2. The positive sequence $a_{0}, a_{1}, \cdots, a_{n}, \cdots$ satisfies $\sqrt{a_{n} a_{n-2}}$ $\sqrt{a_{n-1} a_{n-2}}=2 a_{n-1}(n \geqslant 2)$ and $a_{0}=a_{1}=1$, find the general term.
2. Solution: From $\sqrt{a_{n} a_{n-2}}-\sqrt{a_{n-1} a_{n-2}}=2 a_{n-1}$ we get $\sqrt{\frac{a_{n}}{a_{n-1}}}-2 \sqrt{\frac{a_{n-1}}{a_{n-2}}}=1$, which means $\sqrt{\frac{a_{n}}{a_{n-1}}}+1=$ $2\left(\sqrt{\frac{a_{n-1}}{a_{n-2}}}+1\right)$

Let $b_{n}=\sqrt{\frac{a_{n}}{a_{n-1}}}+1$, then $\left\{b_{n}\right\}$ is a geometric sequence with the first term $\sqrt{\frac{a_{1}}{a_{0}}}+$ $1=2$, and common ratio 2,

So $b_{n}=\sqrt{\frac{a_{n}}{a_{n-1}}}+1=2^{n}$, thus $\frac{a_{n}}{a_{n-1}}=\left(2^{n}-1\right)^{2}$, so $a_{n}=\frac{a_{n}}{a_{n-1}} \cdot \frac{a_{n-1}}{a_{n-2}} \cdots \frac{a_{2}}{a_{1}} \cdot \frac{a_{1}}{a_{0}} \cdot a_{0}$ $=\prod_{k=1}^{n}\left(2^{k}-1\right)^{2}$.
Note: $\prod_{i=1}^{n} C_{i}=C_{1} \cdot C_{2} \cdot \cdots \cdot C_{n}$.