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Question Number 139954 by Ar Brandon last updated on 02/May/21

Answered by mr W last updated on 03/May/21

$$x_{n+1}=\frac{x_n-\frac{1}{\sqrt{2}+1}}{1+x_n\times \frac{1}{\sqrt{2}+1}}$$$$let\tan A_n=x_n,\tan B=\frac{1}{\sqrt{2}+1}=\sqrt{2}-1\Rightarrow B=\frac{\pi }{8}$$$$\tan A_{n+1}=\frac{\tan A_n-\tan B}{1+\tan A_n\tan B}=\tan (A_n-B)$$$$\Rightarrow A_{n+1}=A_n-B\to {it}^{\prime }sA.P.$$$$\Rightarrow A_n=A_1-(n-1)B$$$$\Rightarrow A_n={\tan }^{-1}{x}_1-(n-1)\frac{\pi }{8}$$$$\Rightarrow A_n={\tan }^{-1}2014-\frac{(n-1)\pi }{8}$$$$\Rightarrow x_n=\tan [{\tan }^{-1}2014-\frac{(n-1)\pi }{8}]$$$$\Rightarrow x_n=\frac{2014-\tan \frac{(n-1)\pi }{8}}{1+2014\tan \frac{(n-1)\pi }{8}}$$$$\Rightarrow x_{2015}=\frac{2014-\tan \frac{2014\pi }{8}}{1+2014\tan \frac{2014\pi }{8}}$$$$=\frac{2014-\tan (252\pi -\frac{\pi }{4})}{1+2014\tan (252\pi -\frac{\pi }{4})}$$$$=\frac{2014+1}{1-2014}=-\frac{2015}{2013}$$

Commented by Ar Brandon last updated on 04/May/21

$$\mathrm{Welldone}\mathrm{Sir}!$$

Answered by mr W last updated on 03/May/21

$$anotherway:$$$$letc=\sqrt{2}+1$$$$x_{n+1}+A=\frac{{cx}_n-1}{c+x_n}+A=\frac{(A+c)x_n+Ac-1}{c+x_n}$$$$x_{n+1}+B=\frac{{cx}_n-1}{c+x_n}+B=\frac{(B+c)x_n+Bc-1}{c+x_n}$$$$\frac{x_{n+1}+A}{x_{n+1}+B}=\frac{A+c}{B+c}\times \frac{x_n+\frac{Ac-1}{A+c}}{x_n+\frac{Bc-1}{B+c}}$$$$let\frac{Ac-1}{A+c}=A$$$$-1=A^2$$$$A=i,\Rightarrow B=-i$$$$(orA=-i,\Rightarrow B=i)$$$$$$$$\frac{x_{n+1}+i}{x_{n+1}-i}=\frac{c+i}{c-i}\times \frac{x_n+i}{x_n-i}\to {it}^{\prime }sG.P.$$$$\Rightarrow \frac{x_n+i}{x_n-i}={\left(\frac{c+i}{c-i}\right)}^{n-1}\times \frac{x_1+i}{x_1-i}$$$$e^{\left(2{\tan }^{-1}\frac{1}{x_n}\right)i}=e^{\left(2(n-1){\tan }^{-1}\frac{1}{c}\right)i}{e}^{2\left({\tan }^{-1}\frac{1}{x_1}\right)i}$$$$\Rightarrow {\tan }^{-1}\frac{1}{x_n}=(n-1){\tan }^{-1}\frac{1}{c}+{\tan }^{-1}\frac{1}{x_1}$$$$\Rightarrow \frac{\pi }{2}-{\tan }^{-1}{x}_n=(n-1){\tan }^{-1}\frac{1}{c}+\frac{\pi }{2}-{\tan }^{-1}{x}_1$$$$\Rightarrow {\tan }^{-1}{x}_n={\tan }^{-1}{x}_1-(n-1){\tan }^{-1}\frac{1}{c}$$$$\Rightarrow {\tan }^{-1}{x}_n={\tan }^{-1}2014-(n-1){\tan }^{-1}(\sqrt{2}-1)$$$$\Rightarrow x_n=\tan [{\tan }^{-1}2014-(n-1){\tan }^{-1}(\sqrt{2}-1)]$$$$\Rightarrow x_n=\tan [{\tan }^{-1}2014-\frac{(n-1)\pi }{8}]$$$$\Rightarrow x_{2015}=\tan [{\tan }^{-1}2014-\frac{2014\pi }{8}]$$$$=\tan [{\tan }^{-1}2014+\frac{\pi }{4}]$$$$=\frac{2014+\tan \left(\frac{\pi }{4}\right)}{1-2014\tan \left(\frac{\pi }{4}\right)}$$$$=\frac{2014+1}{1-2014}=-\frac{2015}{2013}$$

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