11. 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 $\qquad$ .
$$
\frac{1}{3}\left(2^{n+2}-n-3\right)
$$
11. 【Analysis and Solution】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, $b_{n+1}=2 b_{n}+\frac{1}{3}, b_{n+1}+\frac{1}{3}=2\left(b_{n}+\frac{1}{3}\right)$, so the sequence $\left\{b_{n}+\frac{1}{3}\right\}$ is a geometric sequence with a common ratio of 2. $b_{n} \frac{1}{3}=2^{n}\left(b_{0}+\frac{1}{3}\right)=2^{n}\left(\frac{1}{a_{0}}+\frac{1}{3}\right)=\frac{1}{3} \times 2^{n+1} \therefore b_{n}=\frac{1}{3}\left(2^{n+1}-1\right)$.
$$
\sum_{i=0}^{n} \frac{1}{a_{i}}=\sum_{i=0}^{n} b_{i}=\sum_{i=0}^{n} \frac{1}{3}\left(2^{i+1}-1\right)=\frac{1}{3}\left[\frac{2\left(2^{n+1}-1\right)}{2-1}-(n+1)\right]=\frac{1}{3}\left(2^{n+2}-n-3\right) .
$$