7. (2004 National High School Mathematics Competition) Given 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 the value of $\sum_{i=0}^{n} \frac{1}{a_{i}}$ is $\qquad$ .
7. $\frac{2^{n+2}-n-3}{3}$.

Brief solution: From the given, we have $\frac{1}{a_{n+1}}+\frac{1}{3}=2\left(\frac{1}{a_{n}}+\frac{1}{3}\right)$, and $\frac{1}{a_{0}}+\frac{1}{3}=\frac{2}{3}$, so $\left\{\frac{1}{a_{n}}+\frac{1}{3}\right\}$ is a geometric sequence with the first term $\frac{2}{3}$ and common ratio 2 $(n \in \mathbf{N})$, hence $\frac{1}{a_{n}}+\frac{1}{3}=\left(\frac{2}{3}\right) \times 2^{n}, \frac{1}{a_{n}}=\frac{2^{n+1}-1}{3}$. Therefore, $\sum_{i=0}^{n} \frac{1}{a_{i}}=$ $\sum_{i=0}^{n} \frac{2^{n+1}-1}{3}=\frac{1}{3}\left[\sum_{i=0}^{n} 2^{i+1}-(n+1)\right]=\frac{2^{n+2}-n-3}{3}$.