Question

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

• A. 2
• B. 4
• C. 3
• D. 1

Answer 1

The correct option is B 4

We observe that

$(\alpha +\beta )({\alpha }^{n-1}-{\beta }^{n-1})={\alpha }^n-{\beta }^n+\alpha \beta ({\alpha }^{n-2}-{\beta }^{n-2})$
$\Rightarrow 6a_{n-1}=a_n-2a_{n-2}$
$\Rightarrow 6=\frac{a_n-2a_{n-2}}{a_{n-1}}$
$\Rightarrow 3=\frac{a_n-2a_{n-2}}{2a_{n-1}},\forall n>2$
Putting $n=10,$we get

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