 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ia/paper-4/6d/c/solution

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

Put

S_{n} = \sum_{k = 0}^{⌊(n - 1) /2 ⌋} (k n - k - 1) .

The initial values are

S_{1} = S_{2} = 1

. Using Pascal's identity and the convention that an out-of-range binomial coefficient is zero,

S_{n + 2} = k \geq 0 \sum (k n + 1 - k) = k \geq 0 \sum (k n - k) + k \geq 1 \sum (k - 1 n - k) = S_{n + 1} + S_{n} .

Thus

(S_{n})

F_{n} = k = 0 \sum ⌊(n - 1) /2 ⌋ (k n - k - 1) .

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