How to prove that (f(n+1)^2=f(n)f(n+2)+(-1)^n where f(n), f(n+1),f(n+2) are fibonaci sequence terms. I have no idea where to start.
No, it should be f(k+2)^2=f(k+1)f(k+3)+(-1)^{k+1}.
You have to use a bit of ingenuity in the induction in order to get the (-1)^k term to change sign. One way is to start by writing
f(k+2)^2 = f(k+2)f(k) + [f(k+2) – f(k)]f(k+2).
Then for the first term on the right-hand side, use the inductive hypothesis in the form f(k+2)f(k) = f(k+1)^2 – (–1)^k. For the second term, use the fact that f(k+2) – f(k) = f(k+1) to write it as f(k+1)f(k+2). You should then be able to complete the inductive step.
Here's how I would try the Induction.
For notational simplicity, let
$\displaystyle F_k=a,\;\;\;F_{k+1}=b,\;\;\;F_{k+2}=c,\;\;\;F_{k+3 }=d$
$\displaystyle a+b=c$
$\displaystyle b+c=d$
P(k)
$\displaystyle b^2=ac+(-1)^k$
P(k+1)
$\displaystyle c^2=bd+(-1)^{k+1}$
Proof
$\displaystyle c^2=b(b+c)+(-1)^k(-1)$
$\displaystyle c^2=b^2+bc-(-1)^k$
$\displaystyle b^2=c^2-bc+(-1)^k=c(c-b)+(-1)^k$
$\displaystyle b^2=c(a)+(-1)^k$
which shows that P(k+1) is certainly true if P(k) is true.
Test for the base case...
Thank you Opalg, thank you danny, thank you Archie meade for getting involved. I am going to verify your suggestions but I have not done it yet. WHAT ABOUT IF I rewrite the equality as:
(f(n+1))^2-f(n)f(n+2)=(-1)^n and then substituting the terms f(n+1)=f(n)+f(n-1) and f(n+2)=f(n)+f(n+1) into the equality they get reduced. It seems that if I use the property of infinite decent, combined with well ordering property the whole left hand can be reduced to(-1)^n Can someone tell me if this will be valid? I will review the suggestions but in case that induction does not work.
Well you could start from
$\displaystyle \left(f_{n+1}\right)^2-f_nf_{n+2}=(-1)^n$
and prove
$\displaystyle \left(f_n\right)^2-f_{n-1}f_{n+1}=-(-1)^n$
showing that the sum inevitably alternatives sign as n increases or decreases by 1.
$\displaystyle \left(f_{n+1}\right)^2-\left(f_{n+1}-f_{n-1}\right)\left(f_n+f_{n+1}\right)=(-1)^n$
$\displaystyle \left(f_{n+1}\right)^2-f_nf_{n+1}-\left(f_{n+1}\right)^2+f_nf_{n-1}+f_{n-1}f_{n+1}=(-1)^n$
$\displaystyle f_n\left(f_{n-1}-f_{n+1}\right)+f_{n-1}f_{n+1}=(-1)^n$
$\displaystyle -f_n\left(f_{n+1}-f_{n-1}\right)+f_{n-1}f_{n+1}=(-1)^n$
$\displaystyle \left(f_n\right)^2-f_{n-1}f_{n+1}=-(-1)^n$
Now, you only need a reference point.
Other solution
Define
$\displaystyle F= \left(\begin{array}{cc}1 & 1 \\ 1 & 0 \\\end{array}\right) $
so
$\displaystyle F^n=\left(\begin{array}{cc}F(n+1) & F(n) \\F(n) & F(n-1) \\ \end{array}\right).$
(by induction)
take the determinant of both sides, then
$\displaystyle Det( F^n)=(DetF)^n=(-1)^n=F(n+1)F(n-1)-[F(n)]^{2}$