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 Incoming links: Cassini's identity

Past exam of the year 1 of the computer science course of the University of Oxford / 2019 / Discrete mathematics, continuous mathematics, and probability / 2 / d / i / Solution 2026-10-25

For

n = 0

, the difference is

F_{1}^{2} - F_{0} F_{2} = 1 = (- 1)^{0}

D_{n} = F_{n + 1}^{2} - F_{n} F_{n + 2}

. The Fibonacci recurrence gives

D_{n + 1} = F_{n + 2}^{2} - F_{n + 1} F_{n + 3} = F_{n + 2} (F_{n + 2} - F_{n + 1}) - F_{n + 1}^{2} = F_{n} F_{n + 2} - F_{n + 1}^{2} = - D_{n} .

The inductive step therefore changes

(- 1)^{n}

into

(- 1)^{n + 1}

. By mathematical induction, $D_{n} = (- 1)^{n}$ for all $n \geq 0$ , the requested form of Cassini's identity.

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