Let and be the roots of x^2+√3x-16=0, and and be the roots of x^2+3x-1=0.

Question

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 ${Q}_{n}={\gamma }^{n}+{\hat{o}}^{n}$, then $\frac{{P}_{25}+\sqrt{3}{P}_{24}}{2{P}_{23}}+\frac{{Q}_{25}-{Q}_{23}}{{Q}_{24}}$ is equal to

Accepted Answer

$$\begin{aligned}
& \Rightarrow \quad P^n=\gamma^n+\delta^n \
& P_{25}-P_{23}=\left(\gamma^{25}-\gamma^{23}\right)+\left(\delta^{25}-\delta^{23}\right) \
& =\gamma^{23}\left(\gamma^2-1\right)+\delta^{23}\left(\delta^2-1\right) \
& =\gamma^{23}(-3 \gamma)+\delta^{23}(-3 \delta)=-3\left[\gamma^{24}+\delta^{24}\right]
\end{aligned}$$

$$\begin{aligned}
&\Rightarrow \frac{P_{25}-P_{23}}{P_{24}}=(-3)\
&\text { Similarly, }
\end{aligned}$$

$$\begin{aligned}
& \Rightarrow Q_{25}+\sqrt{3} Q_{24}=\left(\alpha^{25}+\sqrt{3} \alpha^{24}\right)+\left(\beta^{25}+\sqrt{3} \beta^{24}\right) \
& =\alpha^{23}\left(\alpha^2+\sqrt{3} \alpha\right)+\beta^{23}\left(\beta^2+\sqrt{3} \beta\right) \
& =\alpha^{23}(16)+16 \beta^{23} \
& \Rightarrow \frac{Q_{25}+\sqrt{3} Q_{24}}{2 \cdot Q_{23}}=\frac{16\left(\alpha^{23}+\beta^{23}\right)}{2\left(\alpha^{23}+\beta^{23}\right)}=8 \
& \Rightarrow \frac{Q_{25}+\sqrt{3} Q_{24}}{2 Q_{23}}+\frac{\left(P_{25}-P_{23}\right)}{P_{24}}=8+(-3)=5
\end{aligned}$$

Solutions · 1