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 $F^2_{n+1} - F_nF_{n+2} = (-1)^n$?
 A: Hint: Prove that $$F^2_{n+1} - F_nF_{n+2} = -(F^2_{n} - F_{n-1}F_{n+1})$$
and the result follows immediately by induction.
A: I think I can without  this by induction
Note
$$F_{n+1}F_{n}=(F_{n+2}-F_{n})F_{n}=F_{n+2}F_{n}-F^2_{n}\tag{1}$$
and by other hand we also
$$F_{n}F_{n+1}=(F_{n+1}-F_{n-1})F_{n+1}=F^2_{n+1}-F_{n+1}F_{n-1}\tag{2}$$
with $(1)=(2)$ so 
$$F^2_{n+1}-F_{n}F_{n+2}=-(F^2_{n}-F_{n+1}F_{n-1})$$
so $\{F^2_{n+1}-F_{n}F_{n+2}\}$ is Geometric progression
$$F^2_{n+1}-F_{n}F_{n+2}=(-1)^{n-1}[F^2_{2}-F_{1}F_{3}]=(-1)^n$$
A: $n=1$.
$$F^2_{2}-F_1F_3=1-1.2=-1=(-1)^1$$ 
Correct
Suppose true for $n$, so we assume that $F^2_{n+1} - F_nF_{n+2} = (-1)^n$
So for case $n+1$
$$ F^2_{n+2} - F_{n+1}F_{n+3} =(-1)(F^2_{n+1} - F_{n}F_{n+2})=(-1)(-1)^n=(-1)^{n+1}$$
A: 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) $$det(\begin{bmatrix}
f_{n+2} & f_{n+1}\\ 
f_{n+1}&  f_{n}
\end{bmatrix})=det(\begin{bmatrix}
1 &1 \\ 
1 & 0
\end{bmatrix})^{n+1}) \\f_{n+2}f_{n}-f_{n+1}f_{n+1}=0-(-1)^{n+1}\\f_{n+2}f_{n}-f_{n+1}f_{n+1}=-(-1)(-1)^{n}=(-1)^{n}$$
