Question Number 142207 by 777316 last updated on 27/May/21
Answered by mr W last updated on 28/May/21
$${say}\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}{a}_{{n}} ={L} \\ $$$${L}=\frac{\mathrm{3}{L}+\mathrm{4}}{\mathrm{2}{L}+\mathrm{3}} \\ $$$$\mathrm{2}{L}^{\mathrm{2}} +\mathrm{3}{L}=\mathrm{3}{L}+\mathrm{4} \\ $$$${L}^{\mathrm{2}} =\mathrm{2} \\ $$$$\Rightarrow{L}=\sqrt{\mathrm{2}} \\ $$
Answered by mr W last updated on 27/May/21
![a_(n+1) =((3a_n +4)/(2a_n +3)) a_(n+1) +p=((3a_n +4)/(2a_n +3))+p=(((3+2p)a_n +(4+3p))/(2a_n +3)) a_(n+1) +q=((3a_n +4)/(2a_n +3))+q=(((3+2q)a_n +(4+3q))/(2a_n +3)) ((a_(n+1) +p)/(a_(n+1) +q))=(((3+2p)a_n +(4+3p))/((3+2q)a_n +(4+3q))) ((a_(n+1) +p)/(a_(n+1) +q))=((3+2p)/(3+2q))×((a_n +((4+3p)/(3+2p)))/(a_n +((4+3q)/(3+2q)))) p=((4+3p)/(3+2p)) p^2 =2 ⇒p=(√2) ⇒q=−(√2) ((a_(n+1) +(√2))/(a_(n+1) −(√2)))=((3+2(√2))/(3−2(√2)))×((a_n +(√2))/(a_n −(√2))) with k=((3+2(√2))/(3−2(√2)))=((√2)+1)^4 ((a_(n+1) +(√2))/(a_(n+1) −(√2)))=k×((a_n +(√2))/(a_n −(√2))) ⇒((a_n +(√2))/(a_n −(√2)))=k^(n−1) ×((a_1 +(√2))/(a_1 −(√2)))=k^(n−1) ×((2+(√2))/(2−(√2))) =((√2)+1)^2 k^(n−1) =((√2)+1)^(2(2n−1)) a_n +(√2)=((√2)+1)^(2(2n−1)) a_n −((√2)+1)^(2(2n−1)) (√2) [((√2)+1)^(2(2n−1)) −1]a_n =(√2)[((√2)+1)^(2(2n−1)) +1] ⇒a_n =(((√2)[((√2)+1)^(2(2n−1)) +1])/( ((√2)+1)^(2(2n−1)) −1)) ⇒a_n =(((√2)[1+(1/(((√2)+1)^(2(2n−1)) ))])/( 1−(1/(((√2)+1)^(2(2n−1)) )))) ⇒lim_(n→∞) a_n =(((√2)[1+0])/( 1−0))=(√2)](https://www.tinkutara.com/question/Q142222.png)
$${a}_{{n}+\mathrm{1}} =\frac{\mathrm{3}{a}_{{n}} +\mathrm{4}}{\mathrm{2}{a}_{{n}} +\mathrm{3}} \\ $$$${a}_{{n}+\mathrm{1}} +{p}=\frac{\mathrm{3}{a}_{{n}} +\mathrm{4}}{\mathrm{2}{a}_{{n}} +\mathrm{3}}+{p}=\frac{\left(\mathrm{3}+\mathrm{2}{p}\right){a}_{{n}} +\left(\mathrm{4}+\mathrm{3}{p}\right)}{\mathrm{2}{a}_{{n}} +\mathrm{3}} \\ $$$${a}_{{n}+\mathrm{1}} +{q}=\frac{\mathrm{3}{a}_{{n}} +\mathrm{4}}{\mathrm{2}{a}_{{n}} +\mathrm{3}}+{q}=\frac{\left(\mathrm{3}+\mathrm{2}{q}\right){a}_{{n}} +\left(\mathrm{4}+\mathrm{3}{q}\right)}{\mathrm{2}{a}_{{n}} +\mathrm{3}} \\ $$$$\frac{{a}_{{n}+\mathrm{1}} +{p}}{{a}_{{n}+\mathrm{1}} +{q}}=\frac{\left(\mathrm{3}+\mathrm{2}{p}\right){a}_{{n}} +\left(\mathrm{4}+\mathrm{3}{p}\right)}{\left(\mathrm{3}+\mathrm{2}{q}\right){a}_{{n}} +\left(\mathrm{4}+\mathrm{3}{q}\right)} \\ $$$$\frac{{a}_{{n}+\mathrm{1}} +{p}}{{a}_{{n}+\mathrm{1}} +{q}}=\frac{\mathrm{3}+\mathrm{2}{p}}{\mathrm{3}+\mathrm{2}{q}}×\frac{{a}_{{n}} +\frac{\mathrm{4}+\mathrm{3}{p}}{\mathrm{3}+\mathrm{2}{p}}}{{a}_{{n}} +\frac{\mathrm{4}+\mathrm{3}{q}}{\mathrm{3}+\mathrm{2}{q}}} \\ $$$${p}=\frac{\mathrm{4}+\mathrm{3}{p}}{\mathrm{3}+\mathrm{2}{p}} \\ $$$${p}^{\mathrm{2}} =\mathrm{2} \\ $$$$\Rightarrow{p}=\sqrt{\mathrm{2}} \\ $$$$\Rightarrow{q}=−\sqrt{\mathrm{2}} \\ $$$$\frac{{a}_{{n}+\mathrm{1}} +\sqrt{\mathrm{2}}}{{a}_{{n}+\mathrm{1}} −\sqrt{\mathrm{2}}}=\frac{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}}×\frac{{a}_{{n}} +\sqrt{\mathrm{2}}}{{a}_{{n}} −\sqrt{\mathrm{2}}} \\ $$$${with}\:{k}=\frac{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}}=\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{4}} \\ $$$$\frac{{a}_{{n}+\mathrm{1}} +\sqrt{\mathrm{2}}}{{a}_{{n}+\mathrm{1}} −\sqrt{\mathrm{2}}}={k}×\frac{{a}_{{n}} +\sqrt{\mathrm{2}}}{{a}_{{n}} −\sqrt{\mathrm{2}}} \\ $$$$\Rightarrow\frac{{a}_{{n}} +\sqrt{\mathrm{2}}}{{a}_{{n}} −\sqrt{\mathrm{2}}}={k}^{{n}−\mathrm{1}} ×\frac{{a}_{\mathrm{1}} +\sqrt{\mathrm{2}}}{{a}_{\mathrm{1}} −\sqrt{\mathrm{2}}}={k}^{{n}−\mathrm{1}} ×\frac{\mathrm{2}+\sqrt{\mathrm{2}}}{\mathrm{2}−\sqrt{\mathrm{2}}} \\ $$$$=\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}} {k}^{{n}−\mathrm{1}} =\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}\left(\mathrm{2}{n}−\mathrm{1}\right)} \\ $$$${a}_{{n}} +\sqrt{\mathrm{2}}=\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}\left(\mathrm{2}{n}−\mathrm{1}\right)} {a}_{{n}} −\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}\left(\mathrm{2}{n}−\mathrm{1}\right)} \sqrt{\mathrm{2}} \\ $$$$\left[\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}\left(\mathrm{2}{n}−\mathrm{1}\right)} −\mathrm{1}\right]{a}_{{n}} =\sqrt{\mathrm{2}}\left[\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}\left(\mathrm{2}{n}−\mathrm{1}\right)} +\mathrm{1}\right] \\ $$$$\Rightarrow{a}_{{n}} =\frac{\sqrt{\mathrm{2}}\left[\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}\left(\mathrm{2}{n}−\mathrm{1}\right)} +\mathrm{1}\right]}{\:\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}\left(\mathrm{2}{n}−\mathrm{1}\right)} −\mathrm{1}} \\ $$$$\Rightarrow{a}_{{n}} =\frac{\sqrt{\mathrm{2}}\left[\mathrm{1}+\frac{\mathrm{1}}{\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}\left(\mathrm{2}{n}−\mathrm{1}\right)} }\right]}{\:\mathrm{1}−\frac{\mathrm{1}}{\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}\left(\mathrm{2}{n}−\mathrm{1}\right)} }} \\ $$$$\Rightarrow\underset{{n}\rightarrow\infty} {\mathrm{lim}}{a}_{{n}} =\frac{\sqrt{\mathrm{2}}\left[\mathrm{1}+\mathrm{0}\right]}{\:\mathrm{1}−\mathrm{0}}=\sqrt{\mathrm{2}} \\ $$