FM June 2025 p12 q03
4109
The sequence \(u_1, u_2, u_3, \ldots\) is such that \(u_1 = 5\) and \(u_{n+1} = 6u_n + 5\) for \(n \geq 1\).
(a) Prove by induction that \(u_n = 6^n - 1\) for all positive integers \(n\).
(b) Deduce that \(u_{2n}\) is divisible by \(u_n\) for \(n \geq 1\).
Solution
(a) Base case: For \(n = 1\), \(u_1 = 5 = 6^1 - 1\). Assume it is true for \(n = k\), so \(u_k = 6^k - 1\).
Then \(u_{k+1} = 6(6^k - 1) + 5 = 6^{k+1} - 1\).
So, it is also true for \(n = k+1\). Hence, by induction, \(u_n = 6^n - 1\) is true for all positive integers.
(b) \(\frac{u_{2n}}{u_n} = \frac{6^{2n} - 1}{6^n - 1} = \frac{(6^n - 1)(6^n + 1)}{6^n - 1} = 6^n + 1\).
Thus, \(u_{2n}\) is divisible by \(u_n\).
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