Fibonaci sequence terms.

**joi** · May 6th 2011, 12:59 PM

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.

**dwsmith** · May 6th 2011, 12:59 PM

Originally Posted by joi

(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.

If this is for n > then something, I would try induction.

**joi** · May 7th 2011, 06:04 AM

Thank you dwsmith for answering. You are right: The condition is n>3 or n=3. I forgot to mention that. I will try induction.

**joi** · May 7th 2011, 07:41 AM

Induction does not work. I find f(k+2)^2=f(k)f(k+4)+(-1)^k. It should be f(k+2)^2=f(k+1)f(k+3)+(-1)^k. I am not sure if I am making any mistake.

**Opalg** · May 7th 2011, 12:38 PM

Originally Posted by joi

Induction does not work. I find f(k+2)^2=f(k)f(k+4)+(-1)^k. It should be f(k+2)^2=f(k+1)f(k+3)+(-1)^k.

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.

**Danny** · May 7th 2011, 03:21 PM

I might add that what you're trying to prove is know as Cassini's Identity. A google search will give you lots of places where this identity is proven (and with details :-))

**Archie Meade** · May 7th 2011, 04:06 PM

Here's how I would try the Induction.

For notational simplicity, let

$F_k=a,\;\;\;F_{k+1}=b,\;\;\;F_{k+2}=c,\;\;\;F_{k+3 }=d$

$a+b=c$
$b+c=d$

P(k)

$b^2=ac+(-1)^k$

P(k+1)

$c^2=bd+(-1)^{k+1}$

Proof

$c^2=b(b+c)+(-1)^k(-1)$

$c^2=b^2+bc-(-1)^k$

$b^2=c^2-bc+(-1)^k=c(c-b)+(-1)^k$

$b^2=c(a)+(-1)^k$

which shows that P(k+1) is certainly true if P(k) is true.

Test for the base case...

**joi** · May 7th 2011, 05:31 PM

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.

**Archie Meade** · May 14th 2011, 06:23 AM

Well you could start from

$\left(f_{n+1}\right)^2-f_nf_{n+2}=(-1)^n$

and prove

$\left(f_n\right)^2-f_{n-1}f_{n+1}=-(-1)^n$

showing that the sum inevitably alternatives sign as n increases or decreases by 1.

$\left(f_{n+1}\right)^2-\left(f_{n+1}-f_{n-1}\right)\left(f_n+f_{n+1}\right)=(-1)^n$

$\left(f_{n+1}\right)^2-f_nf_{n+1}-\left(f_{n+1}\right)^2+f_nf_{n-1}+f_{n-1}f_{n+1}=(-1)^n$

$f_n\left(f_{n-1}-f_{n+1}\right)+f_{n-1}f_{n+1}=(-1)^n$

$-f_n\left(f_{n+1}-f_{n-1}\right)+f_{n-1}f_{n+1}=(-1)^n$

$\left(f_n\right)^2-f_{n-1}f_{n+1}=-(-1)^n$

Now, you only need a reference point.

**Renji Rodrigo** · June 4th 2011, 05:32 AM

Other solution

Define
$F= \left(\begin{array}{cc}1 & 1 \\ 1 & 0 \\\end{array}\right)$

so
$F^n=\left(\begin{array}{cc}F(n+1) & F(n) \\F(n) & F(n-1) \\ \end{array}\right).$
(by induction)

take the determinant of both sides, then

$Det( F^n)=(DetF)^n=(-1)^n=F(n+1)F(n-1)-[F(n)]^{2}$

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