Let [] and [] be the roots of x2 - 6x - 2 = 0.

Question

Let $$\alpha$$ and $$\beta$$ be the roots of x2 - 6x - 2 = 0. If an = $$\alpha$$n - $$\beta$$n for n $$ \ge $$ 1, then the value of $${{{a_{10}} - 2{a_8}} \over {3{a_9}}}$$ is :

Accepted Answer

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

$$\begin{pmatrix} {\alpha  + \beta  = 6}  \
   {\alpha \beta  =  - 2}  \

\end{pmatrix} $$

and $${\alpha ^2} - 6\alpha  - 2 = 0 \Rightarrow {\alpha ^2} - 2 = 6\alpha $$

$${\beta ^2} - 6\beta  - 2 = 0 \Rightarrow {\beta ^2} - 2 = 6\beta $$

$${{{a_{10}} - 2{a_8}} \over {3{a_9}}} = {{\left( {{\alpha ^{10}} - {\beta ^{10}}} \right) - 2\left( {{\alpha ^8} - {\beta ^8}} \right)} \over {3\left( {{\alpha ^9} - {\beta ^9}} \right)}}$$

$$ = {{\left( {{\alpha ^{10}} - 2{\alpha ^8}} \right) - \left( {{\beta ^{10}} - 2{\beta ^8}} \right)} \over {3\left( {{\alpha ^9} - {\beta ^9}} \right)}}$$

Now, $$ = {{{\alpha ^8}\left( {{\alpha ^2} - 2} \right) - {\beta ^8}\left( {{\beta ^2} - 2} \right)} \over {3\left( {{\alpha ^9} - {\beta ^9}} \right)}}$$

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

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