Key Concepts and Formulas

• Roots of a Quadratic Equation: If $\alpha$ is a root of the quadratic equation $ax^2 + bx + c = 0$, then $a\alpha^2 + b\alpha + c = 0$. This implies that $\alpha$ satisfies the equation.
• Linear Recurrence Relation: A linear recurrence relation expresses a term of a sequence as a linear combination of previous terms. In the context of roots of a quadratic equation, sequences involving powers of the roots often satisfy a recurrence relation derived from the quadratic.
• Definition of $S_n$: The sequence $S_n$ is defined as $S_n=2023 \alpha^n+2024 \beta^n$.

Step-by-Step Solution

• Step 1: State the Given Information and the Goal

We are given the quadratic equation $x^2 - x - 1 = 0$ with roots $\alpha$ and $\beta$, and the sequence $S_n = 2023\alpha^n + 2024\beta^n$. Our goal is to find a relationship between $S_{10}$, $S_{11}$, and $S_{12}$ and identify the correct option.

• Step 2: Utilize the Root Property to Find a Relationship

Since $\alpha$ and $\beta$ are roots of $x^2 - x - 1 = 0$, they satisfy the equation. Therefore: $\alpha^2 - \alpha - 1 = 0 \implies \alpha^2 = \alpha + 1 \quad (1)$ $\beta^2 - \beta - 1 = 0 \implies \beta^2 = \beta + 1 \quad (2)$ These relationships will be crucial in simplifying expressions involving powers of $\alpha$ and $\beta$.

• Step 3: Derive a Recurrence Relation for $S_n$

We want to find a relationship between $S_n$, $S_{n-1}$, and $S_{n-2}$. Let's consider $S_{n-1} + S_{n-2}$: $S_{n-1} + S_{n-2} = (2023\alpha^{n-1} + 2024\beta^{n-1}) + (2023\alpha^{n-2} + 2024\beta^{n-2})$ Rearrange the terms: $S_{n-1} + S_{n-2} = 2023(\alpha^{n-1} + \alpha^{n-2}) + 2024(\beta^{n-1} + \beta^{n-2})$ Factor out the lowest powers of $\alpha$ and $\beta$: $S_{n-1} + S_{n-2} = 2023\alpha^{n-2}(\alpha + 1) + 2024\beta^{n-2}(\beta + 1)$ Now, substitute $\alpha + 1 = \alpha^2$ and $\beta + 1 = \beta^2$ using equations (1) and (2): $S_{n-1} + S_{n-2} = 2023\alpha^{n-2}(\alpha^2) + 2024\beta^{n-2}(\beta^2)$ Simplify the exponents: $S_{n-1} + S_{n-2} = 2023\alpha^n + 2024\beta^n$ By the definition of $S_n$, we have: $S_{n-1} + S_{n-2} = S_n$ Thus, the recurrence relation is $S_n = S_{n-1} + S_{n-2}$.

• Step 4: Apply the Recurrence Relation for n=12

Substitute $n = 12$ into the recurrence relation: $S_{12} = S_{11} + S_{10}$

• Step 5: Compare the Result with the Given Options

We found that $S_{12} = S_{11} + S_{10}$. Comparing this with the given options: (A) $2 S_{12}=S_{11}+S_{10}$ (B) $S_{12}=S_{11}+S_{10}$ (C) $S_{11}=S_{10}+S_{12}$ (D) $2 S_{11}=S_{12}+S_{10}$

Our result directly matches option (B).

Common Mistakes & Tips

• Avoid Direct Calculation of Roots: Calculating the roots $\alpha$ and $\beta$ using the quadratic formula is unnecessary and complicates the problem. Utilizing the fact that the roots satisfy the given equation is a much more efficient approach.
• Recognize the Recurrence Relation: The recurrence relation $S_n = S_{n-1} + S_{n-2}$ is a key result and should be recognized. This relation is derived from the quadratic equation itself and holds for any linear combination of $\alpha^n$ and $\beta^n$.
• Focus on Algebraic Manipulation: The problem is best solved through algebraic manipulation and substitution, rather than numerical computation.

Summary

By utilizing the property that the roots of the quadratic equation satisfy the equation, we derived the recurrence relation $S_n = S_{n-1} + S_{n-2}$. Substituting $n=12$ into this relation, we found that $S_{12} = S_{11} + S_{10}$, which corresponds to option (B).

Final Answer The final answer is \boxed{S_{12}=S_{11}+S_{10}}, which corresponds to option (B).