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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

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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

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

(∵ \[a_{n} = \alpha^{n} - \beta^{n}\])

⇒ \[a_{10} = 6a_{9} + 2a_{8}\]

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

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