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ધારો કે $\alpha, \beta ; \alpha>\beta$ એ સમીકરણ $x^2-\sqrt{2} x-\sqrt{3}=0$ ના બીજ છે. ધારો કે $\mathrm{P}_n=\alpha^n-\beta^n, n \in \mathbb{N}$. તો $(11 \sqrt{3}-10 \sqrt{2}) \mathrm{P}_{10}+(11 \sqrt{2}+10) \mathrm{P}_{11}-11 \mathrm{P}_{12}=$ .............
$10 \sqrt{2} \mathrm{P}_9$
$10 \sqrt{3} \mathrm{P}_9$
$11 \sqrt{2} \mathrm{P}_9$
$11 \sqrt{3} \mathrm{P}_9$
Solution
$ x^2-\sqrt{2 x}-\sqrt{3}=0\left\langle_\beta^\alpha\right. $
$ \alpha^{n+2}-\sqrt{2} \alpha^{n+1}-\sqrt{3} \alpha^n=0 $
$ \text { and } \beta^{n+2}-\sqrt{2} \beta^{n+1}-\sqrt{3} \beta^n=0$
Subtracting
$ \left(\alpha^{n+2}-\beta^{n+2}\right)-\sqrt{2}\left(\alpha^{n+1}-\beta^{n+1}\right)-\sqrt{3}\left(\alpha^n-\beta^n\right)=0 $
$ \Rightarrow P_{n+2}-\sqrt{2} P_{n+1}-\sqrt{3} P_n=0$
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$
