If alpha, beta are roots of the equation x^{2 | vedclass

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Question

$x^{2}+5 \sqrt{2} x+10=0$

$\& p_{n}=\alpha^{n}-\beta^{n} 	ext { (Given) }$

$	ext { Now } \frac{P_{17} P_{20}+5 \sqrt{2} p_{11} P_{19}}{P_{

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$1$

Vedclass · Answer

$x^{2}+5 \sqrt{2} x+10=0$

$\& p_{n}=\alpha^{n}-\beta^{n} 	ext { (Given) }$

$	ext { Now } \frac{P_{17} P_{20}+5 \sqrt{2} p_{11} P_{19}}{P_{18} P_{19}+5 \sqrt{2} P_{18}^{2}}=\frac{P_{17}\left(P_{20} 5 \sqrt{2} P_{19}ight)}{P_{18}\left(P_{19}+5 \sqrt{2 P}_{18}ight)}$

$\frac{P_{17}\left(\alpha^{20}-\beta^{20}+5 \sqrt{2}\left(\alpha^{19}-\beta^{19}ight)ight)}{P_{18}\left(\alpha^{19}-\beta^{19}+5 \sqrt{2}\left(\alpha^{18}-\beta^{18}ight)ight)}$

$\frac{P_{17}\left(\alpha^{19}(\alpha+5 \sqrt{2})-\beta^{19}(\beta+5 \sqrt{2})ight)}{P_{18}\left(\alpha^{18}(\alpha+5 \sqrt{2})-\beta^{18}(\beta+5 \sqrt{2})ight)}$

$	ext { Since } \alpha+5 \sqrt{2}=-10 / \alpha \ldots \ldots \ldots \ldots . .(1)$

$\&\, \beta+5 \sqrt{2}=-10 / \beta \ldots \ldots \ldots \ldots(2)$

Now put there values in above expression

$=-\frac{10 P_{1} P_{18}}{-10 P_{18} P_{17}}=1$

If $\alpha, \beta$ are roots of the equation $x^{2}+5 \sqrt{2} x+10=0, \alpha\,>\,\beta$ and $P_{n}=\alpha^{n}-\beta^{n}$ for each positive integer $\mathrm{n}$, then the value of $\left(\frac{P_{17} P_{20}+5 \sqrt{2} P_{11} P_{19}}{P_{18} P_{19}+5 \sqrt{2} P_{18}^{2}}\right)$ is equal to $....$

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