Fibonacci numbers/Simpson formula/Exercise/Solution

The initial case for ${}n=2$ is clear because of

${}f_{3}\cdot f_{1}-f_{2}^{2}=2\cdot 1-1^{2}=1=(-1)^{2}\,.$

We assume now that the statement is already known for some ${}n$ and we consider the statement for ${}n+1$. This follows from

${}{\begin{aligned}f_{n+2}f_{n}-f_{n+1}^{2}&={\left(f_{n}+f_{n+1}\right)}f_{n}-f_{n+1}^{2}\\&=f_{n}^{2}+f_{n+1}f_{n}-f_{n+1}^{2}\\&=f_{n}^{2}+f_{n+1}{\left(f_{n}-f_{n+1}\right)}\\&=f_{n}^{2}+f_{n+1}{\left(f_{n}-f_{n}-f_{n-1}\right)}\\&=f_{n}^{2}-f_{n+1}f_{n-1}\\&=(-1){\left(f_{n+1}f_{n-1}-f_{n}^{2}\right)}\\&=(-1)(-1)^{n}\\&=(-1)^{n+1}\end{aligned}}$

To the exercise