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Let $\alpha, \beta$ be the roots of the equation $x^{2}-6 x-2=0$ with $\alpha>\beta$. If $a_{n}=\alpha^{n}-\beta^{n}$ for $n \geq 1$, then the value of $\frac{a_{10}-2 a_{8}}{2 a_{9}}$ is
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$x^{2}-6 x-2=0$
$x^{n}-6 x^{n-1}-2 x^{n-2}=0$
$\Rightarrow x^{n}-2 x^{n-2}=6 x^{n-1}$
for $n=10$
$x^{10}-2 x^{8}=6 x^{9}$
$\alpha^{10}-2 \alpha^{8}=6 \alpha^{9}$
$\beta^{10}-2 \beta^{8}=6 \beta^{9}$
$\left(\alpha^{10}-\beta^{10}\right)-2\left(\alpha^{8}-\beta^{8}\right)=6\left(\alpha^{9}-\beta^{9}\right)$
$\Rightarrow a_{10}-2 a_{8}=6 a_{9}$
$\Rightarrow \frac{a_{10}-2 a_{8}}{2 a_{9}}=3$
$x^{n}-6 x^{n-1}-2 x^{n-2}=0$
$\Rightarrow x^{n}-2 x^{n-2}=6 x^{n-1}$
for $n=10$
$x^{10}-2 x^{8}=6 x^{9}$
$\alpha^{10}-2 \alpha^{8}=6 \alpha^{9}$
$\beta^{10}-2 \beta^{8}=6 \beta^{9}$
$\left(\alpha^{10}-\beta^{10}\right)-2\left(\alpha^{8}-\beta^{8}\right)=6\left(\alpha^{9}-\beta^{9}\right)$
$\Rightarrow a_{10}-2 a_{8}=6 a_{9}$
$\Rightarrow \frac{a_{10}-2 a_{8}}{2 a_{9}}=3$
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