Let $E$ denote the vector space of real sequences $\left(u_{n}\right)_{n \in \mathbb{N}}$ satisfying $: \forall n \in \mathbb{N}, u_{n+2}=$ $u_{n+1}+u_{n}$.

1) Determine the two non-zero geometric sequences, denoted $\left(a_{n}\right)$ and $\left(b_{n}\right)$ with $a_{0}=b_{0}=1$ that belong to $\tilde{\mathrm{A}} E$.
2) Show that the sequences $\left(a_{n}\right)$ and $\left(b_{n}\right)$ form a basis of $E$ and deduce the dimension of $E$.
3) Let $\left(\mathcal{F}_{n}\right)$ be the sequence in $E$ such that : $\mathcal{F}_{0}=0$ and $\mathcal{F}_{1}=1$.

Determine, using the previous questions, the general term of the sequence $\left(\mathcal{F}_{n}\right)$. What sequence is this?
1) Let $\left(u_{n}\right)$ be a non-zero geometric sequence with ratio $q$, belonging to $\tilde{\mathrm{A}} E$ and such that $u_{0}=1$.

We can write: $\forall n \in \mathbb{N}, u_{n}=q^{n}$. Since the sequence $\left(u_{n}\right)$ belongs to $\tilde{\mathrm{A}} E$, we have: $u_{n+2}=u_{n+1}+u_{n} \Rightarrow q^{n+2}=q^{n+1}+q^{n} \Rightarrow q^{2}-q-1=0$ (by simplifying by $q^{n} \neq 0$ since $q \neq 0$), hence: $q=\frac{1-\sqrt{5}}{2}$ or $q=\frac{1+\sqrt{5}}{2}$. Conversely, the sequences $\left(a_{n}\right)$ and $\left(b_{n}\right)$ such that for all natural integers $n, a_{n}=\left(\frac{1+\sqrt{5}}{2}\right)^{n}$ and $b_{n}=\left(\frac{1-\sqrt{5}}{2}\right)^{n}$ belong to $\tilde{\mathrm{A}} E$.

2) Let's first show that the sequences $\left(a_{n}\right)$ and $\left(b_{n}\right)$ are linearly independent. Let: $\forall n \in \mathbb{N}$,
$\lambda a_{n}+\mu b_{n}=0, \lambda, \mu \in \mathbb{R}$.

By taking $n=0$ and then $n=1$, we obtain the following system: $\left\{\begin{array}{l}\lambda+\mu=0 \\ \lambda\left(\frac{1+\sqrt{5}}{2}\right)+\mu\left(\frac{1-\sqrt{5}}{2}\right)=0\end{array}\right.$, solving this system gives us: $\lambda=\mu=0$. Thus, the sequences $\left(a_{n}\right)$ and $\left(b_{n}\right)$ are linearly independent.

Let's show that these two sequences generate $E$. Let $\left(u_{n}\right)$ be a sequence belonging to $\tilde{A} E$.

We prove the following equality by induction (we find this equality by finding $\lambda$ and $\mu \in \mathbb{R}$ such that: $u_{n}=\lambda a_{n}+\mu b_{n}$. To find $\lambda$ and $\mu$, we take $n=0$ and then $n=1$ and solve the obtained system):

$u_{n}=\left(\left(\frac{\sqrt{5}-1}{2 \sqrt{5}}\right) u_{0}+\frac{u_{1}}{\sqrt{5}}\right)\left(\frac{1+\sqrt{5}}{2}\right)^{n}+\left(-\frac{u_{1}}{\sqrt{5}}+\left(\frac{1+\sqrt{5}}{2 \sqrt{5}}\right) u_{0}\right)\left(\frac{1-\sqrt{5}}{2}\right)^{n}$

The induction we perform here is of order 2: to prove this formula (denoted as (*)) at rank $n+2$, we need the formula at rank $n$ and rank $n+1$ (and we will use the fact that: $u_{n+2}=u_{n+1}+u_{n}$). Thus, we proceed as follows: we first verify (*) for $n=0$ and for $n=1$. Then, we assume that we have the equality (*) at ranks $n$ and $n+1$ and we prove by calculation that (*) is true at rank $n+2$. (We do not detail the calculations as they are quite cumbersome).

We have found real numbers $\lambda$ and $\mu$ such that $\forall n \in \mathbb{N}, u_{n}=\lambda a_{n}+\mu b_{n}$, so the sequences $\left(a_{n}\right)$ and $\left(b_{n}\right)$ generate $E$.

Thus, these two sequences form a basis of $E$ and we deduce that $\operatorname{dim} E=2$.

3) The sequence $\left(\mathcal{F}_{n}\right)$ belongs to $\tilde{A} E$ and the sequences $\left(a_{n}\right)$ and $\left(b_{n}\right)$ form a basis of $E$ so there exists a unique pair of real numbers $(\lambda, \mu)$ such that: $\forall n \in \mathbb{N}, \mathcal{F}_{n}=\lambda a_{n}+\mu b_{n}$. By taking $n=0$ and then $n=1$, we obtain a system that we solve and find: $\lambda=\frac{1}{\sqrt{5}}$ and $\mu=-\frac{1}{\sqrt{5}}$. We then recover the general term of the well-known Fibonacci sequence.