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

Let $$\alpha$$ and $$\beta$$ be the roots of $$x^2 + \sqrt{3}x - 16 = 0$$. Therefore, they satisfy the equation:

$$\alpha^2 + \sqrt{3}\alpha - 16 = 0 \implies \alpha^2 + \sqrt{3}\alpha = 16$$

$$\beta^2 + \sqrt{3}\beta - 16 = 0 \implies \beta^2 + \sqrt{3}\beta = 16$$

Multiplying the first equation by $$\alpha^{23}$$ and the second by $$\beta^{23}$$ gives:

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

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

Adding these two equations together:

$$(\alpha^{25} + \beta^{25}) + \sqrt{3}(\alpha^{24} + \beta^{24}) = 16(\alpha^{23} + \beta^{23})$$

Since $$P_n = \alpha^n + \beta^n$$, this simplifies to:

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

Dividing by $$2P_{23}$$ gives the value of the first term:

$$\frac{P_{25} + \sqrt{3}P_{24}}{2P_{23}} = \frac{16P_{23}}{2P_{23}} = 8$$

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Similarly, let $$\gamma$$ and $$\delta$$ be the roots of $$x^2 + 3x - 1 = 0$$. They satisfy:

$$\gamma^2 + 3\gamma - 1 = 0 \implies \gamma^2 - 1 = -3\gamma$$

$$\delta^2 + 3\delta - 1 = 0 \implies \delta^2 - 1 = -3\delta$$

Multiplying by $$\gamma^{23}$$ and $$\delta^{23}$$ respectively:

$$\gamma^{25} - \gamma^{23} = -3\gamma^{24}$$

$$\delta^{25} - \delta^{23} = -3\delta^{24}$$

Adding these two equations together:

$$(\gamma^{25} + \delta^{25}) - (\gamma^{23} + \delta^{23}) = -3(\gamma^{24} + \delta^{24})$$

Since $$Q_n = \gamma^n + \delta^n$$, this simplifies to:

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

Dividing by $$Q_{24}$$ gives the value of the second term:

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

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Combining both results, the required expression is equal to:

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