a-1-3-2-a-n-1-4a-n-3-3a-n-n-1-a-n-

Question Number 157156 by Rasheed.Sindhi last updated on 20/Oct/21

{\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

{\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

\dots . .

\dots

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 \dots 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!