Let fn n012 be the Fibonacci sequence Use determinants and i

Let f_n, n=0,1,2,… be the Fibonacci sequence. Use determinants and induction to show that for any n1, (f_n+1)(f_n-1) (fn)²=(1)ⁿ

Solution

For n = 0, its obvious. (Check!) Now we suppose that F(n-1)*F(n+1) = (-1)^n + (F(n))^2 holds for n. We can rewrite this to F(n-1) * F(n+1) - (F(n))^2 = (-1)^n [*] Now we will replace all n by n+1 at the left-hand side of the equation, and show that it yields (-1)^(n+1). For that we use truth we suppose of [*] as well as application of the Fibonacci rule : F(m+2) = F(m)+F(m+1) for all m: F(n)*F(n+2) - (F(n+1))^2 = F(n)*[F(n+1) + F(n)] - F(n+1)*[F(n) + F(n-1)] = F(n)*F(n+1) + (F(n))^2 - F(n)*F(n+1) - F(n+1)*F(n-1) = (F(n))^2 - F(n+1)*F(n-1) = -(-1)^n = (-1)^(n+1) Hence the proof.