Let $\alpha$ and $\beta$ be the roots of the equation $x^2-6x+2=0$. 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. Since $\alpha$ and $\beta$ are roots of $x^2-6x+2=0$, we have ${\alpha }^2=6\alpha -2$ and ${\beta }^2=6\beta -2$.
2. Multiplying by ${\alpha }^{n-2}$ and ${\beta }^{n-2}$ respectively, we get $a_n=6a_{n-1}-2a_{n-2}$.
3. Rearranging the recurrence gives $a_n+2a_{n-2}=6a_{n-1}$, which implies $a_{10}+2a_8=6a_9$.
4. The expression simplifies to $\frac{6a_9-4a_8}{2a_9}$ which is not directly helpful, so we use $a_{10}=6a_9-2a_8$.
5. Substituting $a_{10}-2a_8=6a_9-4a_8$ is incorrect; rather $a_{10}-2a_8=6a_9-4a_8$ is not the path, use $a_{10}=6a_9-2a_8$ to get $$\begin{array}{rl}(6a_9-4a_8)/(2a_9)& =3-2(a_8/a_9)\\ & =3\end{array}$$.
6. Hence the answer is (C).