China Mathematical Competition (Hainan)

Set $b_n=\frac{1}{a_n}$, $n=0,1,2,\dots$, then $(3-\frac{1}{b_{n+1}})(6+\frac{1}{b_n})=18$, namely, $$3b_{n+1}-6b_n-1=0.$$ Hence $b_{n+1}=2b_n+\frac{1}{3}$, or $b_{n+1}+\frac{1}{3}=2(b_n+\frac{1}{3})$. So $\{b_n+\frac{1}{3}\}$ is a geometric progression with common ratio $2$. Thus $$\begin{gathered} b_n+\frac{1}{3}=2^n\left(b_0+\frac{1}{3}\right)=2^n\left(\frac{1}{a_0}+\frac{1}{3}\right)=\frac{1}{3}\times 2^{n+1}, \\ {b}_n=\frac{1}{3}(2^{n+1}-1). \end{gathered}$$

$$\sum _{i=0}^n\frac{1}{a_i}=\sum _{i=0}^n{b}_i=\sum _{i=0}^n\frac{1}{3}(2^{i+1}-1)=\frac{1}{3}\left[\frac{2(2^{n+1}-1)}{2-1}-(n+1)\right]$$ $$=\frac{1}{3}(2^{n+2}-n-3).$$