FIBONACCI SEQUENCE

The Fibonacci sequence adds consecutive terms:

• 1
• 1
• 1 + 1 = 2
• 2 + 1 = 3
• 3 + 2 = 5
• 5 + 3 = 8
• 8 + 5 = 13
• 13 + 8 = 21
• 21 + 13 = 34
• 34 + 21 = 55
• 55 + 34 = 89

Since the last two terms were 55 and 89, we would add these together to get 89 + 55 = 144.

Then you would add 144 and 89 to make 233, and so on.

I saw a cool pattern involving the Fibonacci sequence recently at the Mathemagical Site:

Fibonacci Triples via Mathemagical Site

This involves Fibonacci triples.

A Fibonacci triple consists of three consecutive numbers from the Fibonacci sequence, such as:

• 1, 1, 2
• 1, 2, 3
• 2, 3, 5
• 3, 5, 8
• 5, 8, 13
• 8, 13, 21
• 13, 21, 34

As shown on the Mathemagical Site, the square of the middle number is always one less or one more than the product of the first and third numbers:

Here are a few examples:

• (2, 3, 5): 3 x 3 = 2 x 5 – 1
• (3, 5, 8): 5 x 5 = 3 x 8 + 1
• (5, 8, 13): 8 x 8 = 5 x 13 – 1

Curious about this, I’ve been working through the algebra, and finally came up with an algebraic proof, which follows.

My proof is algebraic and not necessarily obvious, but since I worked it out, I thought I would share it here. 🙂

We begin with two facts about the Fibonacci sequence:

These are two different ways of saying that if you add two consecutive numbers from the Fibonacci sequence, you get the next number in the sequence.

Now solve for x_n in each sequence:

Multiply these together:

Foil this out:

Recall that

Plug this into the first term on the right-hand side of the previous equation:

Distribute:

Two of these terms cancel (the remaining terms are rearranged):

Believe it or not, this basically concludes the proof. The remainder is basically interpreting this result.

This is a recursion relation that relates the square of the n-th term to the square of the previous term (x_n-1 times itself).

Following is the Fibonacci sequence, labeling values of n:

• n = 1 is 1.
• n = 2 is 1.
• n = 3 is 2.
• n = 4 is 3.
• n = 5 is 5.
• n = 6 is 8.
• n = 7 is 13.
• n = 8 is 21.

Let’s plug in n = 3 and see what happens:

If instead you plug in n = 4, you get:

Now just plug in these two expressions (x₃x₃ – x₄x₂ and x₄x₄ – x₅x₃) into the previous recursion relation and you can prove that all of the Fibonacci triples satisfy one of the following relations:

That is, if x₃x₃ – x₄x₂ = 1 and x₄x₄ – x₅x₃ = -1, the previous recursion relation gives similar expressions for x₅x₅ – x₆x₄, x₆x₆ – x₇x₅, and so on.

CHRIS MCMULLEN, PH.D.

Copyright © 2015 Chris McMullen, author of the Improve Your Math Fluency series of math workbooks

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