Prove $F_{n+1}F_{n-1}-(F_{n})^2=(-1)^n$ without induction I am asked to pove the statement about fibonacci sequence. The task is from the passage about series and sequences. But the proof seems to need induction way, doesn't it? 
Prove the statement $$F_{n+1}F_{n-1}-(F_{n})^2=(-1)^n$$ for all $n\ge 1$.   
How can I prove this by thinking about limit and convergence? 
 A: I think this depends on what is meant by "mathematical induction". Recall that $F_n$ can be solved in the following way. As
\begin{equation}
\begin{pmatrix}F_{n+1}\\F_n\end{pmatrix}
=\begin{pmatrix}1 & 1\\ 1&0\end{pmatrix}
\begin{pmatrix}F_n\\F_{n-1}\end{pmatrix},\tag{1}
\end{equation}
we have
\begin{equation}
\begin{pmatrix}F_{n+1}\\F_n\end{pmatrix}
=\begin{pmatrix}1 & 1\\ 1&0\end{pmatrix}^n
\begin{pmatrix}F_1\\F_0\end{pmatrix}.\tag{2}
\end{equation}
In a broad sense, the derivation of $(2)$ from $(1)$ is mathematical induction, but in normal context, I think this is seldom regarded as such. Now, if this is not considered as mathematical induction, we can solve $(2)$ directly and hence we may and verify your inequality simply by plugging in the solution for $F_n$. (Edit: Or better, as Qiaochu Yuan suggests, one may simply compute the determinant of the matrix $n$-th power in $(2)$.)
A: As, $F_{n+2}=F_{n+1}+F_n,$ the Characteristic equation of the recurrence relation will be $t^2-t-1=0$
If $a,b$ are the roots of the equation, $a+b=1,ab=-1$ and $F_n=Aa^n+Bb^n$ where $A,B$ are arbitrary constants. 
So, 
$$F_{n+1}F_{n-1}-(F_{n})^2=(Aa^{n+1}+Bb^{n+1})(Aa^{n-1}+Bb^{n-1})-(Aa^n+Bb^n)^2$$
$$=AB\{a^{n+1}b^{n-1}+b^{n+1}a^{n-1}-2(ab)^n\}$$
$$=AB(ab)^n\{\frac{a^2+b^2}{ab}-2\}$$
$$=-5AB(-1)^n$$ as $\frac{a^2+b^2}{ab}-2=\frac{(a+b)^2}{ab}-4=-1-4=-5$
Now, $0=F_0=Aa^0+Bb^0\implies B=-A$
and $1=F_1=Aa+Bb=A(a-b)\implies A=\frac1{a-b}$
So,  $AB=-A^2=\frac{-1}{(a-b)^2}=\frac{-1}{(a+b)^2-4ab}=-\frac15$
