elementary-calculus-are-roots-of-equation-of-x-2-6x-2-0-define-t-n-n-n-n-1-then-evaluate-A-t-10-2t-8-2t-9-

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

\dots e l e m e n t a r y c a l c u l u s \dots

:: 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}} = ? ? ?

\dots m . n . 1970 \dots

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 ✓ ✓

\dots m . n . j u l y . 1970 \dots