Let $a_{0},a_{1},a_{2},a_{3},\cdots \cdots a_{n}$ be a sequence of numbers satisying $(3-a_{n+1})(6+a_{n}) = 18$ and $a_{0} = 3.$ then find $\displaystyle \sum^{10}_{i=0}\frac{1}{a_{i}}$
from $18-6a_{n+1}+3a_{n}-a_{n}a_{n+1} = 18.$ So $6a_{n+1}-3a_{n}+a_{n}a_{n+1} =0$
want be able to go further, could some help me, thanks
Dividing the both sides of $$6a_{n+1}-3a_n+a_na_{n+1}=0$$ by $a_na_{n+1}\ (\not=0)$ gives $$b_{n+1}=2b_n+\frac 13\tag1$$ where $b_n=\frac{1}{a_n}$.
Now we have $$(1)\iff b_{n+1}+\frac 13=2\left(b_n+\frac 13\right)$$ I think that you can take it from here.
Hint
Starting from $$6a_{n+1}-3a_{n}+a_{n}a_{n+1} =0$$ divide each term by $a_{n}a_{n+1}$ to get $$\frac 6{a_n}-\frac 3{a_{n+1}}+1=0$$ Now, define $b_n=\frac 1{a_n}$ and solve.
By using the assumption $a_0=3$ and replacing it in your above result we get $6a_1-9+3a_1=0$ so we get $a_1=1$. Now by replacing this value in $a_1$ we deduce $a_2=\frac{7}{3}$. By continuing this process you can calculate all values of $a_i$s then you can find the sum of reciprocals.
From the condition we have $$-6a_{n+1}+3a_n-a_na_{n+1}=0$$ or $$-\frac{6}{a_n}+\frac{3}{a_{n+1}}-1=0$$ or $$-6\left(\frac{1}{a_0}+\frac{1}{a_1}+...+\frac{1}{a_{10}}\right)+3\left(\frac{1}{a_1}+...+\frac{1}{a_{11}}\right)-11=0$$ and for the ending it remains to find a value of $a_{11}$.