 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ia/paper-1/6c/d/ii/solution

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

The recurrence becomes

d_{n} = - 2 d_{n - 1} - d_{n - 2}, d_{0} = 1, d_{1} = - 2.

Its characteristic polynomial is

r^{2} + 2 r + 1 = (r + 1)^{2} .

Therefore

d_{n} = (α + β n) (- 1)^{n}

α = 1

and

β = 1

, so

d_{n} = (- 1)^{n} (n + 1) .

 New to topics? Read the docs here!