Question

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

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

∴   $a_{n} = \alpha^{n} - \beta^{n}$ for n ≥ 1

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

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

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

Now, consider

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

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

[∵𝛼 and β are the roots of x² - 6x - 2 or x² = 6x + 2

⇒ 𝛼² = 6𝛼 + 2 ⇒ 𝛼² - 2 = 6𝛼

and ꞵ² = 6ꞵ + 2 ⇒ ꞵ² - 2 = 6ꞵ]

Alternate Solution

Since 𝛼 and ꞵ are the roots of the equation

$x^{2} - 6x - 2 = 0$.

or

$x^{2} = 6x + 2$

∴ 𝛼² = 6𝛼 + 2

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

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

On Subtracting Eq (ii) from Eq (i), we get

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

(∵ $a_{n} = \alpha^{n} - \beta^{n}$)

⇒ $a_{10} = 6a_{9} + 2a_{8}$

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