Lời giải:Đặt \(\tan \alpha = 2\). Ta có \(\tan \frac{\pi }{8} = \sqrt 2 - 1\). Suy ra \({u_{n + 1}} = \frac{{{u_n} + \tan \frac{\pi }{8}}}{{1 - \tan \frac{\pi }{8}.{u_n}}}\)
Có \(= \frac{{\tan \alpha + \tan \frac{\pi }{8}}}{{1 - \tan \frac{\pi }{8}.\tan {u_n}}} = \tan \left( {\alpha + \frac{\pi }{8}} \right)\).
Bằng quy nạp, ta chứng minh được \({u_n} = \tan \left[ {\alpha + \left( {n - 1} \right)\frac{\pi }{8}} \right]\).
Vậy \({u_{2018}} = \tan \left( {\alpha + \frac{{2017\pi }}{8}} \right) = \tan \left( {\alpha + \frac{\pi }{8}} \right) = \frac{{\tan \alpha + \tan \frac{\pi }{8}}}{{1 - \tan \alpha .\tan \frac{\pi }{8}}} = 7 + 5\sqrt 2 \).
