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

Given: \(x^2 + \sqrt{3}x - 16 = 0\) has roots \(\alpha, \beta\). Define \(P_n = \alpha^n + \beta^n\). The recurrence relation is \(P_n + \sqrt{3}P_{n-1} - 16P_{n-2} = 0\). From the recurrence, \(P_{25} + \sqrt{3}P_{24} - 16P_{23} = 0\). Thus, \(\frac{P_{25} + \sqrt{3}P_{24}}{2P_{23}} = \frac{16P_{23}}{2P_{23}} = 8\). For the second quadratic, \(x^2 + 3x - 1 = 0\) 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 \(\gamma^2 = -3\gamma + 1\), \(\gamma^2 - 1 = -3\gamma\), and similarly \(\delta^2 - 1 = -3\delta\). Therefore, \(Q_{25} - Q_{23} = -3(\gamma^{24} + \delta^{24}) = -3Q_{24}\). This implies \(\frac{Q_{25} - Q_{23}}{Q_{24}} = -3\). Summing both terms: \(\frac{P_{25} + \sqrt{3}P_{24}}{2P_{23}} + \frac{Q_{25} - Q_{23}}{Q_{24}} = 8 + (-3) = 5\).