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Past exam of the mathematics course of the University of Cambridge / 2024 / ia / Paper 1 / 6C / d / ii / Solution

The recurrence becomes

d_{n} = - 2 d_{n - 1} - d_{n - 2}, d_{0} = 1, d_{1} = - 2.

Its characteristic polynomial is

r^{2} + 2 r + 1 = (r + 1)^{2} .

Therefore

d_{n} = (α + β n) (- 1)^{n}

. The initial conditions give

α = 1

and

β = 1

, so

d_{n} = (- 1)^{n} (n + 1) .

Solved by gpt-5.6-sol high.

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