10. (20 points) The sequence $\left\{a_{n}\right\}$ has the sum of the first $n$ terms $S_{n}$ satisfying $a_{n}=\frac{1}{2}+\frac{1}{2} \sqrt{1+8 S_{n-1}}(n \geqslant 2)$.
Also, $a_{1}=1$, find the general term formula of the sequence.
10. From $S_{n}-S_{n-1}=\frac{1}{2}+\frac{1}{2} \sqrt{1+8 S_{n-1}}$
$$
\begin{aligned}
\Rightarrow & 2 S_{n}=2 S_{n-1}+1+\sqrt{1+8 S_{n-1}} \\
\Rightarrow & 8 S_{n}+1=\left(8 S_{n-1}+1\right)+4 \sqrt{1+8 S_{n-1}}+4 \\
& =\left(\sqrt{8 S_{n-1}+1}+2\right)^{2} \\
\Rightarrow & \sqrt{8 S_{n}+1}=\sqrt{8 S_{n-1}+1}+2 .
\end{aligned}
$$

Since $\sqrt{8 S_{1}+1}=\sqrt{8 a_{1}+1}=3$, we have
$$
\sqrt{8 S_{n}+1}=2 n+1 \text{. }
$$

Notice that, when $n=0$, it is agreed that $S_{0}=0$ also holds.
Thus, $S_{n}=\frac{(2 n+1)^{2}-1}{8}=\frac{n(n+1)}{2}$
$$
\Rightarrow a_{n}=S_{n}-S_{n-1}=n .
$$