Cho dãy số \(\left( {{u_n}} \right)\) thỏa mãn \(\left\{ \begin{array}{l}{u_1} = 2\\{u_{n + 1}} = \frac{{{u_n} + \sqrt 2  - 1}}{{1 -

* Hướng dẫn giải

Ta có \(\tan \frac{\pi }{8} = \sqrt 2  - 1\) suy ra \({u_{n + 1}} = \frac{{{u_n} + \tan \frac{\pi }{8}}}{{1 - \tan \frac{\pi }{8}.{u_n}}}\)

Đặt \(\tan \varphi  = 2\) suy ra \({u_1} = \tan \varphi  \to {u_2} = \frac{{{u_1} + \tan \frac{\pi }{8}}}{{1 - \tan \frac{\pi }{8}.{u_1}}} = \frac{{\tan \varphi  + \tan \frac{\pi }{8}}}{{1 - \tan \varphi .\tan \frac{\pi }{8}}} = \tan \left( {\varphi  + \frac{\pi }{8}} \right)\)

Do đó \({u_3}\left( {\tan \varphi  + 2.\frac{\pi }{8}} \right) \to {u_n}\left( {\tan \varphi  + n.\frac{\pi }{8}} \right)\)

Vậy \({u_{2018}} = \tan \left( {\varphi  + 2017.\frac{\pi }{8}} \right) = \tan \left( {\varphi  + \frac{\pi }{8}} \right) = {u_2} = \frac{{2 + \sqrt 2  - 1}}{{1 - 2\left( {\sqrt 2  - 1} \right)}} = 7 + 5\sqrt 2 \)