If α and β are the roots of the equation x² – x – 1 = 0 and S_n = 2024 αⁿ + 2024 βⁿ then S₃ is equal to

The correct answer is 8096.

Let $\alpha$ and $\beta$ be the roots of the quadratic equation:
$x^2-x-1=0$
From the properties of quadratic equations, we can find the sum and product of the roots:
Sum of roots: $\alpha +\beta =1$
Product of roots: $\alpha \beta =-1$

We are given the expression for $S_n$ as:
$S_n=2024{\alpha }^n+2024{\beta }^n=2024({\alpha }^n+{\beta }^n)$

Since $\alpha$ and $\beta$ are the roots of the given equation, they must satisfy it:
${\alpha }^2-\alpha -1=0\Rightarrow {\alpha }^2=\alpha +1$
Multiplying both sides of the equation by ${\alpha }^n$, we get:
${\alpha }^{n+2}={\alpha }^{n+1}+{\alpha }^n$
Similarly, for the root $\beta$:
${\beta }^{n+2}={\beta }^{n+1}+{\beta }^n$

Adding the two equations and multiplying the entire relation by 2024 gives:
$2024({\alpha }^{n+2}+{\beta }^{n+2})=2024({\alpha }^{n+1}+{\beta }^{n+1})+2024({\alpha }^n+{\beta }^n)$
This simplifies to the recurrence relation:
$S_{n+2}=S_{n+1}+S_n$

Setting $n=1$, we find the formula for $S_3$:
$S_3=S_2+S_1$

Now, let's calculate the values of $S_1$ and $S_2$:
$S_1=2024(\alpha +\beta )=2024\left(1\right)=2024$

Using the identity ${\alpha }^2+{\beta }^2={(\alpha +\beta )}^2-2\alpha \beta$:
${\alpha }^2+{\beta }^2={\left(1\right)}^2-2(-1)=1+2=3$
Thus, $S_2$ is:
$S_2=2024({\alpha }^2+{\beta }^2)=2024\left(3\right)=6072$

Substituting these values back into the recurrence equation to find $S_3$:
$S_3=6072+2024=8096$