Prove $F^2_{n+1} - F_nF_{n+2} = (-1)^n$

Question

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 $F^2_{n+1} - F_nF_{n+2} = (-1)^n$?

Answer 1

$n=1$.

$$F^2_{2}-F_1F_3=1-1.2=-1=(-1)^1$$ Correct

Suppose true for $n$, so we assume that $F^2_{n+1} - F_nF_{n+2} = (-1)^n$ So for case $n+1$

$$ F^2_{n+2} - F_{n+1}F_{n+3} =(-1)(F^2_{n+1} - F_{n}F_{n+2})=(-1)(-1)^n=(-1)^{n+1}$$

Answer 2

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.

Answer 3

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 $$F^2_{n+1}-F_{n}F_{n+2}=(-1)^{n-1}[F^2_{2}-F_{1}F_{3}]=(-1)^n$$

Answer 4

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) $$det(\begin{bmatrix} f_{n+2} & f_{n+1}\\ f_{n+1}& f_{n} \end{bmatrix})=det(\begin{bmatrix} 1 &1 \\ 1 & 0 \end{bmatrix})^{n+1}) \\f_{n+2}f_{n}-f_{n+1}f_{n+1}=0-(-1)^{n+1}\\f_{n+2}f_{n}-f_{n+1}f_{n+1}=-(-1)(-1)^{n}=(-1)^{n}$$