Let $\alpha$ and $\beta$ be the roots of $x^2-6x-2=0$ with $\alpha >\beta$ if $a_n={\alpha }^n-{\beta }^n$ for $n\ge 1$ then the value of $\frac{a_{10}-2a_8}{3a_9}=$

The correct option is D. $2$

Given:

$\alpha$ and $\beta$ be the roots of $x^2-6x-2=0$

${\alpha }^2-6\alpha -2=0$

$\Rightarrow {\alpha }^2=6\alpha +2$

$\Rightarrow {\alpha }^2=6(\alpha {)}^{2-1}+2(\alpha {)}^{2-2}$

$\Rightarrow {\alpha }^{10}=6(\alpha {)}^{10-1}+2(\alpha {)}^{10-2}$

$\Rightarrow {\alpha }^{10}=6{\alpha }^9+2{\alpha }^8$-------(1)

${\beta }^2-6\beta -2=0$

$\Rightarrow {\beta }^{10}=6{\beta }^9+2{\beta }^8$-------(2)

substract eq (2) from eq (1)

${\alpha }^{10}-{\beta }^{10}=6({\alpha }^9-{\beta }^9)+2({\alpha }^8-{\beta }^8)$

$\Rightarrow a_{10}=6a_9+2a_8$ [$\because a_n={\alpha }^n-{\beta }^n$]

$\Rightarrow a_{10}-2a_8=6a_9$

$\Rightarrow a_{10}-2a_8=2\times 3a_9$

$\Rightarrow \frac{a_{10}-2a_8}{3a_9}=2$