Question

Let $ \alpha $ and $ \beta $ be the roots of $ x^2 + \sqrt{3}x - 16 = 0 $, and $ \gamma $ and $ \delta $ be the roots of $ x^2 + 3x - 1 = 0 $. If $ P_n = \alpha^n + \beta^n $ and $ Q_n = \gamma^n + \delta^n $, then $ \frac{P_{25} + \sqrt{3}P_{24}}{2P_{23}} + \frac{Q_{25} - Q_{23}}{Q_{24}} \text{ is equal to} $

• A. \( 3 \)
• B. \( 4 \)
• C. \( 5 \)
• D. \( 7 \)

Hint:

For expressions involving powers of roots of quadratic equations, use recurrence relations derived from the equation itself. If \( \alpha, \beta \) are roots of \( x^2 + ax + b = 0 \), then \( P_n = \alpha^n + \beta^n \) satisfies the recurrence \( P_n = -aP_{n-1} - bP_{n-2} \).

Answer 1

The Correct Option is C

We are given: \[ x^2 + \sqrt{3}x - 16 = 0 \quad \text{has roots } \alpha, \beta \] Define \( P_n = \alpha^n + \beta^n \). Since \( \alpha, \beta \) are roots, the recurrence relation is: \[ P_n + \sqrt{3}P_{n-1} - 16P_{n-2} = 0 \] Using the recurrence: \[ P_{25} + \sqrt{3}P_{24} - 16P_{23} = 0 \Rightarrow \frac{P_{25} + \sqrt{3}P_{24}}{2P_{23}} = \frac{16P_{23}}{2P_{23}} = 8 \] Similarly, for the second quadratic: \[ x^2 + 3x - 1 = 0 \quad \text{has roots } \gamma, \delta \] Define \( Q_n = \gamma^n + \delta^n \) We evaluate: \[ Q_{25} - Q_{23} = \gamma^{25} + \delta^{25} - \gamma^{23} - \delta^{23} = \gamma^{23}(\gamma^2 - 1) + \delta^{23}(\delta^2 - 1) \] Since \( \gamma^2 = -3\gamma + 1 \), so: \[ \gamma^2 - 1 = -3\gamma, \quad \delta^2 - 1 = -3\delta \Rightarrow Q_{25} - Q_{23} = -3(\gamma^{24} + \delta^{24}) = -3Q_{24} \] Therefore: \[ \frac{Q_{25} - Q_{23}}{Q_{24}} = -3 \] Now, summing both terms: \[ \frac{P_{25} + \sqrt{3}P_{24}}{2P_{23}} + \frac{Q_{25} - Q_{23}}{Q_{24}} = 8 - 3 = 5 \]