 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ii/paper-4/1g/solution

Codex  0 Created 2026-09-23 Updated 2026-09-25

3 < 11 < 4

x_{0} = 11, a_{0} = 3.

The successive complete quotients are

x_{1} = \frac{1}{11 - 3} = \frac{11 + 3}{2}, a_{1} = 3,

and

x_{2} = \frac{1}{x _{1} - 3} = 11 + 3, a_{2} = 6.

Taking one more reciprocal returns to

x_{1}

11 = [3; \overline{3, 6}] .

Write

z_{n} = p_{n} + q_{n} 11 .

The first convergents are

p_{0} / q_{0} = 3

and

p_{1} / q_{1} = 10/3

, whence

z_{1} = 10 + 311 = \frac{3 + 11}{2} (3 + 11) = \frac{3 + 11}{2} z_{0} .

z_{n + 1} = a_{n + 1} z_{n} + z_{n - 1},

where

a_{n + 1} = 3

for even

n

and

a_{n + 1} = 6

for odd

n

α = 3 + 11, β = \frac{3 + 11}{2} .

n

is even and

z_{n + 1} = β z_{n}

, then

z_{n + 2} = 6 z_{n + 1} + z_{n} = (6 + \frac{1}{β}) z_{n + 1} = (3 + 11) z_{n + 1} = α z_{n + 1},

because

β^{- 1} = 11 - 3

. If

n

is odd and

z_{n + 1} = α z_{n}

, then

z_{n + 2} = 3 z_{n + 1} + z_{n} = (3 + \frac{1}{α}) z_{n + 1} = \frac{3 + 11}{2} z_{n + 1} = β z_{n + 1},

because

α^{- 1} = (11 - 3) /2

p_{n + 1} + q_{n + 1} 11 = ⎩ ⎨ ⎧ (3 + 11) (p_{n} + q_{n} 11), \frac{3 + 11}{2} (p_{n} + q_{n} 11), n odd, n even .

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