11. 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 $\sum_{i=0}^{n} \frac{1}{a_{i}}=$ $\qquad$ .
$$
\begin{array}{l}
\left(3-a_{n+1}\right)\left(6+a_{n}\right)=18+3 a_{n}-6 a_{n+1}-a_{n+1} a_{n}=18 \Rightarrow 3 a_{n}-6 a_{n+1}=a_{n+1} a_{n} \\
\Rightarrow \frac{3}{a_{n+1}}-\frac{6}{a_{n}}=1 \Rightarrow \frac{1}{a_{n+1}}=\frac{2}{a_{n}}+\frac{1}{3} \Rightarrow \frac{1}{a_{n+1}}+\frac{1}{3}=2\left(\frac{1}{a_{n}}+\frac{1}{3}\right)
\end{array}
$$
$\Rightarrow$ The sequence $\left\{\frac{1}{a_{n}}+\frac{1}{3}\right\}$ is a geometric sequence with a common ratio of 2.
Also, $\frac{1}{a_{0}}+\frac{1}{3}=\frac{2}{3} \Rightarrow \frac{1}{a_{n}}+\frac{1}{3}=\frac{1}{3} \cdot 2^{n+1} \Rightarrow \frac{1}{a_{n}}=\frac{1}{3}\left(2^{n+1}-1\right)(n \in \mathbf{N})$.
Thus, $\sum_{i=0}^{n} \frac{1}{a_{i}}=\frac{1}{3} \sum_{i=0}^{n}\left(2^{n+1}-1\right)=\frac{1}{3}\left[\left(2^{n+2}-2\right)-(n+1)\right]=\frac{1}{3}\left(2^{n+2}-n-3\right)$.
