The definition of a Fibonacci number is as follows:
$$F(0)=0\\ F(1)=1\\ F(n)= F(n-2)+F(n-1)\text{ for }n\geq 2$$
Prove the given property of the Fibonacci numbers directly from the definition. $F(n+3)=2F(n+1)+ F(n)$ for $n$ greater than or equal to $0$.
To get started: -I would do a direct proof.
Assume that $F(0)=0$; $F(1)=1$; $F(n)= F(n-2)+F(n-1)$ for $n$ greater than or equal to $2$.
I am lost. Can I have a clue on the nest step?
$$ F(n+3) = \color{red}{F(n+2)} + \color{blue}{F(n+1)} = \color{red}{F(n+1)} + \color{red}{F(n)} + \color{blue}{F(n+1)} = \color{green}{2F(n+1)} + \color{red}{F(n)} $$
First, show that $F(3) = 2F(1) + F(0)$, and that $F(4) = 2F(2) + F(1)$, using the definition directly, given your definition:
$$F(0)=0;\; F(1)=1;\quad F(n)= F(n-2)+F(n-1) \quad \text{for n greater than or equal to 2.}$$
We use the definition to express $F(n+3)$ in terms of $$\begin{align} F((n+3) - 2) & = F(n + 1)\\ \\ F((n+3) - 1) & = F(n+2)\end{align}$$
Now, we "unpack" $F(n+3)$ and express it as a function of $F(n + 1), F(n + 2)$:
$$\begin{align} {\bf F(n+3)} & = {\bf F(n+1)} + \underbrace{\left( F(n+2)\right)}_{\large = F(n) + F(n+1)} \\ \\ & \\ \\ & = {\bf F(n+1)}+ \left( F(n) + {\bf F(n+1)}\right) \\ \\ & = {\bf 2 F(n+1)} + F(n)\end{align}$$
Hint: Use the defining property that you listed to rewrite $F_{n+3}$ in terms of $F_{n+2}$ and $F_{n+1}$.
