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Question Number 120855 by mnjuly1970 last updated on 03/Nov/20

$. . . e l e m e n t a r y c a l c u l u s . . .$ $:: α, β a r e r o o t s o f e q u a t i o n$ $o f : x^{2} - 6 x - 2 = 0$ $d e f i n e :: t_{n} = α^{n} - β^{n} (n ⩾ 1)$ $t h e n e v a l u a t e :$ $A = \frac{t_{10} - 2 t_{8}}{2 t_{9}} = ? ? ?$ $. . . m . n . 1970 . . .$
	Answered by TANMAY PANACEA last updated on 03/Nov/20

	$x^{2} - 6 x - 2 = 0$ $x = \frac{6 \pm \sqrt{36 + 8}}{2} = 3 \pm \sqrt{11}$ $t_{n} = {(3 + \sqrt{11})}^{n} - {(3 - \sqrt{11})}^{n}$
	Answered by mnjuly1970 last updated on 07/Nov/20

	$s o l u t i o n ::$ $α + β = - \frac{b}{a} = 6$ $α β = \frac{c}{a} = - 2$ $t_{n} = α^{n} - β^{n}$ $A = \frac{α^{10} - β^{10} + α β (α^{8} - β^{8})}{2 (α^{9} - β^{9})}$ $= \frac{α^{9} (α + β) - β^{9} (α + β)}{2 (α^{9} - β^{9})}$ $= \frac{α + β}{2} = 6 ✓ ✓$ $. . . m . n . j u l y . 1970 . . .$
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