4. Given the sequence $a_{0}, a_{1}, \cdots, a_{n}, \cdots$ satisfies the relation $\left(3-a_{n+1}\right)\left(6+a_{n}\right)=18$, and $a_{0}=3$. Then
$$
\sum_{i=0}^{n} \frac{1}{a_{i}}=
$$
$\qquad$
(Let $b_{n}=\frac{1}{a_{n}}(n=0,1, \cdots)$.
Then $[3-$
$\left.\frac{1}{b_{n+1}}\right)\left(6+\frac{1}{b_{n}}\right)=18$, i.e., $3 b_{n+1}-6 b_{n}-$
$1=0$. So,
$b_{n+1}=2 b_{n}+\frac{1}{3}$. Using the fixed point method, we can find
$b_{n}$. Therefore, $\sum_{i=0}^{n} \frac{1}{a_{i}}=\frac{1}{3}\left(2^{n+2}-n-3\right)$. )