If $\alpha, \beta$ are roots of $x^{2}-ax+b=0$ and $v_{n}=\alpha^{n}+\beta^{n}$ then which of the following is correct?

Question

Accepted Answer

To determine why Option 1 is correct, we need to analyze the properties of the roots of the quadratic equation and the sequence $ v_n $.

Given that $\alpha$ and $\beta$ are roots of the quadratic equation $x^2 - ax + b = 0$, by Vieta's formulas, we know:
1. $\alpha + \beta = a$
2. $\alpha \beta = b$

The sequence $v_n = \alpha^n + \beta^n$ is defined recursively. We aim to find a recurrence relation for $v_n$.

### Derivation of the Recurrence Relation

1. **Base Cases:**
   - $v_0 = \alpha^0 + \beta^0 = 1 + 1 = 2$
   - $v_1 = \alpha^1 + \beta^1 = \alpha + \beta = a$

2. **Recurrence Relation:**
   - Consider $v_{n+1} = \alpha^{n+1} + \beta^{n+1}$.
   - Using the identity $\alpha^{n+1} = \alpha \cdot \alpha^n$ and $\beta^{n+1} = \beta \cdot \beta^n$, we have:
     $$
     v_{n+1} = \alpha \cdot \alpha^n + \beta \cdot \beta^n
     $$
   - We can express $\alpha^2$ and $\beta^2$ using the quadratic equation:
     $$
     \alpha^2 = a\alpha - b \quad \text{and} \quad \beta^2 = a\beta - b
     $$
   - Therefore, $\alpha^{n+1} = \alpha \cdot \alpha^n = (a\alpha^n - b\alpha^{n-1})$ and similarly for $\beta$.
   - Adding these, we get:
     $$
     v_{n+1} = a(\alpha^n + \beta^n) - b(\alpha^{n-1} + \beta^{n-1}) = av_n - bv_{n-1}
     $$

This confirms that the correct recurrence relation is $v_{n+1} = av_n - bv_{n-1}$, which matches Option 1.

### Explanation of Incorrect Options

- **Option 2: $v_{n-1} = bv_n + av_{n+1}$**
  - This option incorrectly suggests a relationship that does not align with the derived recurrence relation. The terms $bv_n$ and $av_{n+1}$ do not correctly represent the dependencies of the sequence based on the quadratic roots.

- **Option 3: $v_{n+1} = bv_n - av_{n-1}$**
  - This option reverses the roles of $a$ and $b$ in the recurrence relation, which does not match the derived formula. The correct relation involves $a$ multiplying $v_n$ and $b$ multiplying $v_{n-1}$.

- **Option 4: $v_n = av_{n+1} + bv_{n-1}$**
  - This option suggests a forward-looking relation that is inconsistent with the backward-looking nature of the derived recurrence relation. The sequence $v_n$ depends on previous terms, not future terms.

In conclusion, Option 1 is the correct answer as it accurately represents the recurrence relation derived from the properties of the roots of the quadratic equation.