A sequence $a_0,a_1,a_2,\cdots,a_n,\cdots$ satisfies that $a_0=3$, and $(3-a_{n-1})(6+a_n)=18$, then the value of $\sum_{i=0}^{n}\frac{1}{a_i}$ is________.
1. Given the sequence \(a_0, a_1, a_2, \ldots\) with \(a_0 = 3\) and the recurrence relation \((3 - a_{n-1})(6 + a_n) = 18\), we need to find the value of \(\sum_{i=0}^{n} \frac{1}{a_i}\).

2. Let \(x_j = \frac{1}{a_j}\) for all \(j = 0, 1, 2, \ldots\). Substituting \(x_j\) into the given recurrence relation, we get:
   \[
   (3 - \frac{1}{x_{j+1}})(6 + \frac{1}{x_j}) = 18
   \]

3. Expanding and simplifying the equation:
   \[
   (3 - \frac{1}{x_{j+1}})(6 + \frac{1}{x_j}) = 18
   \]
   \[
   18 + \frac{3}{x_j} - \frac{6}{x_{j+1}} - \frac{1}{x_{j+1} x_j} = 18
   \]
   \[
   \frac{3}{x_j} - \frac{6}{x_{j+1}} - \frac{1}{x_{j+1} x_j} = 0
   \]
   \[
   3x_{j+1} - 6x_j x_{j+1} - 1 = 0
   \]
   \[
   3x_{j+1} - 6x_j x_{j+1} = 1
   \]
   \[
   x_{j+1}(3 - 6x_j) = 1
   \]
   \[
   x_{j+1} = \frac{1}{3 - 6x_j}
   \]

4. To simplify further, we notice that:
   \[
   x_{j+1} = 2x_j + \frac{1}{3}
   \]

5. We can express \(x_{j+1}\) in terms of a geometric progression. Let \(y_j = x_j + \frac{1}{3}\). Then:
   \[
   y_{j+1} = 2y_j
   \]

6. Since \(y_j\) is a geometric sequence with ratio \(r = 2\), we have:
   \[
   y_j = y_0 \cdot 2^j
   \]

7. Given \(x_0 = \frac{1}{a_0} = \frac{1}{3}\), we have:
   \[
   y_0 = x_0 + \frac{1}{3} = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}
   \]
   \[
   y_j = \frac{2}{3} \cdot 2^j
   \]

8. Therefore:
   \[
   x_j = y_j - \frac{1}{3} = \frac{2}{3} \cdot 2^j - \frac{1}{3} = \frac{2^{j+1}}{3} - \frac{1}{3} = \frac{2^{j+1} - 1}{3}
   \]

9. Now, we need to find \(\sum_{i=0}^{n} x_i\):
   \[
   \sum_{i=0}^{n} x_i = \sum_{i=0}^{n} \left( \frac{2^{i+1} - 1}{3} \right)
   \]
   \[
   = \frac{1}{3} \sum_{i=0}^{n} (2^{i+1} - 1)
   \]
   \[
   = \frac{1}{3} \left( \sum_{i=0}^{n} 2^{i+1} - \sum_{i=0}^{n} 1 \right)
   \]
   \[
   = \frac{1}{3} \left( 2 \sum_{i=0}^{n} 2^i - (n+1) \right)
   \]

10. The sum of a geometric series \(\sum_{i=0}^{n} 2^i\) is:
    \[
    \sum_{i=0}^{n} 2^i = 2^{n+1} - 1
    \]

11. Substituting this back, we get:
    \[
    \sum_{i=0}^{n} x_i = \frac{1}{3} \left( 2(2^{n+1} - 1) - (n+1) \right)
    \]
    \[
    = \frac{1}{3} \left( 2^{n+2} - 2 - n - 1 \right)
    \]
    \[
    = \frac{1}{3} \left( 2^{n+2} - n - 3 \right)
    \]

The final answer is \(\boxed{\frac{2^{n+2} - n - 3}{3}}\).