Question

If \( \alpha, \beta \) are the roots of the equation \( x^2 - 6x - 2 = 0 \), \( \alpha > \beta \), and \( a_n = \alpha^n - \beta^n, n \geq 1 \), then the value of \( \frac{a_{10} - 2 a_8}{2 a_9} \) is:

• A. \(6 \)
• B. \(4 \)
• C. \(3 \)
• D. \(2 \)

Hint:

For recurrence relations, solve for characteristic roots and use standard formulas for simplifications.

Answer 1

The Correct Option is C

Given the quadratic equation: \[ x^2 - 6x - 2 = 0 \] Let \( \alpha \) and \( \beta \) be the roots of the equation. Using Vieta's formulas: \[ \alpha + \beta = 6 \quad \text{and} \quad \alpha \beta = -2 \] Step 1: Derive the Recurrence Relation We are given: \[ a_n = \alpha^n - \beta^n \] Using the identities for powers of roots, \[ \alpha^n = (\alpha^{n-1}) \alpha \quad \text{and} \quad \beta^n = (\beta^{n-1}) \beta \] Since \( \alpha \) and \( \beta \) satisfy the equation, \[ \alpha^2 = 6\alpha + 2 \quad \text{and} \quad \beta^2 = 6\beta + 2 \] Multiplying both sides by \( \alpha^{n-2} \) and \( \beta^{n-2} \), respectively, \[ \alpha^n = 6\alpha^{n-1} + 2\alpha^{n-2} \] \[ \beta^n = 6\beta^{n-1} + 2\beta^{n-2} \] Subtracting these two, \[ a_n = \alpha^n - \beta^n = 6a_{n-1} + 2a_{n-2} \] Thus, the recurrence relation is: \[ a_{n+2} = 6a_{n+1} + 2a_n \] Step 2: Derivation of Required Expression We need to calculate: \[ \frac{a_{10} - 2a_8}{2a_9} \] Step 3: Compute \( a_{10} \) in Terms of Previous Terms From the recurrence relation: \[ a_{10} = 6a_9 + 2a_8 \] Now substitute this into the required expression: \[ \frac{a_{10} - 2a_8}{2a_9} = \frac{(6a_9 + 2a_8) - 2a_8}{2a_9} \] Simplifying, \[ = \frac{6a_9}{2a_9} \] \[ = 3 \]