Let α and β are the roots of the equation x^2 −6x−2=0. If a_n = α^n −β^n for n ≥1 then the value of ((a_(10) −2a_8 )/(2a


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Question Number 131688 by liberty last updated on 07/Feb/21

$Let α and β are the roots$ $of the equation x^{2} - 6 x - 2 = 0 .$ $If a_{n} = α^{n} - β^{n} for n ⩾ 1$ $then the value of \frac{a_{10} - 2 a_{8}}{2 a_{9}} ?$
	Answered by mathmax by abdo last updated on 07/Feb/21
	$x^2 −6x−2=0 →Δ^′ =9+2=11 ⇒x_1 =3+(√(11)) and x_2 =3−(√(11)) a_n =x_1 ^n −x_2 ^n =(3+(√(11)))^n −(3−(√(11)))^n =Σ_(k=0) ^n C_n ^k 3^(n−k) ((√(11)))^k −Σ_(k=0) ^n C_n ^k 3^(n−k) (−(√(11)))^k =Σ_(k=0) ^n C_n ^k 3^(n−k) { 1−(−1)^k )((√(11)))^k =Σ_(p=0) ^([((n−1)/2)]) C_n ^(2p+1) 3^(n−2p−1) 2.((√(11)))^(2p+1) =2(√(11))Σ_(p=0) ^([((n−1)/2)]) C_n ^(2p+1) 3^(n−2p−1) (11)^p ⇒a_(10) =2(√(11))Σ_(p=0) ^4 C_(10) ^(2p+1) 3^(9−2p) (11)^p a_8 =2(√(11))Σ_(p=0) ^3 C_8 ^(2p+1) 3^(7−2p) (11)^p a_9 =2(√(11))Σ_(p=0) ^4 C_9 ^(2p+1) 3^(8−2p) (11)^p rest to finish the calculus...$
	$x^{2} - 6 x - 2 = 0 \to Δ^{^{'}} = 9 + 2 = 11 \Rightarrow x_{1} = 3 + \sqrt{11} and x_{2} = 3 - \sqrt{11}$ $a_{n} = x_{1}^{n} - x_{2}^{n} = {(3 + \sqrt{11})}^{n} - {(3 - \sqrt{11})}^{n}$ $= \sum_{k = 0}^{n} C_{n}^{k} 3^{n - k} {(\sqrt{11})}^{k} - \sum_{k = 0}^{n} C_{n}^{k} 3^{n - k} {(- \sqrt{11})}^{k}$ $= \sum_{k = 0}^{n} C_{n}^{k} 3^{n - k} {1 - {(- 1)}^{k}) {(\sqrt{11})}^{k}$ $= \sum_{p = 0}^{[\frac{n - 1}{2}]} C_{n}^{2 p + 1} 3^{n - 2 p - 1} 2 . {(\sqrt{11})}^{2 p + 1}$ $= 2 \sqrt{11} \sum_{p = 0}^{[\frac{n - 1}{2}]} C_{n}^{2 p + 1} 3^{n - 2 p - 1} {(11)}^{p}$ $\Rightarrow a_{10} = 2 \sqrt{11} \sum_{p = 0}^{4} C_{10}^{2 p + 1} 3^{9 - 2 p} {(11)}^{p}$ $a_{8} = 2 \sqrt{11} \sum_{p = 0}^{3} C_{8}^{2 p + 1} 3^{7 - 2 p} {(11)}^{p}$ $a_{9} = 2 \sqrt{11} \sum_{p = 0}^{4} C_{9}^{2 p + 1} 3^{8 - 2 p} {(11)}^{p} rest to finish the calculus . . .$
	Answered by bemath last updated on 07/Feb/21
	$x^2 −6x−2=0→ { (α),(β) :} ⇒α^2 −6α−2=0...(×α^8 ) ⇒α^(10) −6α^9 −2α^8 =0⇒2α^8 =α^(10) −6α^9 ⇒β^(10) −6β^9 −2β^9 =0⇒2β^8 =β^(10) −6β^9 (1)−(2)⇒2(α^8 −β^8 )=α^(10) −β^(10) −6α^9 +6β^9 a_(10) = α^(10) −β^(10) ; a_8 =α^8 −β^8 ; a_9 = α^9 −β^9 we find ((a_(10) −2a_8 )/(2a_9 )) = ((α^(10) −β^(10) −2(α^8 −β^8 ))/(2(α^9 −β^9 ))) = ((α^(10) −β^(10) −(α^(10) −β^(10) −6α^9 +6β^9 ))/(2(α^9 −β^9 )))= 3$
	$x^{2} - 6 x - 2 = 0 \to {\begin{cases} α \\ β \end{cases}$ $\Rightarrow α^{2} - 6 α - 2 = 0 . . . (\times α^{8})$ $\Rightarrow α^{10} - 6 α^{9} - 2 α^{8} = 0 \Rightarrow 2 α^{8} = α^{10} - 6 α^{9}$ $\Rightarrow β^{10} - 6 β^{9} - 2 β^{9} = 0 \Rightarrow 2 β^{8} = β^{10} - 6 β^{9}$ $(1) - (2) \Rightarrow 2 (α^{8} - β^{8}) = α^{10} - β^{10} - 6 α^{9} + 6 β^{9}$ $a_{10} = α^{10} - β^{10}; a_{8} = α^{8} - β^{8}; a_{9} = α^{9} - β^{9}$ $we find \frac{a_{10} - 2 a_{8}}{2 a_{9}} = \frac{α^{10} - β^{10} - 2 (α^{8} - β^{8})}{2 (α^{9} - β^{9})}$ $= \frac{α^{10} - β^{10} - (α^{10} - β^{10} - 6 α^{9} + 6 β^{9})}{2 (α^{9} - β^{9})} = 3$
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