Let α and β be the root of x2−6x−2=0 with α>β. If an=αn−βn for n≥1 the value of a10−2a82a9 is

∵α2−6α−2=0⇒α2−2=6αand β2−6β−2=0⇒β2−2=6β∴ a10−2a82a9=α10−β10−2α8−β82α9−β9=α8α2−2−β8β2−22α9−β9=α8⋅6α−β8⋅6β2α9−β9 [from Eqs. (i) and (ii)] =6α9−β92α9−β9=3

Let α and β be the root of x2−6x−2=0 with α>β. If an=αn−βn for n≥1 the value of a10−2a82a9 is

∵α2−6α−2=0⇒α2−2=6αand β2−6β−2=0⇒β2−2=6β∴ a10−2a82a9=α10−β10−2α8−β82α9−β9=α8α2−2−β8β2−22α9−β9=α8⋅6α−β8⋅6β2α9−β9 [from Eqs. (i) and (ii)] =6α9−β92α9−β9=3

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Let $α$ and $β$ be the root of $x^{2} - 6 x - 2 = 0$ with $α > β .$ If $a_{n} = α^{n} - β^{n} for n \geq 1$ the value of $\frac{a_{10} - 2 a_{8}}{2 a_{9}}$ is

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Detailed Solution

$∵ α^{2} - 6 α - 2 = 0 \Rightarrow α^{2} - 2 = 6 α$

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

$\begin{array}{l} ∴ \frac{a_{10} - 2 a_{8}}{2 a_{9}} = \frac{(α^{10} - β^{10}) - 2 (α^{8} - β^{8})}{2 (α^{9} - β^{9})} \\ = \frac{α^{8} (α^{2} - 2) - β^{8} (β^{2} - 2)}{2 (α^{9} - β^{9})} \end{array}$