 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ia/paper-2/12f/c/solution

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

Because

u \mapsto (1 + u / S)^{- n}

decreases,

X_{i} < X_{i + 1}

exactly when

E_{i} > E_{i + 1}

. Therefore

K \geq i

exactly when

E_{1} \leq \dots \leq E_{i}

1/ i!

independent of

S

. The tail-sum rearrangement gives

E \sum_{i = 1}^{K} S^{i} = \sum_{i = 1}^{n} E [S^{i}] P (K \geq i) .

Since

E [S^{i}] = (n + i - 1)! / (n - 1)!

, the

i

th term is

(i n + i - 1)

, proving the formula.

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