Question

Let $\alpha$ and $\beta$ be the roots of 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?

• A. $6$
• B. $-6$
• C. $3$
• D. $-3$

Answer 1

The correct option is C $3$

$\alpha$ and $\beta$ are the roots of the equation

$x^2-6x-2=0$

${\alpha }^2=6\alpha +2$

${\alpha }^{10}=6{\alpha }^9+2{\alpha }^8----------\left(i\right)$

${\beta }^{10}=6{\beta }^9+2{\beta }^8---------------\left(ii\right)$

Subtracting equation (ii) from (i)
$\Rightarrow {\alpha }^{10}-{\beta }^{10}=6({\alpha }^9-{\beta }^9)+2({\alpha }^8-{\beta }^9)$

$\Rightarrow a_{10}=6a_9+2a_8(\therefore a_n=a^n-{\beta }^n)$

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

$\Rightarrow \frac{a_{10}-2a_8}{2a_9}=3$