8. (2004 National High School Competition Question) Given the sequence $a_{0}, a_{1}, \cdots, a_{n}, \cdots$ satisfies $\left(3-a_{n+1}\right)\left(6+a_{n}\right)=18$ and $a_{0}=3$, then $\sum_{i=0}^{n} \frac{1}{a_{i}}$ equals $\qquad$ -
8. $\frac{1}{3}\left(2^{n+2}-n-3\right)$. Let $b_{n}=\frac{1}{a_{n}}$, then $\left(3-\frac{1}{b_{n+1}}\right)\left(6+\frac{1}{b_{n}}\right)=18$, which means $3 b_{n+1}-6 b_{n}-1=0, b_{n+1}+\frac{1}{3}=$ $2\left(b_{n}+\frac{1}{3}\right)$ (here $-\frac{1}{3}$ is the root of $3 x-6 x-1=0$). It can be seen that $b_{n}+\frac{1}{3}$ is a geometric sequence with the first term $b_{0}+\frac{1}{3}=\frac{1}{a_{0}}+\frac{1}{3}=\frac{2}{3}$, and the common ratio is 2, so $b_{n}+\frac{1}{3}=\frac{2}{3} \cdot 2^{n}, b_{n}=\frac{2^{n+1}}{3}-\frac{1}{3}$, thus $\sum_{i=0}^{n} \frac{1}{a_{i}}=\sum_{i=0}^{n} b_{i}=\sum_{i=0}^{n} \frac{2^{n+1}-1}{3}=$ $\frac{1}{3}\left[\frac{2\left(2^{n+1}-1\right)}{2-1}-(n+1)\right]=\frac{1}{3}\left(2^{n+2}-n-3\right)$