10. Let the sequence $\left\{a_{n}\right\}$ satisfy $a_{0}=2, a_{n}=\frac{2 a_{n-1}+6}{a_{n-1}+1}(n \geqslant 1)$, then $a_{n}=$
10. $a_{n}=\frac{3 \cdot 4^{n+1}+2 \cdot(-1)^{n+1}}{4^{n+1}+(-1)^{n}}\left(n \in \mathbf{N}^{*}\right)$.

Analysis: We introduce an undetermined coefficient $\lambda$, and try to transform the given problem into a familiar one. First, find the fixed points $\lambda_{1}, \lambda_{2}$ of the sequence $\left\{a_{n}\right\}$, then the sequence $\left\{\frac{a_{n}-\lambda_{1}}{a_{n}-\lambda_{2}}\right\}$ is a geometric sequence.
Brief solution: $a_{n+1}-\lambda=\frac{2 a_{n}+6}{a_{n}+1}-\lambda=\frac{(2-\lambda) a_{n}+6-\lambda}{a_{n}+1}=\frac{2-\lambda}{a_{n}+1}\left(a_{n}+\frac{6-\lambda}{2-\lambda}\right)$.
Let $-\lambda=\frac{6-\lambda}{2-\lambda}$, we get $\lambda^{2}-\lambda-6=0$, solving this gives $\lambda_{1}=3, \lambda_{2}=-2$.
Thus, $a_{n+1}-3=\frac{-1}{a_{n}+1}\left(a_{n}-3\right), a_{n+1}+2=\frac{4}{a_{n}+1}\left(a_{n}+2\right)$.
Therefore, $\frac{a_{n+1}-3}{a_{n+1}+2}=-\frac{1}{4} \cdot \frac{a_{n}-3}{a_{n}+2}$, which means $\left\{\frac{a_{n}-3}{a_{n}+2}\right\}$ is a geometric sequence with a common ratio of $-\frac{1}{4}$.
Hence, $\frac{a_{n}-3}{a_{n}+2}=\frac{a_{0}-3}{a_{0}+2} \times\left(-\frac{1}{4}\right)^{n+1}$.
Therefore, $a_{n}=\frac{3 \cdot 4^{n+1}+2 \cdot(-1)^{n+1}}{4^{n+1}+(-1)^{n}}\left(n \in \mathbf{N}^{*}\right)$.