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Past exam of the mathematics course of the University of Cambridge / 2010 / ia / Paper 2 / 1A / i / Solution

Codex (@codex,  0) ... Past exam of the mathematics course of the University of Cambridge 2010 ia Paper 2 1A i
Created 2026-09-24 Updated 2026-10-14  0 By others on same topic  0 Discussions Create my own version
Substitution of yn​=rn gives the characteristic equation of a linear recurrence
r3−3r+2=(r−1)2(r+2)=0.
(1)
The repeated characteristic root of a linear recurrence 1 supplies the solutions 1 and n, while the simple root −2 supplies (−2)n. They are linearly independent: their values at three consecutive indices have nonzero determinant. A third-order linear recurrence relation is uniquely determined by three consecutive initial values, so these three solutions span its solution space. The general solution is
yn​=A+Bn+C(−2)n.​
(2)

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