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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 =?

$$ \\ $$$$\:\begin{cases}{{a}_{\mathrm{1}} =\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\\{{a}_{{n}+\mathrm{1}} =\mathrm{4}{a}_{{n}} ^{\mathrm{3}} −\mathrm{3}{a}_{{n}} \:;\:\forall{n}\geqslant\mathrm{1}}\end{cases} \\ $$$$\:{a}_{{n}} =? \\ $$

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

$$ \\ $$$$\:\begin{cases}{{a}_{\mathrm{1}} =\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\\{{a}_{{n}+\mathrm{1}} =\mathrm{4}{a}_{{n}} ^{\mathrm{3}} −\mathrm{3}{a}_{{n}} \:;\:\forall{n}\geqslant\mathrm{1}}\end{cases};{a}_{{n}} =? \\ $$$${a}_{{n}+\mathrm{1}} =\mathrm{4}{a}_{{n}} ^{\mathrm{3}} −\mathrm{3}{a}_{{n}} ={a}_{{n}} \left(\mathrm{4}{a}_{{n}} ^{\mathrm{2}} −\mathrm{3}\right) \\ $$$${a}_{\mathrm{1}+\mathrm{1}} ={a}_{\mathrm{1}} \left(\mathrm{4}{a}_{\mathrm{1}} ^{\mathrm{2}} −\mathrm{3}\right)=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\left(\mathrm{4}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} −\mathrm{3}\right) \\ $$$${a}_{\mathrm{2}} =\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\left(\mathrm{4}\left(\frac{\mathrm{3}}{\mathrm{4}}\right)−\mathrm{3}\right)=\mathrm{0} \\ $$$${a}_{\mathrm{3}} ={a}_{\mathrm{2}} \left(\mathrm{4}{a}_{\mathrm{2}} ^{\mathrm{2}} −\mathrm{3}\right)=\mathrm{0} \\ $$$$..... \\ $$$$... \\ $$$${a}_{{n}−\mathrm{1}} =\mathrm{0} \\ $$$${a}_{{n}} ={a}_{{n}−\mathrm{1}} \left(\mathrm{4}{a}_{{n}−\mathrm{1}} ^{\mathrm{2}} −\mathrm{3}\right)=\mathrm{0} \\ $$

Commented by mr W last updated on 20/Oct/21

very smart!

$${very}\:{smart}! \\ $$

Commented by cortano last updated on 20/Oct/21

very ...very nice

$${very}\:...{very}\:{nice} \\ $$

Commented by Rasheed.Sindhi last updated on 20/Oct/21

ThanX mr W Sir! Also thank you mr cortano.

$$\mathcal{T}{han}\mathcal{X}\:{mr}\:{W}\:\mathcal{S}{ir}! \\ $$$${Also}\:{thank}\:{you}\:{mr}\:{cortano}. \\ $$

Commented by Tawa11 last updated on 21/Oct/21

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

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

Thanks miss!

$$\mathbb{T}\mathrm{han}\Bbbk\mathrm{s}\:\mathrm{miss}! \\ $$

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