Euler-Binet Formula

Theorem
The Fibonacci numbers have a closed-form solution:
 * $\displaystyle F \left({n}\right) = \frac {\phi^n - \left({1 - \phi}\right)^n} {\sqrt 5} = \frac {\phi^n - \left({-1 / \phi}\right)^n} {\sqrt 5}$

where $\phi$ is the golden mean.

Putting $\hat \phi = 1 - \phi = -\dfrac 1 \phi$ this can be written:
 * $\displaystyle F \left({n}\right) = \frac {\phi^n - \hat \phi^n} {\sqrt 5}$

Proof
Proof by induction:

For all $n \in \N$, let $P \left({n}\right)$ be the proposition:
 * $\displaystyle F \left({n}\right) = \frac {\phi^n - \hat \phi^n} {\sqrt 5}$

Basis for the Induction
$P(0)$ is true, as this just says:
 * $\displaystyle \frac {\phi^0 - \hat \phi^0} {\sqrt 5} = \frac {1 - 1} {\sqrt 5} = 0 = F(0)$

$P(1)$ is the case, as follows from the following computation:
 * $\displaystyle \frac {\phi - \hat \phi} {\sqrt 5} = \frac {\left({ \frac {1 + \sqrt {5}} {2} }\right) - \left({ \frac {1 - \sqrt 5} {2} }\right)} {\sqrt 5} =

\frac {\left({1 - 1}\right) + \left({ \sqrt 5 + \sqrt 5 }\right)} {2 \sqrt 5} = 1 = F \left({1}\right)$

This is our basis for the induction.

Induction Hypothesis

 * Now we need to show that, if $P \left({n}\right)$ is true for all $0 \le n \le k+1$, then it logically follows that $P \left({k+2}\right)$ is true.

So this is our induction hypothesis:
 * $\displaystyle F \left({n}\right) = \frac {\phi^n - \hat \phi^n} {\sqrt 5}$ for all $0 \le n \le k+1$.

Then we need to show:
 * $\displaystyle F \left({k+2}\right) = \frac {\phi^{k+2} - \hat \phi^{k+2}} {\sqrt 5}$

Induction Step
This is our induction step:

We observe that we have the following two identities:
 * $\displaystyle \phi^2 = \left({ \frac {1 + \sqrt 5} {2} }\right)^2 = \frac 1 4 \left({6 + 2 \sqrt {5}}\right) = \frac {3 + 2 \sqrt 5} {2} = 1 + \phi$
 * $\displaystyle \hat \phi^2 = \left({ \frac {1 - \sqrt 5} {2} }\right)^2 = \frac 1 4 \left({6 - 2 \sqrt 5}\right) = \frac {3 - 2 \sqrt 5} {2} = 1 + \hat \phi$

Alternatively, this can be deduced from the definition of the golden mean: the fact that $\phi$ and $\hat \phi$ are the solutions to the quadratic equation $x^2 = x + 1$.

This means that we can compute:

The result follows by the Second Principle of Mathematical Induction.

Therefore:
 * $\displaystyle \forall n \in \N: F \left({n}\right) = \frac {\phi^n - \hat \phi^n} {\sqrt 5}$

Alternative Proof
This follows as a direct application of the first Binet form:


 * $U_n = m U_{n-1} + U_{n-2}$

where:

has the closed-form solution:
 * $U_n = \dfrac {\alpha^n - \beta^n} {\Delta}$

where:

where $m=1$.

It is also known as Binet's Formula.

Binet derived it in 1843, but it was already known to Euler, de Moivre and Daniel Bernoulli over a century earlier.

However, it was Binet who derived the more general Binet Form of which this is an elementary application.