Discussion Forum

Let  $\alpha$ and  $\beta$  be thr roots of equation   $x^{2}-6x-2=0$ . If   $a_{n}=\alpha^{n}-\beta^{n}$ , for   $n\geq 1$  , then the value  of   $\frac{a_{10}-2a_{8}}{2a_{9}}$  is equal to

Answer:

Explanation:

 Given,    $\alpha$ and  $\beta$  are the roots of the equation  $x^{2}-6x-2=0$

 $\therefore$    $a_{n}=\alpha^{n}-\beta^{n}$       $n\geq 1$ 

   $\therefore$         $a_{10}=\alpha^{10}-\beta^{10}$

                              $a_{8}=\alpha^{8}-\beta^{8}$

                              $a_{9}=\alpha^{9}-\beta^{9}$

Now, consider

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

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

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

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

{   $\therefore$   $\alpha$  and  $\beta$  are the roots of the equation:

     $x^{2}-6x-2=0$ 

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

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

     and      $\beta^{2}=6\beta-2$

     $\Rightarrow$    $\beta^{2}-2=6\beta$  }

 Alter :

   Since ,  $\alpha$ and  $\beta$  be thr roots of equation   $x^{2}-6x-2=0$ 

   or    $x^{2}=6x+2$

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

   $\Rightarrow$     $\alpha^{10}=6\alpha^{9}+2\alpha^{8}$  ......(i)

Similarly,    $\beta^{10}=6\beta^{9}+2\beta^{8}$.................(ii)

On subtracting  Eq. (ii) from Eq.(i) , we get

     $\alpha^{10}-\beta^{10}=6(\alpha^{9}-\beta^{9})+2(\alpha^{8}-\beta^{8})$  ( $\Rightarrow a_{10}=6a_{9}+2a_{8}  (\because a_{n}=\alpha^{n}-\beta^{n})$ )

 $\Rightarrow a_{10}-2a_{8}=6a_{9}\Rightarrow \frac{a_{10}-2a_{8}}{2a_{9}}=3$