 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2017/ia/paper-1/5a/b/i/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

The identity in part (a), together with

∣ a ∣ = 1

, gives

x_{2} = λ^{2} [a (a \cdot x_{0}) - x_{0}] .

For every

n \geq 1

x_{n}

a

, because it is a scalar multiple of

a \times x_{n - 1}

. Therefore

x_{n + 2} = λ^{2} a \times (a \times x_{n}) = - λ^{2} x_{n} .

Also

∣ x_{1} ∣ = λ ∣ a \times x_{0} ∣

and

∣ x_{2} ∣ = λ^{2} ∣ a \times x_{0} ∣

. Induction using the two-step recurrence yields

∣ x_{n} ∣ = λ^{n} ∣ a \times x_{0} ∣ (n \geq 1) .

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