2. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{0}=0,5 a_{n+1}=4 a_{n}+3 \sqrt{1-a_{n}^{2}}(n \in \mathbf{N}) \text {. }
$$

Let $S_{n}=\sum_{i=0}^{n} a_{i}$. Then $S_{51}=$ $\qquad$
2. 48 .

Let $a_{n}=\sin \alpha_{n}$.
Suppose $\sin \theta=\frac{3}{5}, \cos \theta=\frac{4}{5}$.
Then $30^{\circ}0 ; \\
\sin \left(\alpha_{n}-\theta\right), \cos \alpha_{n}<0 .
\end{array}\right.
$

Thus, $a_{n}=\left\{\begin{array}{l}\sin n \theta, n=0,1 ; \\ \sin 2 \theta, n=2 k\left(k \in \mathbf{Z}_{+}\right) ; \\ \sin 3 \theta, n=2 k+1\left(k \in \mathbf{Z}_{+}\right) .\end{array}\right.$
Therefore, when $n=0$, $S_{n}=0$;
when $n=1$, $S_{n}=\frac{3}{5}$.
Hence, $S_{51}=\frac{3}{5}+25 \sin 2 \theta+25 \sin 3 \theta=48$.