Five. (20 points) Let $P_{1}, P_{2}, \cdots, P_{n}, \cdots$ be a sequence of points on the curve $y=\sqrt{x}$, and $Q_{1}, Q_{2}, \cdots, Q_{n}, \cdots$ be a sequence of points on the positive half of the $x$-axis. The triangles $\triangle O Q_{1} P_{1}, \triangle Q_{1} Q_{2} P_{2}$, $\triangle Q_{2} Q_{3} P_{3}, \cdots, \triangle Q_{n} Q_{n+1} P_{n+1}, \cdots$ are all equilateral triangles, and their side lengths are $a_{1}, a_{2}, a_{3}$, $\cdots, a_{n}, \cdots$, respectively. Let $S_{n}$ be the sum of the first $n$ terms of the sequence $a_{1}, a_{2}, \cdots, a_{n}$. Find $S_{n}$.
As shown in Figure 3, the point \( Q_{n-1}\left(a_{1}+a_{2}+\cdots+a_{n-1}, 0\right) \) is the point \( Q_{n-1}\left(S_{n-1}, 0\right) \). Therefore, the equation of the line \( Q_{n-1} P_{n} \) is
\[
y=\sqrt{3}\left(x-S_{n-1}\right).
\]

Thus, the coordinates of point \( P_{n} \) satisfy
\[
\left\{\begin{array}{l}
y=\sqrt{3}\left(x-S_{n-1}\right), \\
y=\sqrt{x}.
\end{array}\right.
\]

Eliminating \( x \) gives \( \sqrt{3} y^{2}-y-\sqrt{3} S_{n-1}=0 \).
Since \( y>0 \), we have \( y=\frac{1+\sqrt{1+12 S_{n-1}}}{2 \sqrt{3}} \).
Also, \( a_{n}=\frac{y}{\sin 60^{\circ}} \), so \( 3 a_{n}=1+\sqrt{1+12 S_{n-1}} \).
Thus, \( \left(3 a_{n}-1\right)^{2}=1+12 S_{n-1} \), which simplifies to
\[
3 a_{n}^{2}-2 a_{n}=4 S_{n-1}.
\]

From equation (1), we have
\[
3 a_{n+1}^{2}-2 a_{n+1}=4 S_{n}.
\]
Subtracting (1) from (2) gives
\[
\begin{array}{l}
3\left(a_{n+1}^{2}-a_{n}^{2}\right)-2\left(a_{n+1}-a_{n}\right)=4 a_{n} \\
\Rightarrow 3\left(a_{n+1}+a_{n}\right)\left(a_{n+1}-a_{n}\right)-2\left(a_{n+1}+a_{n}\right)=0 \\
\Rightarrow\left(a_{n+1}+a_{n}\right)\left[3\left(a_{n+1}-a_{n}\right)-2\right]=0.
\end{array}
\]

Given \( a_{n}>0 \), we have \( a_{n+1}-a_{n}=\frac{2}{3} \).
Therefore, \( \left\{a_{n}\right\} \) is an arithmetic sequence with a common difference of \( \frac{2}{3} \).
It is easy to find that \( a_{1}=\frac{2}{3} \).
Thus, \( S_{n}=n a_{1}+\frac{n(n-1)}{2} d=\frac{1}{3} n(n+1) \).