If α, β are the roots of the equation, x² – x – 1 and S_n = 2023 αⁿ + 2024 βⁿ, then :

The correct option is S₁₂ = S₁₁ + S₁₀.

Here is the step-by-step mathematical derivation:

Step 1: Understand the properties of the roots of the quadratic equation.
The given quadratic equation is:
$x^2-x-1=0$
Since $\alpha$ and $\beta$ are the roots of this equation, they must satisfy it. Therefore, we have:
${\alpha }^2-\alpha -1=0\Rightarrow {\alpha }^2=\alpha +1$
and
${\beta }^2-\beta -1=0\Rightarrow {\beta }^2=\beta +1$

Step 2: Find a general recurrence relation for the powers of the roots.
Multiply the equation for $\alpha$ by ${\alpha }^n$ on both sides:
${\alpha }^{n+2}={\alpha }^{n+1}+{\alpha }^n$
Similarly, multiply the equation for $\beta$ by ${\beta }^n$ on both sides:
${\beta }^{n+2}={\beta }^{n+1}+{\beta }^n$

Step 3: Relate these equations to the definition of S_n.
We are given the definition:
$S_n=2023{\alpha }^n+2024{\beta }^n$
Let us write the expression for $S_{n+2}$:
$S_{n+2}=2023{\alpha }^{n+2}+2024{\beta }^{n+2}$
Substitute the relations derived in Step 2:
$S_{n+2}=2023({\alpha }^{n+1}+{\alpha }^n)+2024({\beta }^{n+1}+{\beta }^n)$
Group the terms by their powers:
$S_{n+2}=(2023{\alpha }^{n+1}+2024{\beta }^{n+1})+(2023{\alpha }^n+2024{\beta }^n)$
Using the definition of S_n, this simplifies to the recurrence relation:
$S_{n+2}=S_{n+1}+S_n$

Step 4: Substitute the value n = 10.
By substituting $n=10$ into our recurrence relation, we get:
$S_{12}=S_{11}+S_{10}$