Induction with Fibonacci Numbers

I am trying to figure out how to show that f_(n+1)f_(n-1)-(f_n)^2=(-1)^n when n is a positive integer.

__My Work__

P(n) is f_(n+1)f_(n-1)-(f_n)^2=(-1)^n for n>=1

Basis step - P(1) is true because f_2f_0-(f_1)^2=1*0-1=-1=(-1)^1

Inductive step - Assume P(k)=f_(k+1)f_(k-1)-(f_k)^2=(-1)^k is true, then P(k+1)=f_(k+2)f_(k)-(f_(k+1))^2=(-1)^(k+1) is true

P(k+1)=[f_(k+1)f_(k-1)-(f_k)^2]+[f_(k+2)f_(k)-(f_(k+1))^2]=(-1)^k+[f_(k+2)f_(k)-(f_(k+1))^2]

I can't figure out how to get this equation to equal (-1)^(k+1)

My Work So Far With Your Help

I am trying to get the following equation to equal $\displaystyle {-1}^{k+1}$

$\displaystyle P(k+1)=[f_{k+1}f_{k-1}-f_{k}^{2}]+[f_{k+2}f_{k}-f_{k+1}^{2}]={-1}^{k}+[f_{k+2}f_{k}-f_{k+1}^{2}]$

__Steps__

$\displaystyle P(k+1)={-1}^{k}+[f_{k+2}f_{k}-f_{k+1}^{2}]$

$\displaystyle ={-1}^{k}+(f_{k}+f_{k+1})f_{k}-(f_{k-1}+f_{k})f_{k+1}$

$\displaystyle ={-1}^{k}+f_{k}^{2}+f_{k}f_{k+1}-f_{k-1}f_{k+1}-f_{k}f_{k+1}$

$\displaystyle ={-1}^{k}+f_{k}^2-f_{k+1}f_{k-1}$

$\displaystyle

={-1}^{k}+f_{k}(f_{k-1}+f_{k-2})-(f_{k}+f_{k-1})f_{k-1}

$

$\displaystyle

={-1}^{k}+f_{k}f_{k-1}+f_{k}f_{k-2}-f_{k}f_{k-1}-f_{k-1}f_{k-1}

$

$\displaystyle

={-1}^{k}+f_{k}f_{k-2}-f_{k-1}f_{k-1}

$

$\displaystyle

={-1}^{k}

$

Does anyone know how to get the extra -1 to turn $\displaystyle {-1}^{k}$ into $\displaystyle {-1}^{k+1}$?

Now I understand that part

Thank you! I'm concentrating so hard that I'm missing the obvious!

You simply switched the terms around to get an extra negative on the right side of the equation? and you simplified the P(k+1) equation which made it the negative of the P(k) equation, correct?

Also, am I incorrect to think you have to make the following equation equal http://www.mathhelpforum.com/math-he...e0cc649a-1.gif, not just the $\displaystyle f_{k+2}f_{k}-f_{k+1}^{2}={-1}^{k}$?:

http://www.mathhelpforum.com/math-he...09abcb66-1.gif