 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ia/paper-2/6a/a/solution

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

The homogeneous characteristic equation is

r^{2} + r - 6 = (r - 2) (r + 3) = 0.

u_{n} = an + b

gives

- 4 an + 11 a - 4 b = n,

a = - 1/4

and

b = - 11/16

. Thus

u_{n} = A 2^{n} + B (- 3)^{n} - \frac{n}{4} - \frac{11}{16} .

The initial data yield

A = 1

and

B = - 5/16

, hence

u_{n} = 2^{n} - \frac{5}{16} (- 3)^{n} - \frac{n}{4} - \frac{11}{16} .

x^{n}

and sum for

n \geq 2

. Using

u_{0} = 0, u_{1} = 2

gives

(1 + x - 6 x^{2}) G (x) = x + \frac{x}{( 1 - x ) ^{2}} .

Equivalently,

G (x) = \frac{1}{1 - 2 x} - \frac{5}{16 ( 1 + 3 x )} - \frac{x}{4 ( 1 - x ) ^{2}} - \frac{11}{16 ( 1 - x )} .

Expanding each geometric series gives exactly the displayed formula for

u_{n}

, verifying consistency.

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