Ta có: $u_1=\sqrt{3}=tan\frac{\pi}{3}$$u_2=\frac{tan\frac{\pi}{4}+tan\frac{\pi}{3}}{1-tan\frac{\pi}{3}tan\frac{\pi}{4}}=tan(\frac{\pi}{3}+\frac{\pi}{4})$
$u_3=\frac{tan\frac{\pi}{4}+tan\frac{7\pi}{12}}{1-tan\frac{\pi}{4}.tan\frac{7\pi}{12}}=tan(\frac{\pi}{4}+\frac{\pi}{4}+\frac{\pi}{3})=tan(\frac{\pi}{3}+2.\frac{\pi}{4})$
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$u_n=tan(\frac{\pi}{3}+(n-1)\frac{\pi}{4})$
Ta sẽ chứng minh quy nạp $(u_n)=tan(\frac{\pi}{3}+(n-1)\frac{\pi}{4})$ $(*)$
Thật vậy, theo nguyên lý quy nạp thì $(*)$ đúng $\forall n\geq 1$
$\Rightarrow u_{2014}=tan(\frac{\pi}{3}+2013\frac{\pi}{4})=tan(-\frac{5\pi}{12})=-2-\sqrt{3}$