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The Fibonacci sequence is expressed in a recurrent way as:
Here are its first terms:
The general term of the Fibonacci sequence \((F_n)\) is worth:
The Fibonacci sequence is expressed in a recurrent way as:
We know that the general term of a sequence of order 2, linear combination of its preceding terms can be determined by doing this:
In our case: \((p = q = 1)\).
So,
Where the coefficients \((\alpha, \ \beta)\) are solutions for \((E_c)\):
So, we have to solve this:
Since we are dealing with a quadratic equation, we first calculate the discriminant \((\Delta)\):
As \((\Delta > 0)\), we do have two distinct roots:
For the sake of simplicity, let us set down the golden ratio :
So that:
So, the general term of the sequence \((F_n)\) can be written as:
Let's now determine the coefficients \((A, \ B)\) thanks to the first two terms of the sequence. We know that:
So,
Which leads us to solve the following system \((\mathcal{S})\):
By subtracting the values of the two lines, we get:
Now, taking the first part of the original system, we inject the value of \(B\):
Hence, the general term of the Fbonacci sequence \((F_n)\) is worth:
Starting the characteristic equation \((E_c)\) seen above:
As the golden ratio \((\varphi)\) is solution, we can write the following relation for \(\varphi^2\) :
Furthermore, it is true that \((1 - \varphi)\) is also a solution.
But upon further development, we realize that it amounts to the same thing:
Multiplying both members by \(\varphi\), we do obtain that:
We can see that the recurrence relation is quite obvious.
Let us then attempt to demonstrate, using a double induction, that:
For \((k = 0) \), it is the previous formula, which has already been proven \((\varphi^2)\):
For \((k = 1) \), we do want to prove that \((P_1)\) is true:
So, both propositions \((P_0)\) and \((P_1)\) are true.
From the proposition \((P_k)\):
The recurrence relation is direct when multiplying by \(\varphi\):
So, both propositions \((P_{k + 1})\) et \((P_{k + 2})\) are true.
The proposition \((P_n)\) is true for its two first terms, \(n_0 = 0\) and \(n_1 = 1\), and it is hereditary from terms to terms for all \(k \in \mathbb{N}\).
By the double recurrence principle, that statement is true for all \(n \in \mathbb{N}\).
Hence,
Retrieving the expression \((\varphi^3)\), which has been checked above:
If we do replace \(\varphi^2\) by its other value:
Similarly, if we continue in this way with \(\varphi^4\):
The terms of the Fibonacci sequence then appears:
Specifically, in \((\varphi^3)^*\) and \((\varphi^4)^*\) where:
A second time, one is tempted to demonstrate that recurrence is such that:
For \((k = 0) \):
So, the proposition \((Q_0)\) is true.
We wish to establish the following relationship:
Let's start from \((Q_k)\) and let us multiply by \(\varphi\):
We have clearly shown that \((Q_{k})\) is true, then \((Q_{k + 1})\) was too.
The proposition \((P_n)\) is true for its first term \(n_0 = 0\), and it is hereditary from terms to terms for all \(k \in \mathbb{N}\).
By the recurrence principle, that statement is true for all \(n \in \mathbb{N}\).
Hence,
By retrieving the previous expression of \(\varphi^2\), which has already been proved:
Taking the square root of it:
As \(\varphi\) is expressed according to itself, it can be self-injected:
And so on...
As well as above, we start again from \(\varphi^2\):
Dividing by \(\varphi\), we do have now:
Hence:
Again, we express \(\varphi\) according to itself and we auto-inject it:
And so on...