Menu Close

Given-a-0-1-a-n-1-1-2-3a-n-5a-n-2-4-n-0-n-I-find-a-n-

Tinku Tara 0 Comments
Pin




Question Number 184160 by cortano1 last updated on 03/Jan/23
Given { ((a_0 =1)),((a_(n+1) =(1/2)(3a_n +(√(5a_n ^2 −4)) ))) :} ∀n≥0 , n∈I find a_n .
$$\:\:{Given}\:\begin{cases}{{a}_{\mathrm{0}} =\mathrm{1}}\\{{a}_{{n}+\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{3}{a}_{{n}} +\sqrt{\mathrm{5}{a}_{{n}} ^{\mathrm{2}} −\mathrm{4}}\:\right)}\end{cases} \\ $$$$\:\forall{n}\geqslant\mathrm{0}\:,\:{n}\in{I}\: \\ $$$$\:\:{find}\:{a}_{{n}} . \\ $$
Commented by mr W last updated on 04/Jan/23
solution see Q184280
$${solution}\:{see}\:{Q}\mathrm{184280} \\ $$
Answered by a.lgnaoui last updated on 03/Jan/23
a_(n+1) =((3a_n )/2)+(1/2)(√(5a_n ^2 −4)) (√(5a_n ^2 −4))=2a_(n+1) −3a_n 5a_n ^2 −4=4a_(n+1) ^2 +9a_n ^2 −12a_n a_(n+1) 4a_(n+1) ^2 −12a_n a_(n+1) +4a_n ^2 +4=0 a_(n+1) ^2 −3a_n a_(n+1) +a_n ^2 +1=0 (a_(n+1) −((3a_n )/2))^2 −(((5a_n ^2 −4)/4)) (a_(n+1) −((3a_n )/2)−((√(5a_n ^2 −4))/2))(a_(n+1) −((3a_n )/2)+((√(5a_n ^2 −4))/2)) { ((a_(n+1) =(1/2)(3a_n +(√(5a_n ^2 −4)) ))),((a_(n+1) =(1/2)(3a_n −(√(5a_n ^2 −4)) ))) :} a_(n+1) =(3/2)a_n ⇒ a_n =(3/2)a_(n−1)
$$ \\ $$$${a}_{{n}+\mathrm{1}} =\frac{\mathrm{3}{a}_{{n}} }{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{5}{a}_{{n}} ^{\mathrm{2}} −\mathrm{4}}\: \\ $$$$\sqrt{\mathrm{5}{a}_{{n}} ^{\mathrm{2}} −\mathrm{4}}=\mathrm{2}{a}_{{n}+\mathrm{1}} −\mathrm{3}{a}_{{n}} \\ $$$$\mathrm{5}{a}_{{n}} ^{\mathrm{2}} −\mathrm{4}=\mathrm{4}{a}_{{n}+\mathrm{1}} ^{\mathrm{2}} +\mathrm{9}{a}_{{n}} ^{\mathrm{2}} −\mathrm{12}{a}_{{n}} {a}_{{n}+\mathrm{1}} \\ $$$$\mathrm{4}{a}_{{n}+\mathrm{1}} ^{\mathrm{2}} −\mathrm{12}{a}_{{n}} {a}_{{n}+\mathrm{1}} +\mathrm{4}{a}_{{n}} ^{\mathrm{2}} +\mathrm{4}=\mathrm{0} \\ $$$${a}_{{n}+\mathrm{1}} ^{\mathrm{2}} −\mathrm{3}{a}_{{n}} {a}_{{n}+\mathrm{1}} +{a}_{{n}} ^{\mathrm{2}} +\mathrm{1}=\mathrm{0} \\ $$$$\left({a}_{{n}+\mathrm{1}} −\frac{\mathrm{3}{a}_{{n}} }{\mathrm{2}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{5}{a}_{{n}} ^{\mathrm{2}} −\mathrm{4}}{\mathrm{4}}\right) \\ $$$$ \\ $$$$\left({a}_{{n}+\mathrm{1}} −\frac{\mathrm{3}{a}_{{n}} }{\mathrm{2}}−\frac{\sqrt{\mathrm{5}{a}_{{n}} ^{\mathrm{2}} −\mathrm{4}}}{\mathrm{2}}\right)\left({a}_{{n}+\mathrm{1}} −\frac{\mathrm{3}{a}_{{n}} }{\mathrm{2}}+\frac{\sqrt{\mathrm{5}{a}_{{n}} ^{\mathrm{2}} −\mathrm{4}}}{\mathrm{2}}\right) \\ $$$$\begin{cases}{{a}_{{n}+\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{3}{a}_{{n}} +\sqrt{\mathrm{5}{a}_{{n}} ^{\mathrm{2}} −\mathrm{4}}\:\right)}\\{{a}_{{n}+\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{3}{a}_{{n}} −\sqrt{\mathrm{5}{a}_{{n}} ^{\mathrm{2}} −\mathrm{4}}\:\right)}\end{cases} \\ $$$$\:\:\:\:\:\:\:\:{a}_{{n}+\mathrm{1}} =\frac{\mathrm{3}}{\mathrm{2}}{a}_{{n}} \:\:\:\Rightarrow \\ $$$$\:\:\:\:\:\:\:\:\boldsymbol{{a}}_{\boldsymbol{{n}}} =\frac{\mathrm{3}}{\mathrm{2}}\boldsymbol{{a}}_{\boldsymbol{{n}}−\mathrm{1}} \:\:\: \\ $$$$ \\ $$
Commented by SEKRET last updated on 03/Jan/23
thanks sir. and you
$$\boldsymbol{\mathrm{thanks}}\:\:\boldsymbol{\mathrm{sir}}.\:\:\boldsymbol{\mathrm{and}}\:\:\boldsymbol{\mathrm{you}} \\ $$
Commented by SEKRET last updated on 03/Jan/23
(√(5a_n ^2 −4)) = 2a_(n+1) −3a_n 5a_n ^2 −4 = 4a_(n+1) ^2 +9a_n ^2 −12a_n a_(n+1) 4a_(n+1) ^2 (+)4a_n ^2 −12a_n a_(n+1) +4=0
