19. Let $O$ be the origin, $A_{1}, A_{2}, A_{3}, \ldots$ be points on the curve $y=\sqrt{x}$ and $B_{1}, B_{2}, B_{3}, \ldots$ be points on the positive $x$-axis such that the triangles $O B_{1} A_{1}, B_{1} B_{2} A_{2}, B_{2} B_{3} A_{3}, \ldots$ are all equilateral, with side lengths $l_{1}, l_{2}, l_{3}, \ldots$ respectively. Find the value of $l_{1}+l_{2}+l_{3}+\cdots+l_{2005}$.
(2 marks)
Let $O$ be the origin, $A_{1}, A_{2}, A_{3}, \ldots$ be points on the curve $y=\sqrt{x}$ and $B_{1}, B_{2}, B_{3}, \ldots$ be points on the positive $x$-axis such that the triangles $O B_{1} A_{1}, B_{1} B_{2} A_{2}, B_{2} B_{3} A_{3}, \ldots$ are all equilateral, with side lengths $l_{1}, l_{2}, l_{3}, \ldots$ respectively. Find the value of $l_{1}+l_{2}+l_{3}+\cdots+l_{2005}$.
19. $\frac{4022030}{3}$
19. Note that $B_{n}=\left(l_{1}+l_{2}+l_{3}+\cdots+l_{n}, 0\right)$. Set $S_{n}=l_{1}+l_{2}+\cdots+l_{n}$. The equation of $B_{n-1} A_{n}$ is $y=\sqrt{3}\left(x-S_{n-1}\right)$. Hence the $y$-coordinate of $A_{n}$ satisfies
$$
y_{n}=\sqrt{3}\left(y_{n}^{2}-S_{n-1}\right) \text {. }
$$

Solving under the condition $y_{n}>0$, we get
$$
y_{n}=\frac{1+\sqrt{1+12 S_{n-1}}}{2 \sqrt{3}}
$$

Since $l_{n}=\frac{y_{n}}{\sin 60^{\circ}}$, we have
$$
\frac{\sqrt{3}}{2} l_{n}=\frac{1+\sqrt{1+12 S_{n-1}}}{2 \sqrt{3}},
$$
i.e. $\left(3 l_{n}-1\right)^{2}=1+12 S_{n-1}$. Upon simplification, we have
$$
\begin{array}{l}
3 l_{n}^{2}-2 l_{n}=4 S_{n-1} \\
3 l_{n+1}^{2}-2 l_{n+1}=4 S_{n}
\end{array}
$$
and hence
$$
\begin{aligned}
3 l_{n+1}^{2}-2 l_{n+1} & =4 S_{n} \\
3\left(l_{n+1}^{2}-l_{n}^{2}\right)-2\left(l_{n+1}-l_{n}\right) & =4\left(S_{n}-S_{n-1}\right) \\
3\left(l_{n+1}+l_{n}\right)\left(l_{n+1}-l_{n}\right)-2\left(l_{n+1}-l_{n}\right) & =4 l_{n} \\
3\left(l_{n+1}+l_{n}\right)\left(l_{n+1}-l_{n}\right)-2\left(l_{n+1}+l_{n}\right) & =0
\end{aligned}
$$

Since $l_{n} \neq l_{n+1}$, we have
$$
l_{n+1}-l_{n}=\frac{2}{3}
$$

Thus $\left\{l_{n}\right\}$ is an arithmetic sequence with common difference $\frac{2}{3}$. Furthermore,
$$
\frac{\sqrt{3}}{2} l_{1}=\frac{1+\sqrt{1+12 S_{0}}}{2 \sqrt{3}},
$$
where $S_{0}$ is taken to be zero. This gives $l_{1}=\frac{2}{3}$. It follows that
$$
l_{1}+l_{2}+\cdots+l_{2005}=\frac{2}{3}(1+2+\cdots 2005)=\frac{2}{3} \cdot \frac{2005 \cdot 2006}{2}=\frac{4022030}{3} .
$$

Remark. The condition that $B_{1}, B_{2}, \ldots$ are distinct is missing in the original question. This is necessary to ensure that $l_{n} \neq l_{n+1}$ so that we can cancel out the term $l_{n+1}-l_{n}$.