 Solution

ID: 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

For

n = 0

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

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

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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