Let alpha and beta be the roots of x^2+sqrt{3 | vedclass

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(C) For the equation $x^2+\sqrt{3}x-16=0$,since $\alpha$ and $\beta$ are roots,we have $\alpha^2+\sqrt{3}\alpha-16=0$ and $\beta^2+\sqrt{3}\beta-16=0$

Vedclass · Accepted Answer

$5$

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(C) For the equation $x^2+\sqrt{3}x-16=0$,since $\alpha$ and $\beta$ are roots,we have $\alpha^2+\sqrt{3}\alpha-16=0$ and $\beta^2+\sqrt{3}\beta-16=0$. Multiplying by $\alpha^{n-2}$ and $\beta^{n-2}$ respectively and adding,we get $P_n+\sqrt{3}P_{n-1}-16P_{n-2}=0$. For $n=25$,$P_{25}+\sqrt{3}P_{24}=16P_{23}$. Thus,$\frac{P_{25}+\sqrt{3}P_{24}}{2P_{23}} = \frac{16P_{23}}{2P_{23}} = 8$. For the equation $x^2+3x-1=0$,since $\gamma$ and $\delta$ are roots,we have $\gamma^2+3\gamma-1=0$ and $\delta^2+3\delta-1=0$,which implies $\gamma^2-1=-3\gamma$ and $\delta^2-1=-3\delta$. Then $Q_{25}-Q_{23} = (\gamma^{25}+\delta^{25})-(\gamma^{23}+\delta^{23}) = \gamma^{23}(\gamma^2-1)+\delta^{23}(\delta^2-1) = \gamma^{23}(-3\gamma)+\delta^{23}(-3\delta) = -3(\gamma^{24}+\delta^{24}) = -3Q_{24}$. Thus,$\frac{Q_{25}-Q_{23}}{Q_{24}} = -3$. Finally,$8-3=5$.

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

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