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