$$\:\:\:\sqrt{\mathrm{5}\boldsymbol{{a}}_{\boldsymbol{{n}}} ^{\mathrm{2}} −\mathrm{4}}\:\:\:=\:\mathrm{2}\boldsymbol{{a}}_{\boldsymbol{{n}}+\mathrm{1}} −\mathrm{3}\boldsymbol{{a}}_{\boldsymbol{{n}}} \\ $$$$\:\:\:\:\:\:\mathrm{5}\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{n}}} ^{\mathrm{2}} −\mathrm{4}\:=\:\mathrm{4}\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{n}}+\mathrm{1}} ^{\mathrm{2}} +\mathrm{9}\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{n}}} ^{\mathrm{2}} −\mathrm{12}\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{n}}} \boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{n}}+\mathrm{1}} \\ $$$$\:\:\:\:\:\mathrm{4}\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{n}}+\mathrm{1}} ^{\mathrm{2}} \left(+\right)\mathrm{4}\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{n}}} ^{\mathrm{2}} −\mathrm{12}\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{n}}} \boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{n}}+\mathrm{1}} +\mathrm{4}=\mathrm{0} \\ $$
Commented by a.lgnaoui last updated on 03/Jan/23
yes look at rectification thanks ; happy new year
$${yes}\:\:{look}\:{at}\:{rectification} \\ $$$${thanks}\:\:;\:{happy}\:{new}\:{year} \\ $$
Commented by SEKRET last updated on 03/Jan/23
your answer is wrong
$$\:\:\boldsymbol{\mathrm{your}}\:\:\boldsymbol{\mathrm{answer}}\:\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{wrong}} \\ $$
Answered by SEKRET last updated on 03/Jan/23
a_n =a_(n−1) +Σ_(k=0) ^(n−1) a_k
$$\:\:\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{n}}} =\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{n}}−\mathrm{1}} +\underset{\boldsymbol{\mathrm{k}}=\mathrm{0}} {\overset{\boldsymbol{\mathrm{n}}−\mathrm{1}} {\sum}}\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{k}}} \\ $$
Answered by SEKRET last updated on 03/Jan/23
3a_n =a_(n−1) +a_(n+1)
$$\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\mathrm{3}\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{n}}} \:=\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{n}}−\mathrm{1}} +\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{n}}+\mathrm{1}} \\ $$
Answered by aleks041103 last updated on 03/Jan/23
a_(n+1) =((−(−3a_n )+(√((−3a_n )^2 −4a_n ^2 −4)))/2) a_(n+1) =((−(−3a_n )+(√((−3a_n )^2 −4.1.(a_n ^2 +1))))/(2.1)) ⇒a_(n+1) is a solution to x^2 −3a_n x+a_n ^2 +1=0 ⇒a_(n+1) ^2 +a_n ^2 −3a_(n+1) a_n +1=0 i′ll continue later
$${a}_{{n}+\mathrm{1}} =\frac{−\left(−\mathrm{3}{a}_{{n}} \right)+\sqrt{\left(−\mathrm{3}{a}_{{n}} \right)^{\mathrm{2}} −\mathrm{4}{a}_{{n}} ^{\mathrm{2}} −\mathrm{4}}}{\mathrm{2}} \\ $$$${a}_{{n}+\mathrm{1}} =\frac{−\left(−\mathrm{3}{a}_{{n}} \right)+\sqrt{\left(−\mathrm{3}{a}_{{n}} \right)^{\mathrm{2}} −\mathrm{4}.\mathrm{1}.\left({a}_{{n}} ^{\mathrm{2}} +\mathrm{1}\right)}}{\mathrm{2}.\mathrm{1}} \\ $$$$\Rightarrow{a}_{{n}+\mathrm{1}} \:{is}\:{a}\:{solution}\:{to} \\ $$$${x}^{\mathrm{2}} −\mathrm{3}{a}_{{n}} {x}+{a}_{{n}} ^{\mathrm{2}} +\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{a}_{{n}+\mathrm{1}} ^{\mathrm{2}} +{a}_{{n}} ^{\mathrm{2}} −\mathrm{3}{a}_{{n}+\mathrm{1}} {a}_{{n}} +\mathrm{1}=\mathrm{0} \\ $$$${i}'{ll}\:{continue}\:{later} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *