Question 5
Find all positive integers $ a_{1}, a_{2}, \ldots, a_{n} $ such that
$$ \frac{99}{100}=\frac{a_{0}}{a_{1}}+\frac{a_{1}}{a_{2}}+\cdots+\frac{a_{n-1}}{a_{n}}, $$
where $ a_{0}=1 $ and $ (a_{k+1}-1)a_{k-1}\geq a_{k}^{2}(a_{k}-1) $ for $ k=1,2,\ldots,n-1 $.
$$ \frac{99}{100}=\frac{a_{0}}{a_{1}}+\frac{a_{1}}{a_{2}}+\cdots+\frac{a_{n-1}}{a_{n}}, $$
where $ a_{0}=1 $ and $ (a_{k+1}-1)a_{k-1}\geq a_{k}^{2}(a_{k}-1) $ for $ k=1,2,\ldots,n-1 $.