Prove $F^2_{n+1} - F_nF_{n+2} = (-1)^n$ This is a question about Fibonacci sequences, a sequence in which the previous terms build up upon the current term (e.g. $F_1 + F_2 = F_3$ where $F_1 = F_2 = 1$). How would I go about proving $F^2_{n+1} - F_nF_{n+2} = (-1)^n$?

you can see $$\begin{bmatrix}
f_{n+2} & f_{n+1}\\ 
f_{n+1}&  f_{n}
\end{bmatrix}=(\begin{bmatrix}
1 &1 \\ 
1 & 0
\end{bmatrix})^{n+1}$$for example put n=1 $$\begin{bmatrix}
f_{3} & f_{2}\\ 
f_{2}&  f_{1}
\end{bmatrix}=(\begin{bmatrix}
1 &1 \\ 
1 & 0
\end{bmatrix})^{2}=\begin{bmatrix}
1 &1 \\ 
1 & 0
\end{bmatrix}\begin{bmatrix}
1 &1 \\ 
1 & 0
\end{bmatrix}=\begin{bmatrix}
2 &1 \\ 
1 & 1
\end{bmatrix} $$if you put n=2 $$\begin{bmatrix}
f_{4} & f_{3}\\ 
f_{3}&  f_{2}
\end{bmatrix}=(\begin{bmatrix}
1 &1 \\ 
1 & 0
\end{bmatrix})^{3}=\begin{bmatrix}
2 &1 \\ 
1 & 0
\end{bmatrix}\begin{bmatrix}
1 &1 \\ 
1 & 0
\end{bmatrix}=\begin{bmatrix}
3 &2 \\ 
2 & 1
\end{bmatrix} $$f(1)=1  ,f(2)=1  ,f(3)=2,  f(4)=3 and so on ...
so now det(l.h.s)=det(R.h.s) $$det(\begin{bmatrix}
f_{n+2} & f_{n+1}\\ 
f_{n+1}&  f_{n}
\end{bmatrix})=det(\begin{bmatrix}
1 &1 \\ 
1 & 0
\end{bmatrix})^{n+1}) \\f_{n+2}f_{n}-f_{n+1}f_{n+1}=0-(-1)^{n+1}\\f_{n+2}f_{n}-f_{n+1}f_{n+1}=-(-1)(-1)^{n}=(-1)^{n}$$
