Question 31
The sequence $\{a_n\}_{n\in\mathbb N}$ has first term $a_1=3$ and satisfies
$$
(3-a_{n+1})(6+a_n)=18
$$
for every $n\ge1$. Prove that for every positive integer $n$,
$$
\sum_{k=1}^n\frac1{a_k}=\frac{2^{n+1}}3-\frac{n+2}3.
$$
$$
(3-a_{n+1})(6+a_n)=18
$$
for every $n\ge1$. Prove that for every positive integer $n$,
$$
\sum_{k=1}^n\frac1{a_k}=\frac{2^{n+1}}3-\frac{n+2}3.
$$