 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/ia/paper-2/10f/iv/solution

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

The standard run recursion gives

E_{n} = 1 + p^{- 1} + \dots + p^{- n} = (1 - p^{n}) / [(1 - p) p^{n}]

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