 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2010/ia/paper-2/1a/ii/solution

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

Let

L [y]_{n} = y_{n + 3} - 3 y_{n + 1} + 2 y_{n}

. For the exponential trial sequence

D 2^{n}

L [D 2^{n}] = (8 - 6 + 2) D 2^{n} = 4 D 2^{n} .

Thus

D = 1/4

gives a particular solution. The difference of any two solutions of the inhomogeneous linear recurrence satisfies the homogeneous linear recurrence relation in part (i). Therefore

y_{n} = A + B n + C (- 2)^{n} + \frac{2 ^{n}}{4} .

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