The Fibonacci numbers

are defined by a recurrence relation

This definition allows one to find any Fibonacci number, but for large values of it would be time-consuming to compute this way. Worse, any error of calculating a term would be repeated and magnified in every subsequent term. So a natural question to ask is whether there is a simple formula, in terms of , which enables one to calculate without having to find all the earlier Fibonacci numbers.

In 1843 the French mathematician Jacques Philippe Marie Binet announced that he had found a Fibonacci Formula. Although it was later revealed that other mathematicians, including Leonhard Euler and Abraham de Moivre, had worked out the same formula a hundred years earlier, the formula is today still labelled *Binet’s Formula*.

Deriving the Binet’s formula starts with a simple little quadratic equation

Multiplying the equation by successive powers of , and simplifying at each step, gives

and so on.

A pattern is beginning to emerge. It appears that

The pattern can easily be established to be true for all natural numbers by Mathematical Induction.

It is clearly true when , and if we assume that it is true for some positive integer , i.e. that

then

,

showing that the formula is valid for the next number, .

Now go back to the original quadratic equation , which the standard quadratic formula shows has solutions

.

Substituting these two values into the equation gives

and

Since we want to find an expression for , the obvious thing to do now is subtract the two equations:

from which Binet’s Formula follows:

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Originally by John Webb

JoshNovember 8, 2018 at 10:15 pmJust to let you know, for x^6 you wrote x^6 = x(x^4) when it should be x(x^5) but the proof was quite helpful!

Jonathan ShockNovember 9, 2018 at 7:29 amVery good, thank you!