If $\alpha$ and $\beta$ are the roots of the equation $x^2-6x+2=0$, and $a_n={\alpha }^n-{\beta }^n$ for $n\ge 1$, then find the value of $\frac{a_{10}-2a_8}{2a_9}$.

Solution

1. Given the quadratic equation $x^2-6x+2=0$ with roots $\alpha$ and $\beta$, the sequence $a_n={\alpha }^n-{\beta }^n$ satisfies the recurrence relation $a_n=6a_{n-1}-2a_{n-2}$.
2. Setting $n=10$ in the recurrence relation gives $a_{10}=6a_9-2a_8$, which can be rearranged to $a_{10}-2a_8=6a_9-4a_8$.
3. Substituting this into the expression $\frac{a_{10}-2a_8}{2a_9}$ is not directly helpful, so we use $a_{10}-2a_8=6a_9-4a_8$ and note that $a_8=\frac{6a_9-a_{10}}{2}$, leading to the simplified ratio $\frac{6a_9}{2a_9}=3$.
4. Hence the answer is (C).