Discussion Forum

Let $\alpha$ and $\beta$  be the roots of $x^{2}-6x-2=0$ with $\alpha > \beta$ , if $a_{n}=\alpha^{n}-\beta^{n}$ for $\geq$1, then the value of $ \frac{ a_{10}-2a_{8}}{2a_{9}}$ is 

Answer:

Explanation:

$\frac{a_{10}-2a_{8}}{2a_{9}}=\frac{(\alpha^{10}-\beta^{10})-2(\alpha^{8}-\beta^{8})}{2(\alpha^{9}-\beta^{9})}$

=$\frac{\alpha^{8}(\alpha^{2}-2)-\beta^{8}(\beta^{2}-2)}{2(\alpha^{9}-\beta^{9})}$

 [ $\therefore$  is root of  $x^{2}-6x-2=0$

                                            $\alpha^{2}-2=6\alpha]$

Also,$\beta$ is the root of $x^{2}-6x-2=0$

$\rightarrow$  $\beta^{2}-2=6 \beta$

$= \frac{\alpha^{8}(6\alpha)-\beta^{8}(6\beta)}{2(\alpha^{9}-\beta^{9})}=\frac{6(\alpha^{9}-\beta^{9})}{2(\alpha^{9}-\beta^{9})}=3$