Solve Fibonacci Squares: Formula for Difference of Squares

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In summary: F_{1}In summary, the problem asks for a formula for the difference of the squares of two Fibonacci numbers.
  • #1
pleasehelpsos
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I need help with this problem...

By experimenting with numerous examples in search of a pattern, determine a simple formula for (Fn+1)2−(Fn−1)2; that is, a formula for the difference of the squares of two Fibonacci numbers.What does this question want? What is it asking for?
 
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  • #2
Hi pleasehelpsos and welcome to MHB! :D

pleasehelpsos said:
...(Fn+1)2−(Fn−1)2...

By the above, did you intend

$$F_{n+1}^2-F_{n-1}^2$$

?
 
  • #3
I think I would begin with:

\(\displaystyle F_{n+1}=F_{n}+F_{n-1}\)

What do you get when you square both sides?
 
  • #4
How do I pick which sequence number it goes along with?

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greg1313 said:
Hi pleasehelpsos and welcome to MHB! :D
By the above, did you intend

$$F_{n+1}^2-F_{n-1}^2$$

?

Yes
 
  • #5
The Fibonacci sequence defined by $F_n=F_{n-1}+F_{n-2}$ has initial terms $0,1,1,2,3,5,8,13,21...$ Pick the 1st and 3rd and apply the rule, the 2nd and 4th and apply the rule, the 3rd and 5th and apply the rule and so on. Then look for a pattern in the results.
 
  • #6
I honestly still do not understand
 
  • #7
like i got zero when i had tried to do n=1 n=2 n=3 and n=4

for the first one i tried to sub in f1 but i don't know what i am doing
 
  • #8
Okay; let's do some more terms of the Fibonacci sequence:

$$0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144...$$

Apply the rule to them:

$$1^2-0^2=1,2^2-1^2=3,3^2-1^2=8,5^2-2^2=21,8^2-3^2=55,13^2-5^2=169-25=144...$$

Now look at those results and at the Fibonacci sequence above them. What do you notice?
 
  • #9
Consider the closed form for the $n$th Fibonacci number:

\(\displaystyle F_n=\frac{\varphi^n-(-\varphi)^{-n}}{\sqrt{5}}\) where \(\displaystyle \varphi=\frac{1+\sqrt{5}}{2}\)

Hence:

\(\displaystyle F_n^2=\frac{\varphi^{2n}-2(-1)^n+(-\varphi)^{-2n}}{5}\)

And so:

\(\displaystyle F_{n+1}^2-F_{n-1}^2=\frac{\varphi^{2(n+1)}-2(-1)^{n+1}+(-\varphi)^{-2(n+1)}}{5}-\frac{\varphi^{2(n-1)}-2(-1)^{n-1}+(-\varphi)^{-2(n-1)}}{5}\)

\(\displaystyle F_{n+1}^2-F_{n-1}^2=\frac{\varphi^{2n}\left(\varphi^2-\varphi^{-2}\right)+(-\varphi)^{-2n}\left((-\varphi)^{-2}-(-\varphi)^{2}\right)}{5}\)

Observing that:

\(\displaystyle \varphi^2-\varphi^{-2}=\sqrt{5}\)

\(\displaystyle (-\varphi)^{-2}-(-\varphi)^{2}=-\sqrt{5}\)

We have:

\(\displaystyle F_{n+1}^2-F_{n-1}^2=\frac{\varphi^{2n}-(-\varphi)^{-2n}}{\sqrt{5}}=F_{2n}\)
 

FAQ: Solve Fibonacci Squares: Formula for Difference of Squares

What is the Fibonacci sequence?

The Fibonacci sequence is a mathematical sequence in which each number is the sum of the two preceding numbers, starting with 0 and 1. The sequence goes as follows: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, etc.

What is the formula for the difference of squares?

The formula for the difference of squares is (a+b)(a-b) = a2 - b2, where a and b are any two numbers.

How do you apply the formula for the difference of squares to solve Fibonacci squares?

To solve Fibonacci squares using the formula for the difference of squares, you need to express the Fibonacci numbers as the sum of two squares. This can be done by finding the two consecutive Fibonacci numbers that have a difference of 1, and then expressing those numbers as the sum of two squares. For example, 8 and 9 have a difference of 1, and can be expressed as (32)2 and (32+1)2. Plugging these into the formula, we get (32+1)2 - (32)2 = 13.

What is the significance of solving Fibonacci squares?

Solving Fibonacci squares can help in understanding and analyzing the properties of the Fibonacci sequence. It can also be used in various mathematical and scientific applications, such as generating fractal patterns and analyzing the growth patterns of plants and animals.

Are there any other methods for solving Fibonacci squares?

Yes, there are other methods for solving Fibonacci squares, such as using matrices and binomial coefficients. However, the formula for the difference of squares is the most commonly used method as it is simple and efficient.

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