**The Fibonacci Sequence**

The Fibonacci sequence is the sequence of integers 0, 1, 1, 2, 3, 5, 8, 13, 21,… or 1, 1, 2, 3, 5, 8, 13, 21, … It is a sequence of numbers that starts with 0 (or 1) and each number is the sum of the previous two. The sequence first appeared in *Liber Abaci*, a book written by Leonardo of Pisa, more popularly known as Fibonacci.

The sequence appear in many branches as well as in many form. Take for instance the rectangle above. You can create a rectangle whose sides are consecutive numbers of the Fibonacci Sequence. The Fibonacci Sequence also appears in the Pascal’s Triangle.

In this post, we discuss another interesting characteristics of Fibonacci Sequence.

**The Four Consecutive Numbers**

Take any four consecutive numbers in the sequence. Multiply the outer numbers, then multiply the inner numbers. Subtract them. The difference is 1.

*Example 1*

0, 1, 1, 2, 3, 5, 8, 13, 21

8(2) – 5(3) = 1

*Example 2*

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55

21(5) – 8(13) = 1

*Example 3*

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55

34(8) – 21(13) = 1

Now the question is, is this always true?

***

*Source: Math Fun Facebook Page*

This trick is not always true.

The one sequence it doesn’t work for is 0, 1, 1, 2.

2*0= 0

1*1=1

0-1=-1

This trick does not work for all four-number sequences in the Fibonacci Sequence.

Thanks you SV.

I think there is a mistake in example 3, which helps explain SV’s observation. As you go up the sequence, the result actually alternates between +1 and -1:

(0*2) – (1*1) = 0 – 1 = -1

(1*3) – (1*2) = 3 – 2 = +1

(1*5) – (2*3) = 5 – 6 = -1

(2*8) – (3*5) = 16 – 15 = +1

…

(34*8) – (21*13) = 272 – 273 = -1

…

If we like, we can show the general rule via proof by induction. Let’s call the nth Fibonacci number f(n), so

f(1) = 0,

f(2) = 1,

f(3) = 1,

f(4) = 2

and so on. We can then write four consecutive terms starting from the nth term as

f(n), f(n+1), f(n+2), f(n+3)

or, by using the defining property of the sequence (the next term is the sum of the two previous terms),

f(n), f(n+1), f(n) + f(n+1), f(n) + 2*f(n+1)

Then we can call the difference of multiples starting from the nth term D(n), so

D(n) = [f(n) * f(n+3)] – [f(n+1) * f(n+2)]

= [f(n) * (f(n) + 2*f(n+1))] – [f(n+1) * (f(n) + f(n+1))]

= f(n)^2 + 2*f(n)*f(n+1) – f(n)*f(n+1) – f(n+1)^2

= f(n)^2+f(n)*f(n+1) – f(n+1)^2

Using this last line, and then the defining Fibonacci relation to remove the f(n+2)’s, we can write

D(n+1) = f(n+1)^2 + f(n+1)*f(n+2) – f(n+2)^2

= f(n+1)^2 + f(n+1)*[f(n) + f(n+1)] – [f(n) + f(n+1)]^2

= f(n+1)^2 + f(n+1)*f(n) + f(n+1)^2 -f(n)^2 – 2*f(n)*f(n+1) – f(n+1)^2

= f(n+1)^2 – f(n)*f(n+1) – f(n)^2

= – D(n)

We already saw that D(1) = -1, so we can now see that D(n) = (-1)^n.

Thanks Michael for the correction. And thanks for the proof