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Let $\alpha$ and $\beta$ be the roots of $x^2-6 x-2=0$ 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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The correct answer is:
3
3
$\frac{a_{10}-2 a_8}{2 a_9}=\frac{\left(\alpha^{10}-\beta^{10}\right)-2\left(\alpha^8-\beta^8\right)}{2\left(\alpha^9-\beta^9\right)}$ $=\frac{\alpha^8\left(\alpha^2-2\right)-\beta^8\left(\beta^2-2\right)}{2\left(\alpha^9-\beta^9\right)}$ $\left[\because \alpha\right.$ is root of $x^2-6 x-2=0$ $\left.\alpha^2-2=6 \alpha\right]$
Also, $\beta$ is root of $x^2-6 x-2=0$
$$
\begin{aligned}
& \Rightarrow \quad \beta^2-2=6 \beta \\
& =\frac{\alpha^8(6 \alpha)-\beta^8(6 \beta)}{2\left(\alpha^9-\beta^9\right)}=\frac{6\left(\alpha^9-\beta^9\right)}{2\left(\alpha^9-\beta^9\right)}=3
\end{aligned}
$$
Also, $\beta$ is root of $x^2-6 x-2=0$
$$
\begin{aligned}
& \Rightarrow \quad \beta^2-2=6 \beta \\
& =\frac{\alpha^8(6 \alpha)-\beta^8(6 \beta)}{2\left(\alpha^9-\beta^9\right)}=\frac{6\left(\alpha^9-\beta^9\right)}{2\left(\alpha^9-\beta^9\right)}=3
\end{aligned}
$$
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