Let $$\alpha \,\,{\text{and }}\beta $$   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 equal to:

Solution :
$$\eqalign{ & \alpha ,\beta = \frac{{6 \pm \sqrt {36 + 8} }}{2} = 3 \pm \sqrt {11} \cr & \alpha = 3 + \sqrt {11} ,\beta = 3 - \sqrt {11} \cr & \therefore \,\,{a_n} = {\left( {3 + \sqrt {11} } \right)^n} - {\left( {3 - \sqrt {11} } \right)^n} \cr & \frac{{{a_{10}} - 2{a_8}}}{{2{a_9}}} \cr & = \frac{{{{\left( {3 + \sqrt {11} } \right)}^{10}} - {{\left( {3 - \sqrt {11} } \right)}^{10}} - 2{{\left( {3 + \sqrt {11} } \right)}^8} + 2{{\left( {3 - \sqrt {11} } \right)}^8}}}{{2\left[ {{{\left( {3 + \sqrt {11} } \right)}^9} - {{\left( {3 - \sqrt {11} } \right)}^9}} \right]}} \cr & = \frac{{{{\left( {3 + \sqrt {11} } \right)}^8}\left[ {{{\left( {3 + \sqrt {11} } \right)}^2} - 2} \right] + {{\left( {3 - \sqrt {11} } \right)}^8}\left[ {2 - {{\left( {3 - \sqrt {11} } \right)}^2}} \right]}}{{2\left[ {{{\left( {3 + \sqrt {11} } \right)}^9} - {{\left( {3 - \sqrt {11} } \right)}^9}} \right]}} \cr & = \frac{{{{\left( {3 + \sqrt {11} } \right)}^8}\left( {9 + 11 + 6\sqrt {11} - 2} \right) + {{\left( {3 - \sqrt {11} } \right)}^8}\left( {2 - 9 - 11 + 6\sqrt {11} } \right)}}{{2\left[ {{{\left( {3 + \sqrt {11} } \right)}^9} - {{\left( {3 - \sqrt {11} } \right)}^9}} \right]}} \cr & = \frac{{6{{\left( {3 + \sqrt {11} } \right)}^9} - 6{{\left( {3 - \sqrt {11} } \right)}^9}}}{{2\left[ {{{\left( {3 + \sqrt {11} } \right)}^9} - {{\left( {3 - \sqrt {11} } \right)}^9}} \right]}} = \frac{6}{2} = 3 \cr} $$