Let α&β be the roots of x2−6x−2=0 , with α>β . If an=αn−βn for n≥1 , then the value of a10−2a86a9= ___________

a10−2a86a9=(α10−β10)−2(α8−β8)6(α9−β9) =α8(α2−2)−β8(β2−2)6(α9−β9) (∴α is root of x2−6x−2=0⇒α2−2=6α) (∴Also, β is root of x2−6x−2=0⇒β2−2=6β) ⇒α8(6α−β8(6β))6(α9−β9) =6(α9−β9)6(α9−β9)=1

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Let $α & β$ be the roots of $x^{2} - 6 x - 2 = 0$ , with $α > β$ . If $a_{n} = α^{n} - β^{n}$ for $n \geq 1$ , then the value of $\frac{a_{10} - 2 a_{8}}{6 a_{9}} =$ ___________

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Detailed Solution

$\frac{a_{10} - 2 a_{8}}{6 a_{9}} = \frac{(α^{10} - β^{10}) - 2 (α^{8} - β^{8})}{6 (α^{9} - β^{9})} = \frac{α^{8} (α^{2} - 2) - β^{8} (β^{2} - 2)}{6 (α^{9} - β^{9})} (∴ α is root of x^{2} - 6 x - 2 = 0 \Rightarrow α^{2} - 2 = 6 α) (∴ Also, β is root of x^{2} - 6 x - 2 = 0 \Rightarrow β^{2} - 2 = 6 β) \Rightarrow \frac{α^{8} (6 α - β^{8} (6 β))}{6 (α^{9} - β^{9})} = \frac{6 (α^{9} - β^{9})}{6 (α^{9} - β^{9})} = 1$