Let $\alpha, \beta ; \alpha\beta$, be the roots of the equation $x^2-\sqrt{2} x-\sqrt{3}=0$. Let $\mathrm{P}_n=\alpha^n-\beta^n, n \in \mathrm{N}$. Then $(11 \sqrt{3}-10 \sqrt{2}) \mathrm{P}_{10}+(11 \sqrt{2}+10) \mathrm{P}_{11}-11 \mathrm{P}_{12}$ is equal to

Question

Let $\alpha, \beta ; \alpha>\beta$, be the roots of the equation $x^2-\sqrt{2} x-\sqrt{3}=0$. Let $\mathrm{P}_n=\alpha^n-\beta^n, n \in \mathrm{N}$. Then $(11 \sqrt{3}-10 \sqrt{2}) \mathrm{P}_{10}+(11 \sqrt{2}+10) \mathrm{P}_{11}-11 \mathrm{P}_{12}$ is equal to

Aicharya · Accepted Answer

$\begin{aligned} & \mathrm{x}^2-\sqrt{2 \mathrm{x}}-\sqrt{3}=0\left\langle_\beta^\alpha\right. \\ & \alpha^{\mathrm{n}+2}-\sqrt{2} \alpha^{\mathrm{n}+1}-\sqrt{3} \alpha^{\mathrm{n}}=0 \\ & \text { and } \beta^{\mathrm{n}+2}-\sqrt{2} \beta^{\mathrm{n}+1}-\sqrt{3} \beta^{\mathrm{n}}=0 \end{aligned}$
Subtracting $\begin{aligned} & \left(\alpha^{\mathrm{n}+2}-\beta^{\mathrm{n}+2}\right)-\sqrt{2}\left(\alpha^{\mathrm{n}+1}-\beta^{\mathrm{n}+1}\right)-\sqrt{3}\left(\alpha^{\mathrm{n}}-\beta^{\mathrm{n}}\right)=0 \\ & \Rightarrow \mathrm{P}_{\mathrm{n}+2}-\sqrt{2} \mathrm{P}_{\mathrm{n}+1}-\sqrt{3} \mathrm{P}_{\mathrm{n}}=0 \end{aligned}$ $\begin{aligned} & \text { Put } \mathrm{n}=10 \\ & \mathrm{P}_{12}-\sqrt{2} \mathrm{P}_{11}-\sqrt{3} \mathrm{P}_{10}=0 \\ & \mathrm{n}=9 \\ & \mathrm{P}_{11}-\sqrt{2} \mathrm{P}_{10}-\sqrt{3} \mathrm{P}_9=0 \\ & 11\left(\sqrt{3} \cdot \mathrm{P}_{10}+\sqrt{2} \mathrm{P}_{11}-\mathrm{P}_{11}\right)-10\left(\sqrt{2} \mathrm{P}_{10}-\mathrm{P}_{11}\right) \\ & =0-10\left(-\sqrt{3} \mathrm{P}_9\right)=10 \sqrt{3} \mathrm{P}_9\end{aligned}$

Let $\alpha, \beta ; \alpha>\beta$, be the roots of the equation $x^2-\sqrt{2} x-\sqrt{3}=0$. Let…

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