Question Number 98280 by I want to learn more last updated on 12/Jun/20
$$\boldsymbol{\mathrm{Let}}\:\:\left\{\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{n}}} \right\}\:\:\boldsymbol{\mathrm{be}}\:\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{sequence}}\:\boldsymbol{\mathrm{such}}\:\boldsymbol{\mathrm{that}}\:\:\:\boldsymbol{\mathrm{a}}_{\mathrm{1}} \:=\:\:\mathrm{2}, \\ $$$$\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{n}}\:\:+\:\:\mathrm{1}} \:\:=\:\:\frac{\mathrm{3}\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{n}}} \:+\:\:\mathrm{4}}{\mathrm{2}\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{n}}} \:\:+\:\:\mathrm{3}},\:\:\:\:\:\boldsymbol{\mathrm{n}}\:\geqslant\:\mathrm{1},\:\:\:\:\:\boldsymbol{\mathrm{find}}\:\:\:\:\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{n}}} \\ $$
Answered by mr W last updated on 12/Jun/20
![a_(n+1) =((3a_n +4)/(2a_n +3)) a_(n+1) +A=((3a_n +4)/(2a_n +3))+A=(((3+2A)a_n +(4+3A))/(2a_n +3)) a_(n+1) +B=((3a_n +4)/(2a_n +3))+B=(((3+2B)a_n +(4+3B))/(2a_n +3)) ((a_(n+1) +A)/(a_(n+1) +B))=((3+2A)/(3+2B))×((a_n +((4+3A)/(3+2A)))/(a_n +((4+3B)/(3+2B)))) set A=((4+3A)/(3+2A)) ⇒A^2 =2 ⇒A=(√2) ⇒B=−(√2) ((3+2A)/(3+2B))=((3+2(√2))/(3−2(√2)))=(3+2(√2))^2 with b_n =((a_n +(√2))/(a_n −(√2))) b_(n+1) =(3+2(√2))^2 b_n ← G.P. ⇒b_n =b_1 (3+2(√2))^(2(n−1)) b_1 =((2+(√2))/(2−(√2)))=3+2(√2) ⇒b_n =(3+2(√2))^(2n−1) a_n =(((√2)(b_n +1))/(b_n −1))=(√2)(1+(2/(b_n −1))) ⇒a_n =(√2)[1+(2/((3+2(√2))^(2n−1) −1))] we get: a_1 =2 a_2 =((10)/7) a_3 =((58)/(41)) a_4 =((338)/(239))](https://www.tinkutara.com/question/Q98295.png)
$${a}_{{n}+\mathrm{1}} =\frac{\mathrm{3}{a}_{{n}} +\mathrm{4}}{\mathrm{2}{a}_{{n}} +\mathrm{3}} \\ $$$${a}_{{n}+\mathrm{1}} +{A}=\frac{\mathrm{3}{a}_{{n}} +\mathrm{4}}{\mathrm{2}{a}_{{n}} +\mathrm{3}}+{A}=\frac{\left(\mathrm{3}+\mathrm{2}{A}\right){a}_{{n}} +\left(\mathrm{4}+\mathrm{3}{A}\right)}{\mathrm{2}{a}_{{n}} +\mathrm{3}} \\ $$$${a}_{{n}+\mathrm{1}} +{B}=\frac{\mathrm{3}{a}_{{n}} +\mathrm{4}}{\mathrm{2}{a}_{{n}} +\mathrm{3}}+{B}=\frac{\left(\mathrm{3}+\mathrm{2}{B}\right){a}_{{n}} +\left(\mathrm{4}+\mathrm{3}{B}\right)}{\mathrm{2}{a}_{{n}} +\mathrm{3}} \\ $$$$\frac{{a}_{{n}+\mathrm{1}} +{A}}{{a}_{{n}+\mathrm{1}} +{B}}=\frac{\mathrm{3}+\mathrm{2}{A}}{\mathrm{3}+\mathrm{2}{B}}×\frac{{a}_{{n}} +\frac{\mathrm{4}+\mathrm{3}{A}}{\mathrm{3}+\mathrm{2}{A}}}{{a}_{{n}} +\frac{\mathrm{4}+\mathrm{3}{B}}{\mathrm{3}+\mathrm{2}{B}}} \\ $$$${set}\:{A}=\frac{\mathrm{4}+\mathrm{3}{A}}{\mathrm{3}+\mathrm{2}{A}} \\ $$$$\Rightarrow{A}^{\mathrm{2}} =\mathrm{2} \\ $$$$\Rightarrow{A}=\sqrt{\mathrm{2}} \\ $$$$\Rightarrow{B}=−\sqrt{\mathrm{2}} \\ $$$$\frac{\mathrm{3}+\mathrm{2}{A}}{\mathrm{3}+\mathrm{2}{B}}=\frac{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}}=\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$${with}\:{b}_{{n}} =\frac{{a}_{{n}} +\sqrt{\mathrm{2}}}{{a}_{{n}} −\sqrt{\mathrm{2}}} \\ $$$${b}_{{n}+\mathrm{1}} =\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} {b}_{{n}} \:\leftarrow\:{G}.{P}. \\ $$$$\Rightarrow{b}_{{n}} ={b}_{\mathrm{1}} \left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\right)^{\mathrm{2}\left({n}−\mathrm{1}\right)} \\ $$$${b}_{\mathrm{1}} =\frac{\mathrm{2}+\sqrt{\mathrm{2}}}{\mathrm{2}−\sqrt{\mathrm{2}}}=\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$\Rightarrow{b}_{{n}} =\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\right)^{\mathrm{2}{n}−\mathrm{1}} \\ $$$${a}_{{n}} =\frac{\sqrt{\mathrm{2}}\left({b}_{{n}} +\mathrm{1}\right)}{{b}_{{n}} −\mathrm{1}}=\sqrt{\mathrm{2}}\left(\mathrm{1}+\frac{\mathrm{2}}{{b}_{{n}} −\mathrm{1}}\right) \\ $$$$\Rightarrow{a}_{{n}} =\sqrt{\mathrm{2}}\left[\mathrm{1}+\frac{\mathrm{2}}{\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\right)^{\mathrm{2}{n}−\mathrm{1}} −\mathrm{1}}\right] \\ $$$${we}\:{get}: \\ $$$${a}_{\mathrm{1}} =\mathrm{2} \\ $$$${a}_{\mathrm{2}} =\frac{\mathrm{10}}{\mathrm{7}} \\ $$$${a}_{\mathrm{3}} =\frac{\mathrm{58}}{\mathrm{41}} \\ $$$${a}_{\mathrm{4}} =\frac{\mathrm{338}}{\mathrm{239}} \\ $$
Commented by I want to learn more last updated on 12/Jun/20
$$\mathrm{Great}\:\mathrm{sir},\:\:\mathrm{thanks} \\ $$