{ ((a_1 =((√3)/2))),((a_(n+1) =4a_n ^3 −3a_n ; ∀n≥1)) :} a


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Question Number 157156 by Rasheed.Sindhi last updated on 20/Oct/21
${ ((a_1 =((√3)/2))),((a_(n+1) =4a_n ^3 −3a_n ; ∀n≥1)) :} a_n =?$
${\begin{cases} a_{1} = \frac{\sqrt{3}}{2} \\ a_{n + 1} = 4 a_{n}^{3} - 3 a_{n}; \forall n ⩾ 1 \end{cases}$ $a_{n} = ?$
	Answered by Rasheed.Sindhi last updated on 20/Oct/21
	${ ((a_1 =((√3)/2))),((a_(n+1) =4a_n ^3 −3a_n ; ∀n≥1)) :};a_n =? a_(n+1) =4a_n ^3 −3a_n =a_n (4a_n ^2 −3) a_(1+1) =a_1 (4a_1 ^2 −3)=((√3)/2)(4(((√3)/2))^2 −3) a_2 =((√3)/2)(4((3/4))−3)=0 a_3 =a_2 (4a_2 ^2 −3)=0 ..... ... a_(n−1) =0 a_n =a_(n−1) (4a_(n−1) ^2 −3)=0$
	${\begin{cases} a_{1} = \frac{\sqrt{3}}{2} \\ a_{n + 1} = 4 a_{n}^{3} - 3 a_{n}; \forall n ⩾ 1 \end{cases}; a_{n} = ?$ $a_{n + 1} = 4 a_{n}^{3} - 3 a_{n} = a_{n} (4 a_{n}^{2} - 3)$ $a_{1 + 1} = a_{1} (4 a_{1}^{2} - 3) = \frac{\sqrt{3}}{2} (4 {(\frac{\sqrt{3}}{2})}^{2} - 3)$ $a_{2} = \frac{\sqrt{3}}{2} (4 (\frac{3}{4}) - 3) = 0$ $a_{3} = a_{2} (4 a_{2}^{2} - 3) = 0$ $. . . . .$ $. . .$ $a_{n - 1} = 0$ $a_{n} = a_{n - 1} (4 a_{n - 1}^{2} - 3) = 0$
	Commented by mr W last updated on 20/Oct/21

	$v e r y s m a r t!$
	Commented by cortano last updated on 20/Oct/21

	$v e r y . . . v e r y n i c e$
	Commented by Rasheed.Sindhi last updated on 20/Oct/21

	$T h a n X m r W S i r!$ $A l s o t h a n k y o u m r c o r t a n o .$
	Commented by Tawa11 last updated on 21/Oct/21

	$Great sir$
	Commented by Rasheed.Sindhi last updated on 22/Oct/21

	$T han k s miss!$
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