 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2015/discrete-mathematics-continuous-mathematics-and-probability/2/d/solution

n = 1

F_{0} F_{2} - F_{1}^{2} = 0 \cdot 1 - 1 = - 1 = (- 1)^{1} .

Suppose

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

F_{n - 1} = F_{n + 1} - F_{n}

and hence

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

The induction proves

F_{n - 1} F_{n + 1} - F_{n}^{2} = (- 1)^{n} (n \geq 1) .

 New to topics? Read the docs here!