FM June 2022 p12 q03
4234
The sequence of positive numbers \(u_1, u_2, u_3, \ldots\) is such that \(u_1 > 4\) and, for \(n \geq 1\),
\(u_{n+1} = \frac{u_n^2 + u_n + 12}{2u_n}.\)
(a) By considering \(u_{n+1} - 4\), or otherwise, prove by mathematical induction that \(u_n > 4\) for all positive integers \(n\). [5]
(b) Show that \(u_{n+1} < u_n\) for \(n \geq 1\). [3]
Solution
(a) Base case: \(u_1 > 4\) (given).
Inductive step: Assume \(u_k > 4\) for some positive integer \(k\).
Then \(u_{k+1} - 4 = \frac{u_k^2 + u_k + 12}{2u_k} - 4 = \frac{u_k^2 - 7u_k + 12}{2u_k}\).
\(u_{k+1} - 4 = \frac{(u_k - 3)(u_k - 4)}{2u_k} > 0\).
Hence, by induction, \(u_n > 4\) for all positive integers \(n\).
(b) \(u_{n+1} - u_n = \frac{-u_n^2 + u_n + 12}{2u_n} = \frac{-(u_n - 4)(u_n + 3)}{2u_n}\).
So \(u_n > 4 \Rightarrow u_{n+1} - u_n < 0\).
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