 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ia/paper-2/10f/iii/solution

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

Using the mean

1/ p

of a geometric distribution and the decomposition in part (ii),

E (T_{n}) = j = 1 \sum n \frac{n}{j} = n j = 1 \sum n \frac{1}{j} = n H_{n} .

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