Question

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

Answer 1

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

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

∵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