11. The sequence $a_{0}, a_{1}, a_{2}, \cdots, a_{n}, \cdots$, satisfies the relation (3$\left.a_{n+1}\right)\left(6+a_{n}\right)=18$, and $a_{0}=3$. Then $\sum_{i=0}^{n} \frac{1}{a_{i}}=$ $\qquad$ .
11. $\frac{1}{3}\left(2^{n+2}-n-3\right)$.

Let $b_{n}=\frac{1}{a_{n}}, n=0,1,2, \cdots$.
From the given condition, we have $3 b_{n+1}-6 b_{n}-1=0$. Therefore,
$$
b_{n+1}+\frac{1}{3}=2\left(b_{n}+\frac{1}{3}\right) \text {. }
$$

Thus, the sequence $\left\{b_{n}+\frac{1}{3}\right\}$ is a geometric sequence with a common ratio of 2.
$$
b_{n}+\frac{1}{3}=2^{n}\left(b_{0}+\frac{1}{3}\right)=2^{n}\left(\frac{1}{a_{0}}+\frac{1}{3}\right)=\frac{1}{3} \times 2^{n+1}
$$

That is, $b_{n}=\frac{1}{3}\left(2^{n+1}-1\right)$.
$$
\begin{array}{l}
\sum_{i=0}^{n} \frac{1}{a_{i}}=\sum_{i=0}^{n} b_{i}=\sum_{i=0}^{n} \frac{1}{3}\left(2^{i+1}-1\right) \\
=\frac{1}{3}\left[\frac{2\left(2^{n+1}-1\right)}{2-1}-(n+1)\right]=\frac{1}{3}\left(2^{n+2}-n-3\right) .
\end{array}
$$