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Question Number 98280 by I want to learn more last updated on 12/Jun/20

$$\mathbf{Let}\left\{{\mathbf{a}}_{\mathbf{n}}\right\}\mathbf{be}\mathbf{a}\mathbf{sequence}\mathbf{such}\mathbf{that}{\mathbf{a}}_1=2,$$$${\mathbf{a}}_{\mathbf{n}+1}=\frac{3{\mathbf{a}}_{\mathbf{n}}+4}{2{\mathbf{a}}_{\mathbf{n}}+3},\mathbf{n}⩾1,\mathbf{find}{\mathbf{a}}_{\mathbf{n}}$$

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

$$a_{n+1}=\frac{3a_n+4}{2a_n+3}$$$$a_{n+1}+A=\frac{3a_n+4}{2a_n+3}+A=\frac{(3+2A)a_n+(4+3A)}{2a_n+3}$$$$a_{n+1}+B=\frac{3a_n+4}{2a_n+3}+B=\frac{(3+2B)a_n+(4+3B)}{2a_n+3}$$$$\frac{a_{n+1}+A}{a_{n+1}+B}=\frac{3+2A}{3+2B}\times \frac{a_n+\frac{4+3A}{3+2A}}{a_n+\frac{4+3B}{3+2B}}$$$$setA=\frac{4+3A}{3+2A}$$$$\Rightarrow A^2=2$$$$\Rightarrow A=\sqrt{2}$$$$\Rightarrow B=-\sqrt{2}$$$$\frac{3+2A}{3+2B}=\frac{3+2\sqrt{2}}{3-2\sqrt{2}}={(3+2\sqrt{2})}^2$$$$with{b}_n=\frac{a_n+\sqrt{2}}{a_n-\sqrt{2}}$$$$b_{n+1}={(3+2\sqrt{2})}^2{b}_n\leftarrow G.P.$$$$\Rightarrow b_n=b_1{(3+2\sqrt{2})}^{2(n-1)}$$$$b_1=\frac{2+\sqrt{2}}{2-\sqrt{2}}=3+2\sqrt{2}$$$$\Rightarrow b_n={(3+2\sqrt{2})}^{2n-1}$$$$a_n=\frac{\sqrt{2}(b_n+1)}{b_n-1}=\sqrt{2}(1+\frac{2}{b_n-1})$$$$\Rightarrow a_n=\sqrt{2}[1+\frac{2}{{(3+2\sqrt{2})}^{2n-1}-1}]$$$$weget:$$$$a_1=2$$$$a_2=\frac{10}{7}$$$$a_3=\frac{58}{41}$$$$a_4=\frac{338}{239}$$

Commented by I want to learn more last updated on 12/Jun/20

$$\mathrm{Great}\mathrm{sir},\mathrm{thanks}$$

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