Question

Let $\alpha$ and $\beta$ be the roots of $x^2-6x-2=0$. If $a_n={\alpha }^n-{\beta }^n$ for $n\ge 1,$ then the value of $\frac{a_{10}-2a_8}{3a_9}$ is :

• A. 2
• B. 1
• C. 4
• D. 3

Answer 1

Answer: (A)

${\alpha }^2-6\alpha -2=0$
${\alpha }^{10}-6{\alpha }^9-2{\alpha }^8=0$
Similarly $\frac{{\beta }^{10}-6{\beta }^9-2{\beta }^8=0}{\left({\alpha }^{10}-{\beta }^{10}\right)-6\left({\alpha }^9-{\beta }^9\right)-2\left({\alpha }^8-{\beta }^8\right)}=0$
$\Rightarrow a_{10}-6a_9-2a_8=0$
$\Rightarrow \frac{a_{10}-2a_8}{3a_9}=2$