Question

Let $\alpha$ and $\beta$ be the roots of equation $x^2-6x-2=0$ . If $a_n={\alpha }^n{\beta }^n$, for $n\le 1$, then the value of $\frac{a_{10}-2a_8}{2a_9}$ is equal to :

• A. 3
• B. -3
• C. 6
• D. -6

Answer 1

The correct option is A 3

$\alpha ,\beta =\frac{6\pm \sqrt{36+8}}{2}=3\pm \sqrt{11}\alpha =3+\sqrt{11}.\beta =3-\sqrt{11}\therefore a_n=(3+\sqrt{11}{)}^n-(3-\sqrt{11}{)}^n\frac{a_{10}-2a_8}{2a_9}=\frac{(3+\sqrt{11}{)}^{10}-(3-\sqrt{11}{)}^{10}-2(3+\sqrt{11}{)}^8+2(3-\sqrt{11}{)}^8}{2\left[\right(3+{\sqrt{11}}^9-(3-){\sqrt{11}}^9\left]\right)}=\frac{(3+{\sqrt{11}}^8[(3+){\sqrt{11}}^2-2]+)(3-{\sqrt{11}}^8[2-\left(3\right)-{\sqrt{11}}^2\left]\right)}{2\left[\right(3+{\sqrt{11}}^9-(3-){\sqrt{11}}^9]})=\frac{6(3+{\sqrt{11}}^9-6(3-\sqrt{11}{)}^9}{2\left[\right(3+{\sqrt{11}}^9-(3-){\sqrt{11}}^9\left]\right)}=\frac{6}{2}=3$