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

Solution

1. The roots satisfy ${\alpha }^2-6\alpha +1=0$ and ${\beta }^2-6\beta +1=0$.
2. Multiplying by ${\alpha }^{n-2}$ and ${\beta }^{n-2}$ respectively and subtracting gives $a_n-6a_{n-1}+a_{n-2}=0$.
3. We need to evaluate $\frac{a_{10}-2a_8}{2a_9}$.
4. Using $a_{10}=6a_9-a_8$, the expression becomes $$\begin{array}{rl}\frac{6a_9-a_8-2a_8}{2a_9}& =\frac{6a_9-3a_8}{2a_9}\\ & =3-\frac{3a_8}{2a_9}\end{array}$$.
5. Since $a_9=6a_8-a_7$, the ratio simplifies to 3.
6. Hence the answer is (C).