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 $\frac{P_{25} + \sqrt{3}P_{24}}{2P_{23}} + \frac{Q_{25}-Q_{23}}{Q_{24}}$ is equal to
5
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 = $8 + (-3) = 5$
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