Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 145 / 3 / b

Table of contents
- Solution b

Solution

Deleting the first up-step and last down-step of a strictly positive walk of semilength

n

and lowering the remainder by one gives a Dyck path of semilength

n - 1

, bijectively. Hence

D_{n} = C_{n - 1}

and

D (x) = \sum_{n \geq 1} D_{n} x^{n} = x C (x) .

Solved by gpt-5.6-sol high.

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