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

Let α and β be the roots of x² - 6x - 2 = 0, with α > β. If aₙ = αⁿ - βⁿ for n ≥ 1, then the value of (a₁₀ - 2a₈)/2a₉ is

$\text{Given: } \alpha,\beta \text{ roots of } x^{2}-6x-2=0,\ \alpha>\beta$

$\alpha+\beta = 6, \qquad \alpha\beta = -2$

$a_{n} = \alpha^{n}-\beta^{n}, \quad n \geq 1$

$\text{To find: } \frac{a_{10}-2a_{8}}{2a_{9}}$

$\text{Options: } 1,\ 2,\ 3,\ 4$

$\alpha^{2} = 6\alpha+2, \qquad \beta^{2} = 6\beta+2$

$\alpha^{n} = 6\alpha^{n-1}+2\alpha^{n-2}$

$\beta^{n} = 6\beta^{n-1}+2\beta^{n-2}$

$a_{n} = 6a_{n-1}+2a_{n-2}$

$a_{1} = \alpha-\beta, \qquad a_{2} = \alpha^{2}-\beta^{2} = 6a_{1}$

$a_{10} = 6a_{9}+2a_{8}$

$a_{10}-2a_{8} = (6a_{9}+2a_{8})-2a_{8}$

$= 6a_{9}$

$\frac{a_{10}-2a_{8}}{2a_{9}} = \frac{6a_{9}}{2a_{9}} = 3$

$\boxed{3} \quad \text{(option 3)}$