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

• A. 6
• B. -6
• C. 3
• D. -3

Answer 1

The correct option is C 3

Given, $\alpha$ and $\beta$ are the roots of the equation $x^2-6x-2=0$.
$\therefore a_n=a^n-{\beta }^{n}forn\ge 1a_{10}={\alpha }^{10}-{\beta }^{10}{a}_8={\alpha }^8-{\beta }^8{a}_9={\alpha }^9-{\beta }^9$
Now, consider
$\frac{a_{10}-2a_8}{2a_9}=\frac{{\alpha }^{10}-{\beta }^{10}-2({\alpha }^8-{\beta }^8)}{2({\alpha }^9-{\alpha }^9)}=\frac{{\alpha }^8({\alpha }^2-2)-{\beta }^8({\beta }^2-2)}{2({\alpha }^9-{\beta }^9)}=\frac{{\alpha }^8.6\alpha -{\beta }^{8}6\beta }{2({\alpha }^9-{\beta }^9)}=\frac{6{\alpha }^9-6{\beta }^9}{2({\alpha }^9-6{\beta }^9)}=\frac{6}{2}=3$
$\left[\begin{array}{c}\because \alpha and\beta aretherootsof\\ x^2-6x-2=0or{x}^2=6x+2\\ \Rightarrow {\alpha }^2=6\alpha +2\Rightarrow {\alpha }^2-2=6\alpha \\ and{\beta }^2=6\beta +2\Rightarrow {\beta }^2-2=6\beta \end{array}\right]$
Alternate Solution
Since, $\alpha$ and $\beta$ are the roots of the equation
$x^2-6x-2=0$
or $x^2=6x+2$
$\therefore {\alpha }^2=6\alpha +2\Rightarrow {\alpha }^{10}=6{\alpha }^9+2{\alpha }^8....\left(i\right)$
Similarly, ${\beta }^{10}=6{\beta }^9+2{\beta }^8....\left(ii\right)$
On subtracting Eq. (ii) from Eq. (i), we get
${\alpha }^{10}-{\beta }^{10}=6({\alpha }^9-{\beta }^9)+2({\alpha }^8-{\beta }^8)(\because a_n={\alpha }^n-{\beta }^n)\Rightarrow a_{10}=6a_9+2a_8\Rightarrow a_{10}-2a_8=6a_9\Rightarrow \frac{a_{10}-2a_8}{2a_9}=3$