By using $n=1,$ options can be eliminated easily and there are methods like solving linear homogeneous recurrence
by making characteristic equation, generating function etc.
As answer is added here by converting $L_n-L_{n-1}-L_{n-2}=0$ to characteristic equation $r^2-r-1=0$
by assuming solution $L_n=cr^n; c>0$
But the question is “why” we assume so ?
So, we can use an approach of Linear Algebra and prove that our assumption is correct because it is difficult to assume
something while solving this question first time because we might find difficulty to assume it without having any reason.
So,this method is long but answer your question of $\textbf{"Why ?"}$
Here, we have given a difference equation of order two as $L_n = L_{n-1}+L_{n-2} \ \ ; n\geq 3$
The idea is, we will make the system of two simultaneous difference equations of order one from this linear difference equation
of order 2 and then write it in the matrix form and solve this system.
Assume, $L_{n-2}=M_{n-1}$ or $M_n = L_{n-1}$
And this way, we have a system of two difference equations of order one as :
$$L_n = L_{n-1}+ M_{n-1} \ \ \ \ (1)$$
$$M_n = L_{n-1} + 0M_{n-1}\ \ \ \ (2)$$
Now, if we write this system in the matrix form as:
$\begin{pmatrix}
L_n \\
M_n
\end{pmatrix} = \begin{pmatrix}
1 & 1 \\
1 & 0
\end{pmatrix}\begin{pmatrix}
L_{n-1} \\
M_{n-1}
\end{pmatrix}$
Now, we replace $\begin{pmatrix}
L_{n-1} \\
M_{n-1}
\end{pmatrix}$ by $\begin{pmatrix}
1 & 1 \\
1 & 0
\end{pmatrix}\begin{pmatrix}
L_{n-2} \\
M_{n-2}
\end{pmatrix}$
So, $\begin{pmatrix}
L_n \\
M_n
\end{pmatrix} = \begin{pmatrix}
1 & 1 \\
1 & 0
\end{pmatrix}^2\begin{pmatrix}
L_{n-2} \\
M_{n-2}
\end{pmatrix}$
In this way, we get,
$\begin{pmatrix}
L_n \\
M_n
\end{pmatrix} = \begin{pmatrix}
1 & 1 \\
1 & 0
\end{pmatrix}^{n-2}\begin{pmatrix}
L_{2} \\
M_{2}
\end{pmatrix}$
Hence,
$\begin{pmatrix}
L_n \\
M_n
\end{pmatrix} = \begin{pmatrix}
1 & 1 \\
1 & 0
\end{pmatrix}^{n-2}\begin{pmatrix}
3 \\
1
\end{pmatrix} \ \ \ \ \ (3)$
Now, say, $A=\begin{pmatrix}
1 & 1 \\
1 & 0
\end{pmatrix}$
Now, to find $A^{n-2},$ we need to find $A^n$ and for that we can use the idea of Cayley-Hamilton theorem.
($A^n$ can be computed with the use of Diagonalization concept by finding eigenvalues and eigenvectors for the matrix $A.$)
Suppose, $\lambda$ is an eigenvalue for the matrix $A.$ So characteristic equation will be:
$\det (A \ - \ \lambda I) = 0$
So, characteristic equation will be:
$\lambda^2 \ - \lambda \ - \ 1 = 0$
This is the same characteristic equation which we got while solving linear homogeneous recurrence and
so, you can say, this is a proof for making that characteristic equation for linear homogeneous recurrence with
constant coefficient without assuming solution of the recurrence
in the form of $c\lambda^n$ or $\lambda^n$.
Now, According to the Cayley-Hamilton theorem, every square matrix $A$ satisfies its characteristic equation and so,
$A^2 - A - I = 0$
$A^2 = A+I$
$A^3= A^2 + A = 2A + I$
$A^4 = 2A^2 + A = 3A+2I$
Hence, you can write $A^n = aA + bI$ for some arbitrary constant $a$ and $b.$
Hence,
$\lambda_1^n = a\lambda_1 + b$ and
$\lambda_2^n = a\lambda_2 + b$
So, $a= \dfrac{\lambda_1^n - \lambda_2^n}{\lambda_1-\lambda_2}$ and
$b = \lambda_1^n - a\lambda_1$
Since, $\lambda^2 \ - \lambda \ - \ 1 = 0$ gives
$\lambda_1 = \dfrac{1+\sqrt{5}}{2}$ and $\lambda_2 = \dfrac{1-\sqrt{5}}{2}$
So, we get,
$a = \dfrac{1}{\sqrt{5}}\left(\dfrac{1+\sqrt{5}}{2}\right)^n - \dfrac{1}{\sqrt{5}}\left(\dfrac{1-\sqrt{5}}{2}\right)^n$
$b = \dfrac{\sqrt{5}-1}{2\sqrt{5}}\left(\dfrac{1+\sqrt{5}}{2}\right)^n - \dfrac{\sqrt{5}+1}{2\sqrt{5}}\left(\dfrac{1-\sqrt{5}}{2}\right)^n$
Now, put $a$ and $b$ in $A^n = aA+bI,$ we get:
$A^n = \begin{pmatrix}
\dfrac{\sqrt{5}+1}{2\sqrt{5}}\left(\dfrac{1+\sqrt{5}}{2}\right)^n + \dfrac{\sqrt{5}-1}{2\sqrt{5}}\left(\dfrac{1-\sqrt{5}}{2}\right)^n & \dfrac{1}{\sqrt{5}}\left(\dfrac{1+\sqrt{5}}{2}\right)^n - \dfrac{1}{\sqrt{5}}\left(\dfrac{1-\sqrt{5}}{2}\right)^n \\
... & ...
\end{pmatrix}$
And so,
$A^{n-2} = \begin{pmatrix}
\dfrac{2}{\sqrt{5}(\sqrt{5}+1)}\left(\dfrac{1+\sqrt{5}}{2}\right)^n + \dfrac{2}{\sqrt{5}(\sqrt{5}-1)}\left(\dfrac{1-\sqrt{5}}{2}\right)^n & \dfrac{4}{\sqrt{5}(1+\sqrt{5})^2}\left(\dfrac{1+\sqrt{5}}{2}\right)^n - \dfrac{4}{\sqrt{5}(1-\sqrt{5})^2}\left(\dfrac{1-\sqrt{5}}{2}\right)^n \\
... & ...
\end{pmatrix}$
Now, finally, by $(3),$ we get:
$L_n = \dfrac{2\times 3}{\sqrt{5}(\sqrt{5}+1)}\left(\dfrac{1+\sqrt{5}}{2}\right)^n + \dfrac{2 \times 3}{\sqrt{5}(\sqrt{5}-1)}\left(\dfrac{1-\sqrt{5}}{2}\right)^n + \dfrac{4}{\sqrt{5}(1+\sqrt{5})^2}\left(\dfrac{1+\sqrt{5}}{2}\right)^n - \dfrac{4}{\sqrt{5}(1-\sqrt{5})^2}\left(\dfrac{1-\sqrt{5}}{2}\right)^n$
After solving it, we get,
$L_n = \left(\dfrac{1+\sqrt{5}}{2}\right)^n + \left(\dfrac{1-\sqrt{5}}{2}\right)^n$