10、The sequence $\left\{a_{n}\right\}$ satisfies $a_{0}=0, a_{n+1}=\frac{8}{5} a_{n}+\frac{6}{5} \sqrt{4^{n}-a_{n}^{2}}(n \in N)$, and let $S_{n}=\sum_{i=0}^{n} a_{i}$, find $S_{n}$
10. Analysis: $a_{0}=1, a_{n+1}=2\left(\frac{4}{5} a_{n}+\frac{3}{5} \sqrt{4^{n}-a_{n}^{2}}\right)$, let $a_{n}=2^{n} \sin \alpha_{n}$. Set $\sin \theta=\frac{3}{5}, \cos \theta=\frac{4}{5}, 00 \\ \sin \left(\alpha_{n}-\theta\right), \cos \alpha_{n}<0\end{array}\right.$,
thus $a_{n}=\left\{\begin{array}{l}2^{n} \sin n \theta, n=0,1 \\ 2^{n} \sin 2 \theta, n=2 k, k \in N^{*} \\ 2^{n} \sin 3 \theta, n=2 k+1, k \in N^{*}\end{array}\right.$, Therefore, when $n=0$, $S_{n}=0$; when $n=1$, $S_{n}=\frac{6}{5}$; when $n=2 k, k \in N^{*}$:
$$
S_{n}=\frac{6}{5}+\left(4+4^{2}+\mathrm{L}+4^{k}\right) \sin 2 \theta+2\left(4+4^{2}+\mathrm{L}+4^{k-1}\right) \sin 3 \theta=\frac{238}{125} \times 2^{n}-\frac{322}{125}
$$

When $n=2 k+1, k \in N^{*}$, $S_{n}=S_{n-1}+a_{n}=\frac{238}{125} \times 2^{n-1}-\frac{342}{125}+\frac{117}{125} \times 2^{n}=\frac{236}{125} \times 2^{n}-\frac{322}{125}$,
Therefore, $S_{n}=\left\{\begin{array}{l}0, n=0 \\ \frac{238}{125} \times 2^{n}-\frac{322}{125}, n=2 k, k \in N^{*} \\ \frac{236}{125} \times 2^{n}-\frac{322}{125}, n=2 k+1, k \in N\end{array}\right.$