 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2010/ia/paper-1/4e/i/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

Pairing the first and last terms of the arithmetic progression gives

\sum_{k = 1}^{n} k = n (n + 1) /2

\frac{1 + 2 + \dots + n}{n ^{2}} = \frac{1}{2} + \frac{1}{2 n} ⟶ \frac{1}{2} .

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