Cho dãy số un thỏa mãn

A

- Hướng dẫn giải

Đáp án A

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

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

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