Let $$\alpha$$ and $$\beta$$ be the roots of equation $$x^2 - 6x - 2 = 0$$. If $$a_n = \alpha^n - \beta^n$$, $$\forall n \geq 1$$, then the value of $$\frac{a_{10} - 2a_8}{2a_9}$$ is equal to

We start with the quadratic equation $$x^2-6x-2=0$$ whose roots are denoted by $$\alpha$$ and $$\beta$$. From the standard relation between coefficients and roots of a quadratic, we immediately have

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

For every integer $$n\ge 1$$ the sequence $$a_n$$ is defined by

$$a_n=\alpha^n-\beta^n.$$

Before substituting any particular index, we first derive a recurrence relation for $$a_n$$. We write the expression for $$a_n$$ and use the identities for $$\alpha+\beta$$ and $$\alpha\beta$$:

$$\alpha^n-\beta^n=(\alpha+\beta)(\alpha^{\,n-1}-\beta^{\,n-1})-\alpha\beta(\alpha^{\,n-2}-\beta^{\,n-2}).$$

This rearrangement is obtained by adding and subtracting the same term so that the coefficients match, and it is valid for all $$n\ge 2$$. Now we recognize the right-hand side in terms of earlier members of the sequence:

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

Substituting these into the previous formula gives

$$a_n=(\alpha+\beta)a_{n-1}-\alpha\beta\,a_{n-2}.$$

Now we plug in the numerical values $$\alpha+\beta=6$$ and $$\alpha\beta=-2$$:

$$a_n=6a_{n-1}-(-2)a_{n-2}.$$

Because $$-(-2)=+2$$, the recurrence simplifies to

$$a_n=6a_{n-1}+2a_{n-2}\qquad(n\ge 2).$$

With this powerful relation in hand, we proceed to compute the quantity requested in the problem, namely

$$\frac{a_{10}-2a_8}{2a_9}.$$

First we express $$a_{10}$$ through the recurrence. Taking $$n=10$$ gives

$$a_{10}=6a_9+2a_8.$$

Now we form the numerator of our fraction:

$$a_{10}-2a_8=(6a_9+2a_8)-2a_8=6a_9.$$

Thus the entire fraction becomes

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

Because $$a_9$$ is non-zero (since $$\alpha\neq\beta$$), we can cancel it safely:

$$\frac{6a_9}{2a_9}=\frac{6}{2}=3.$$

Hence, the correct answer is Option D.