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. $4$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is D $3$

$\alpha$ and $\beta$ be the roots of $x^2-6x-2=0$. So,
${\alpha }^2-6\alpha -2=0\Rightarrow {\alpha }^2-2=6\alpha$
${\beta }^2-6\beta -2=0\Rightarrow {\beta }^2-2=6\beta$
Now,
$\frac{a_{10}-2a_8}{3a_9}=\frac{({\alpha }^{10}-{\beta }^{10})-2({\alpha }^8-{\beta }^8)}{3({\alpha }^9-{\beta }^9)}$
$=\frac{({\alpha }^{10}-2{\alpha }^8)-({\beta }^{10}-2{\beta }^8)}{3({\alpha }^9-{\beta }^9)}$
$=\frac{{\alpha }^8({\alpha }^2-2)-{\beta }^8({\beta }^2-2)}{3({\alpha }^9-{\beta }^9)}$
$=\frac{{\alpha }^8\left(6\alpha \right)-{\beta }^8\left(6\beta \right)}{3({\alpha }^9-{\beta }^9)}=\frac{6({\alpha }^9-{\beta }^9)}{3({\alpha }^9-{\beta }^9)}=\frac{6}{3}=2$