Analyzing the Recurrence Relations

We are given two quadratic equations and definitions for sequences based on their roots.

• Equation 1: $x^2 + \sqrt{3}x - 16 = 0$ with roots $\alpha, \beta$. The sequence is $P_n = \alpha^n + \beta^n$.
• Equation 2: $x^2 + 3x - 1 = 0$ with roots $\gamma, \delta$. The sequence is $Q_n = \gamma^n + \delta^n$.

For a quadratic equation $ax^2 + bx + c = 0$ with roots $r_1, r_2$, the sum of powers $S_n = r_1^n + r_2^n$ satisfies the recurrence relation $aS_n + bS_{n-1} + cS_{n-2} = 0$. This can be written as $aS_n = -bS_{n-1} - cS_{n-2}$.

Calculating the First Term

For the first equation, $x^2 + \sqrt{3}x - 16 = 0$, we have $a=1, b=\sqrt{3}, c=-16$. The recurrence relation for $P_n$ is:

$P_n + \sqrt{3}P_{n-1} - 16P_{n-2} = 0$

Rearranging this, we get:

$P_n + \sqrt{3}P_{n-1} = 16P_{n-2}$

Let $n=25$. Substituting this into the equation gives:

$P_{25} + \sqrt{3}P_{24} = 16P_{23}$

The first term of the expression is $\frac{P_{25} + \sqrt{3}P_{24}}{2P_{23}}$. Substituting the result from the recurrence relation:

First Term = $\frac{16P_{23}}{2P_{23}} = 8$

Calculating the Second Term

For the second equation, $x^2 + 3x - 1 = 0$, we have $a=1, b=3, c=-1$. The recurrence relation for $Q_n$ is:

$Q_n + 3Q_{n-1} - Q_{n-2} = 0$

Rearranging this, we get:

$Q_n - Q_{n-2} = -3Q_{n-1}$

Let $n=25$. Substituting this into the equation gives:

$Q_{25} - Q_{23} = -3Q_{24}$

The second term of the expression is $\frac{Q_{25}-Q_{23}}{Q_{24}}$. Substituting the result from the recurrence relation:

Second Term = $\frac{-3Q_{24}}{Q_{24}} = -3$

Final Calculation

The required expression is the sum of the two calculated terms:

Expression = First Term + Second Term

Expression = $8 + (-3) = 5$