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# Let $\alpha$and $\beta$ be the roots of ${x^2} - 6x - 2 = 0$, with $\alpha > \beta$. 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

Last updated date: 20th Mar 2023
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$\frac{a_{10} - 2a_{8}}{2a_{9}}$ = $\frac{(\alpha^{10} - \beta^{10}) - 2(\alpha^{8} - \beta^{8})}{2(\alpha^{9} - \beta^{9})}$

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

∵𝛼 is root of x² - 6x - 2 = 0 ⇒ 𝛼² - 2 = 6𝛼

[ and β is root of x² - 6x - 2 = 0 ⇒ ꞵ² - 2 = 6ꞵ]

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