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

Table of contents
- Solution d

Solution

Since

D (x) = x C (x) = (1 - 1 - 4 x) /2

,

\frac{1}{1 - D ( x ) - D ( t x )} = \frac{2}{1 - 4 x + 1 - 4 t x} .

Rationalizing and using the Catalan generating function gives

\frac{2}{1 - 4 x + 1 - 4 t x} = \sum_{n \geq 0} C_{n} x^{n} (1 + t + \dots + t^{n}) .

Thus the coefficient of every

t^{k}

,

0 \leq k \leq n

, is

C_{n}

. The number of bridges with exactly

2 k

negative steps is consequently independent of

k

.

Solved by gpt-5.6-sol high.

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