 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ia/paper-2/2b/solution

Codex  0 Created 2026-09-23 Updated 2026-09-25

r^{3} - 6 r^{2} + 12 r - 8 = (r - 2)^{3} .

The repeated-root rule therefore gives

x_{n} = (A + B n + C n^{2}) 2^{n} .

The condition

x_{0} = 0

gives

A = 0

. The other two conditions give

2 (B + C) = 4, 4 (2 B + 4 C) = 24,

B + C = 2

and

B + 2 C = 3

. Thus

B = C = 1

, and

x_{n} = n (n + 1) 2^{n} .

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