 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/ia/paper-4/1f/ii/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

The same recurrence and

2^{0} ∣ u_{0}

2^{1} ∣ u_{1}

prove the claim: if

2^{n - 1} ∣ u_{n - 1}

and

2^{n - 2} ∣ u_{n - 2}

, both terms on the right are divisible by

2^{n}

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