$u_1=\sqrt3$
$u_2=\dfrac{u_1-\sqrt{12-3u_1^2}}{2}=\dfrac{\sqrt3-\sqrt{12-3.3}}{2}=0$
$u_3=\dfrac{u_2-\sqrt{12-3u_2^2}}{2}=\dfrac{0-\sqrt{12}}{2}=-\sqrt3$
$u_4=\dfrac{u_3-\sqrt{12-3u_3^2}}{2}=\dfrac{-\sqrt3-\sqrt{12-3.3}}{2}=-\sqrt3$
Suy ra: $u_n=-\sqrt3\forall n\ge 3$
$\to u_5=u_6=...u_{2018}=-\sqrt3$
Tổng $2018$ số hạng đầu:
$\sum\limits_{i=0}^{2018}u_i=u_1+u_2+...+u_{2018}=\sqrt3+0-2016.\sqrt3=-2015\sqrt3$
