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
- (1) 3
- (2) 4
- (3) 5
- (4) 7
Solution:
$$x^{2}+\sqrt{3}x-16=0$$
$$P_{n}=\alpha^{n}+\beta^{n}$$
$$P_{n}+\sqrt{3}P_{n-1}-16P_{n-2}=0$$
$$P_{25}+\sqrt{3}P_{24}-16P_{23}=0$$
$$P_{25}+\sqrt{3}P_{24}=16P_{23}$$
$$\frac{P_{25}+\sqrt{3}P_{24}}{2P_{23}} = \frac{16P_{23}}{2P_{23}} = 8$$
$$x^{2}+3x-1=0$$
$$Q_{n}=\gamma^{n}+\delta^{n}$$
$$Q_{25}-Q_{23} = \gamma^{25}+\delta^{25}-\gamma^{23}-\delta^{23}$$
$$= \gamma^{23}(\gamma^{2}-1)+\delta^{23}(\delta^{2}-1)$$
$$\gamma^{2}+3\gamma-1=0 \Rightarrow \gamma^{2}-1=-3\gamma$$
$$\delta^{2}+3\delta-1=0 \Rightarrow \delta^{2}-1=-3\delta$$
$$= \gamma^{23}(-3\gamma)+\delta^{23}(-3\delta)$$
$$= -3(\gamma^{24}+\delta^{24})$$
$$= -3Q_{24}$$
$$\frac{Q_{25}-Q_{23}}{Q_{24}} = \frac{-3Q_{24}}{Q_{24}} = -3$$
$$\frac{P_{25}+\sqrt{3}P_{24}}{2P_{23}}+\frac{Q_{25}-Q_{23}}{Q_{24}} = 8 – 3 = 5$$
Ans. (3)