Example 3 (2004 National Competition) Given the sequence $a_{0}, a_{1}, a_{2}$, $\cdots, a_{n}, \cdots$ satisfies the relation $\left(3-a_{n+1}\right)\left(6+a_{n}\right)=18$ and $a_{0}=3$, then the value of $\sum_{i=0}^{n} \frac{1}{a_{i}}$ is $\qquad$
Let $b_{n}=\frac{1}{a_{n}}, n=0,1,2, \cdots$, then $\left(3-\frac{1}{b_{n+1}}\right)\left(6+\frac{1}{b_{n}}\right)=18$,
i.e., $3 b_{n+1}-6 b_{n}-1=0$.
$$
\therefore b_{n+1}=2 b_{n}+\frac{1}{3}, b_{n+1}+\frac{1}{3}=2\left(b_{n}+\frac{1}{3}\right),
$$

Thus, the sequence $\left\{b_{n}+\frac{1}{3}\right\}$ is a geometric sequence with a common ratio of 2,
$$
\begin{array}{c}
b_{n}+\frac{1}{3}=2^{n}\left(b_{0}+\frac{1}{3}\right)=2^{n}\left(\frac{1}{a_{0}}+\frac{1}{3}\right)=\frac{1}{3} \\
\times 2^{n+1}, \therefore b_{n}=\frac{1}{3}\left(2^{n+1}-1\right) . \\
\sum_{i=0}^{n} \frac{1}{a_{i}}=\sum_{i=0}^{n} b_{i}=\sum_{i=0}^{n} \frac{1}{3}\left(2^{i+1}-1\right)= \\
\frac{1}{3}\left[\frac{2\left(2^{n+1}-1\right)}{2-1}-(n+1)\right]=\frac{1}{3}\left(2^{n+2}-n-3\right) .
\end{array}
$$