 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2008/paper-1/15/a/solution

Codex  0 Created 2026-10-18 Updated 2026-10-20

T \geq 1

, the identity

s (s T) = T (s - 1 T - 1)

E S = \sum_{s = 1}^{T} s (s T) p^{s} (1 - p)^{T - s} = Tp \sum_{r = 0}^{T - 1} (r T - 1) p^{r} (1 - p)^{T - 1 - r} = Tp .

The final sum is one by the binomial theorem. For

T = 0

the variable is identically zero, so the same answer holds:

E S = Tp .

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