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 $\dfrac{P_{25}+\sqrt{3}P_{24}}{2P_{23}}+\dfrac{Q_{25}-Q_{23}}{Q_{24}}$ is equal to
Step-by-Step Solution
Key Concept: Use the recurrence $P_n=(\alpha+\beta)P_{n-1}-\alpha\beta P_{n-2}$ for each quadratic to simplify the given ratios.
For roots $\alpha,\beta$ of $x^2+\sqrt{3}x-16=0$, we have $\alpha+\beta=-\sqrt{3}$ and $\alpha\beta=-16$. Hence $$P_n=(\alpha+\beta)P_{n-1}-\alpha\beta P_{n-2}=-\sqrt{3}P_{n-1}+16P_{n-2}.$$ So $$P_n+\sqrt{3}P_{n-1}=16P_{n-2}.$$ Putting $n=25$, $$P_{25}+\sqrt{3}P_{24}=16P_{23},$$ and therefore $$\dfrac{P_{25}+\sqrt{3}P_{24}}{2P_{23}}=8.$$ For roots $\gamma,\delta$ of $x^2+3x-1=0$, $\gamma+\delta=-3$ and $\gamma\delta=-1$. Thus $$Q_n=-3Q_{n-1}+Q_{n-2},$$ so $$Q_n-Q_{n-2}=-3Q_{n-1}.$$ Putting $n=25$, $$Q_{25}-Q_{23}=-3Q_{24},$$ and hence $$\dfrac{Q_{25}-Q_{23}}{Q_{24}}=-3.$$ Therefore the required value is $$8-3=5.$$
Correct Answer: 3