Theory of Equations 1 Question 20
21. Let $\alpha$ and $\beta$ be the roots of $x^{2}-6 x-2=0$, with $\alpha>\beta$. If $a _n=\alpha^{n}-\beta^{n}$ for $n \geq 1$, then the value of $\frac{a _{10}-2 a _8}{2 a _9}$ is
(a) 1
(b) 2
(c) 3
(d) 4 (2011)
Show Answer
Solution:
- $\frac{a _{10}-2 a _8}{2 a _9}=\frac{\left(\alpha^{10}-\beta^{10}\right)-2\left(\alpha^{8}-\beta^{8}\right)}{2\left(\alpha^{9}-\beta^{9}\right)}$
$$ =\frac{\alpha^{8}\left(\alpha^{2}-2\right)-\beta^{8}\left(\beta^{2}-2\right)}{2\left(\alpha^{9}-\beta^{9}\right)} $$
$\because \alpha$ is root of $x^{2}-6 x-2=0 \Rightarrow \alpha^{2}-2=6 \alpha$
[and $\beta$ is root of $x^{2}-6 x-2=0 \Rightarrow \beta^{2}-2=6 \beta$ ]
$$ =\frac{\alpha^{8}(6 \alpha)-\beta^{8}(6 \beta)}{2\left(\alpha^{9}-\beta^{9}\right)}=\frac{6\left(\alpha^{9}-\beta^{9}\right)}{2\left(\alpha^{9}-\beta^{9}\right)}=3 $$
