 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/ii/paper-3/11g/b/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

Apply the formula to the complete quotient and use the determinant identity to express

b_{n}

. Eliminating the complete quotient from the consecutive expressions yields the stated quadratic for

q_{n + 1} / q_{n}

. Taking the compatible signs for adjacent indices and using

c_{n} \leq b_{j}

gives

2 c_{n} a_{n + 1} \leq 2 b_{n} a_{n + 1} \leq 1 - 4 b_{n} b_{n + 1} + 1 - 4 b_{n - 1} b_{n} \leq 2 1 - 4 c_{n}^{2} .

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