathrm{1}} ={a}_{{n}} +{a}_{{n}−\mathrm{1}} \\ $$$$\frac{{a}_{{n}+\mathrm{1}} }{{a}_{{n}} }=\mathrm{1}+\frac{\mathrm{1}}{\frac{{a}_{{n}} }{{a}_{{n}−\mathrm{1}} }} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{a}_{{n}+\mathrm{1}} }{{a}_{{n}} }=\mathrm{1}+\frac{\mathrm{1}}{\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{a}_{{n}} }{{a}_{{n}−\mathrm{1}} }} \\ $$$${L}=\mathrm{1}+\frac{\mathrm{1}}{{L}}\:>\mathrm{1} \\ $$$${L}^{\mathrm{2}} −{L}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{L}=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$

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