Prove: $n \in Z^{\geq 2}$, $f_nf_{n+1} - f_{n-1}f_{n+2} = (-1)^{n+1}$

Question

$n \in Z^{\geq 2}$, $f_nf_{n+1} - f_{n-1}f_{n+2} = (-1)^{n+1}$.

How do you do the inductive step of this proof, every time I do it I cannot find a way to use the definition of a Fibonacci sequence to simplify the right side of the equation enough.

Answer 1

Just keep using Fibonacci recursion for "large" numbers like $n+1,$ $n+2$ and $n+3:$ $$ \begin{aligned} f_{n+1}f_{n+2}-f_nf_{n+3}&=f_{n+1}(f_{n+1}+f_n)-f_n(f_{n+2}+f_{n+1})\\ \\&=f_{n+1}(f_{n+1}+f_n)-f_n(2f_{n+1}+f_n)\\ \\&=(f_n+f_{n-1})(f_{n+1}+f_n)-f_n(2f_{n+1}+f_n)\\ \\&=f_nf_{n+1}+f_n^2+f_{n-1}f_{n+1}+f_{n-1}f_n-2f_nf_{n+1}-f_n^2\\ \\&=f_nf_{n+1}+f_{n-1}f_{n+1}+f_{n-1}f_n-2f_nf_{n+1}\\ \\&=f_{n-1}(f_n+f_{n+1})-f_nf_{n+1}\\ \\&=f_{n-1}f_{n+2}-f_nf_{n+1} \end{aligned} $$

Answer 2

Here is an approach to how you might discover this result.

Notice that $$ f_nf_{n+1} - f_{n-1}f_{n+2} = \det \begin{pmatrix}f_{n}&f_{n+2}\\f_{n-1}&f_{n+1}\end{pmatrix} $$

Using the definition of the Fibonacci sequence, we have $$ \begin{pmatrix}f_{n}&f_{n+2}\\f_{n-1}&f_{n+1}\end{pmatrix} = \begin{pmatrix}1&1\\1&0\end{pmatrix} \begin{pmatrix}f_{n-1}&f_{n+1}\\f_{n-2}&f_{n}\end{pmatrix} $$ and so, by induction, $$ \begin{pmatrix}f_{n}&f_{n+2}\\f_{n-1}&f_{n+1}\end{pmatrix} = \begin{pmatrix}1&1\\1&0\end{pmatrix}^{n-1} \begin{pmatrix}f_{1}&f_{3}\\f_{0}&f_{2}\end{pmatrix} = \begin{pmatrix}1&1\\1&0\end{pmatrix}^{n-1} \begin{pmatrix}1&2\\0&1\end{pmatrix} $$

Now take determinants, noting that $(-1)^{n-1}=(-1)^{n+1}$.