 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ii/paper-1/2g/b/solution

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

The assertion is clear for

T_{1} = x

. If

T_{n}

has degree

n

and leading coefficient

2^{n - 1}

, the recurrence shows that the leading term of

T_{n + 1}

2 x \cdot 2^{n - 1} x^{n} = 2^{n} x^{n + 1};

the lower-degree polynomial

T_{n - 1}

cannot cancel it. Hence

lc (T_{n}) = 2^{n - 1} (n \geq 1) .

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