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Question Number 142207 by 777316 last updated on 27/May/21

Answered by mr W last updated on 28/May/21

$$say\underset{n\to \mathrm{\infty }}{\lim }{a}_n=L$$$$L=\frac{3L+4}{2L+3}$$$$2L^2+3L=3L+4$$$$L^2=2$$$$\Rightarrow L=\sqrt{2}$$

Answered by mr W last updated on 27/May/21

$$a_{n+1}=\frac{3a_n+4}{2a_n+3}$$$$a_{n+1}+p=\frac{3a_n+4}{2a_n+3}+p=\frac{(3+2p)a_n+(4+3p)}{2a_n+3}$$$$a_{n+1}+q=\frac{3a_n+4}{2a_n+3}+q=\frac{(3+2q)a_n+(4+3q)}{2a_n+3}$$$$\frac{a_{n+1}+p}{a_{n+1}+q}=\frac{(3+2p)a_n+(4+3p)}{(3+2q)a_n+(4+3q)}$$$$\frac{a_{n+1}+p}{a_{n+1}+q}=\frac{3+2p}{3+2q}\times \frac{a_n+\frac{4+3p}{3+2p}}{a_n+\frac{4+3q}{3+2q}}$$$$p=\frac{4+3p}{3+2p}$$$$p^2=2$$$$\Rightarrow p=\sqrt{2}$$$$\Rightarrow q=-\sqrt{2}$$$$\frac{a_{n+1}+\sqrt{2}}{a_{n+1}-\sqrt{2}}=\frac{3+2\sqrt{2}}{3-2\sqrt{2}}\times \frac{a_n+\sqrt{2}}{a_n-\sqrt{2}}$$$$withk=\frac{3+2\sqrt{2}}{3-2\sqrt{2}}={(\sqrt{2}+1)}^4$$$$\frac{a_{n+1}+\sqrt{2}}{a_{n+1}-\sqrt{2}}=k\times \frac{a_n+\sqrt{2}}{a_n-\sqrt{2}}$$$$\Rightarrow \frac{a_n+\sqrt{2}}{a_n-\sqrt{2}}=k^{n-1}\times \frac{a_1+\sqrt{2}}{a_1-\sqrt{2}}=k^{n-1}\times \frac{2+\sqrt{2}}{2-\sqrt{2}}$$$$={(\sqrt{2}+1)}^2{k}^{n-1}={(\sqrt{2}+1)}^{2(2n-1)}$$$$a_n+\sqrt{2}={(\sqrt{2}+1)}^{2(2n-1)}{a}_n-{(\sqrt{2}+1)}^{2(2n-1)}\sqrt{2}$$$$[{(\sqrt{2}+1)}^{2(2n-1)}-1]a_n=\sqrt{2}[{(\sqrt{2}+1)}^{2(2n-1)}+1]$$$$\Rightarrow a_n=\frac{\sqrt{2}[{(\sqrt{2}+1)}^{2(2n-1)}+1]}{{(\sqrt{2}+1)}^{2(2n-1)}-1}$$$$\Rightarrow a_n=\frac{\sqrt{2}[1+\frac{1}{{(\sqrt{2}+1)}^{2(2n-1)}}]}{1-\frac{1}{{(\sqrt{2}+1)}^{2(2n-1)}}}$$$$\Rightarrow \underset{n\to \mathrm{\infty }}{\lim }{a}_n=\frac{\sqrt{2}[1+0]}{1-0}=\sqrt{2}$$

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