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Question Number 80645 by mr W last updated on 05/Feb/20
if a_1 =2 a_(n+1) =a_n ^2 −1 find a_n =?
$${if} \\ $$$${a}_{\mathrm{1}} =\mathrm{2} \\ $$$${a}_{{n}+\mathrm{1}} ={a}_{{n}} ^{\mathrm{2}} −\mathrm{1} \\ $$$${find}\:{a}_{{n}} =? \\ $$
Commented by jagoll last updated on 05/Feb/20
a_2 = 3 , a_3 = 8 , a_4 = 63 a_5 = 63^2 −1
$${a}_{\mathrm{2}} =\:\mathrm{3}\:,\:{a}_{\mathrm{3}} \:=\:\mathrm{8}\:,\:{a}_{\mathrm{4}} \:=\:\mathrm{63} \\ $$$${a}_{\mathrm{5}} \:=\:\mathrm{63}^{\mathrm{2}} −\mathrm{1} \\ $$
Commented by mr W last updated on 05/Feb/20
it is to find a formula for a_n in terms of n.
$${it}\:{is}\:{to}\:{find}\:{a}\:{formula}\:{for}\:{a}_{{n}} \:{in} \\ $$$${terms}\:{of}\:{n}. \\ $$
Commented by jagoll last updated on 05/Feb/20
i have not yet got an idea to finish
$${i}\:{have}\:{not}\:{yet}\:{got}\:{an}\:{idea}\:{to} \\ $$$${finish} \\ $$
Commented by jagoll last updated on 05/Feb/20
how to solve it sir?
$${how}\:{to}\:{solve}\:{it}\:{sir}? \\ $$
Commented by mind is power last updated on 05/Feb/20
i will try
$${i}\:{will}\:{try} \\ $$$$ \\ $$
Commented by mr W last updated on 06/Feb/20
this question is too hard, i have no idea how to solve. but it were much easier, if the question were a_(n+1) =a_n ^2 −2.
$${this}\:{question}\:{is}\:{too}\:{hard},\:{i}\:{have}\:{no}\:{idea} \\ $$$${how}\:{to}\:{solve}. \\ $$$${but}\:{it}\:{were}\:{much}\:{easier},\:{if}\:{the}\:{question} \\ $$$${were}\:{a}_{{n}+\mathrm{1}} ={a}_{{n}} ^{\mathrm{2}} −\mathrm{2}. \\ $$
Commented by jagoll last updated on 06/Feb/20
what the answer sih if? a_(n+1) = a_n ^2 −2
$${what}\:{the}\:{answer}\:{sih}\:{if}? \\ $$$${a}_{{n}+\mathrm{1}} \:=\:{a}_{{n}} ^{\mathrm{2}} −\mathrm{2} \\ $$
Commented by mr W last updated on 06/Feb/20
say a_1 =3, a_(n+1) =a_n ^2 −2 a_n =p^2^n +(1/p^(2n) ) a_n ^2 =(p^2^n +(1/p^(2n) ))^2 =p^2^(n+1) +2+(1/p^2^(n+1) ) a_n ^2 −2=p^2^(n+1) +(1/p^2^(n+1) )=a_(n+1) a_1 =p^2 +(1/p^2 )=3 (p−(1/p))^2 =1 p−(1/p)=±1 p^2 ±p−1=0 p=((±1±(√5))/2) ⇒a_n =(((1+(√5))/2))^2^n +(((1−(√5))/2))^2^n
$${say}\:{a}_{\mathrm{1}} =\mathrm{3},\:{a}_{{n}+\mathrm{1}} ={a}_{{n}} ^{\mathrm{2}} −\mathrm{2} \\ $$$$ \\ $$$${a}_{{n}} ={p}^{\mathrm{2}^{{n}} } +\frac{\mathrm{1}}{{p}^{\mathrm{2}{n}} } \\ $$$${a}_{{n}} ^{\mathrm{2}} =\left({p}^{\mathrm{2}^{{n}} } +\frac{\mathrm{1}}{{p}^{\mathrm{2}{n}} }\right)^{\mathrm{2}} ={p}^{\mathrm{2}^{{n}+\mathrm{1}} } +\mathrm{2}+\frac{\mathrm{1}}{{p}^{\mathrm{2}^{{n}+\mathrm{1}} } } \\ $$$${a}_{{n}} ^{\mathrm{2}} −\mathrm{2}={p}^{\mathrm{2}^{{n}+\mathrm{1}} } +\frac{\mathrm{1}}{{p}^{\mathrm{2}^{{n}+\mathrm{1}} } }={a}_{{n}+\mathrm{1}} \\ $$$${a}_{\mathrm{1}} ={p}^{\mathrm{2}} +\frac{\mathrm{1}}{{p}^{\mathrm{2}} }=\mathrm{3} \\ $$$$\left({p}−\frac{\mathrm{1}}{{p}}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$${p}−\frac{\mathrm{1}}{{p}}=\pm\mathrm{1} \\ $$$${p}^{\mathrm{2}} \pm{p}−\mathrm{1}=\mathrm{0} \\ $$$${p}=\frac{\pm\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\Rightarrow{a}_{{n}} =\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{2}^{{n}} } +\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{2}^{{n}} } \\ $$
Commented by jagoll last updated on 08/Feb/20
how to get a_n = p^(2n) +(1/p^(2n) ) sir
$${how}\:{to}\:{get}\:{a}_{{n}} \:=\:{p}^{\mathrm{2}{n}} +\frac{\mathrm{1}}{{p}^{\mathrm{2}{n}} }\:{sir} \\ $$
Commented by mr W last updated on 08/Feb/20
not a_n =p^(2n) +(1/p^(2n) ), but a_n =p^2^n +(1/p^2^n ). intuitive we know the terms must be in the form a_1 =k+(1/k), since we have a_1 ^2 −2=k^2 +2+(1/k^2 )−2=k^2 +(1/k^2 )=a_2 . and a_2 ^2 −2=k^4 +2+(1/k^4 )−2=k^4 +(1/k^4 )=a_3 . the rest is clear, a_n =k^2^(n−1) +(1/k^2^(n−1) )=p^2^n +(1/p^2^n ).
$${not}\:{a}_{{n}} ={p}^{\mathrm{2}{n}} +\frac{\mathrm{1}}{{p}^{\mathrm{2}{n}} },\:{but}\:{a}_{{n}} ={p}^{\mathrm{2}^{{n}} } +\frac{\mathrm{1}}{{p}^{\mathrm{2}^{{n}} } }. \\ $$$${intuitive}\:{we}\:{know}\:{the}\:{terms}\:{must}\:{be} \\ $$$${in}\:{the}\:{form}\:{a}_{\mathrm{1}} ={k}+\frac{\mathrm{1}}{{k}},\:{since}\:{we}\:{have} \\ $$$${a}_{\mathrm{1}} ^{\mathrm{2}} −\mathrm{2}={k}^{\mathrm{2}} +\mathrm{2}+\frac{\mathrm{1}}{{k}^{\mathrm{2}} }−\mathrm{2}={k}^{\mathrm{2}} +\frac{\mathrm{1}}{{k}^{\mathrm{2}} }={a}_{\mathrm{2}} .\:{and} \\ $$$${a}_{\mathrm{2}} ^{\mathrm{2}} −\mathrm{2}={k}^{\mathrm{4}} +\mathrm{2}+\frac{\mathrm{1}}{{k}^{\mathrm{4}} }−\mathrm{2}={k}^{\mathrm{4}} +\frac{\mathrm{1}}{{k}^{\mathrm{4}} }={a}_{\mathrm{3}} .\:{the} \\ $$$${rest}\:{is}\:{clear},\:{a}_{{n}} ={k}^{\mathrm{2}^{{n}−\mathrm{1}} } +\frac{\mathrm{1}}{{k}^{\mathrm{2}^{{n}−\mathrm{1}} } }={p}^{\mathrm{2}^{{n}} } +\frac{\mathrm{1}}{{p}^{\mathrm{2}^{{n}} } }. \\ $$
Commented by mr W last updated on 08/Feb/20
that′s why i said it is only possible for a_(n+1) =a_n ^2 −2. you can not apply this for a_(n+1) =a_n ^2 −1.
$${that}'{s}\:{why}\:{i}\:{said}\:{it}\:{is}\:{only}\:{possible}\:{for} \\ $$$${a}_{{n}+\mathrm{1}} ={a}_{{n}} ^{\mathrm{2}} −\mathrm{2}.\:{you}\:{can}\:{not}\:{apply}\:{this}\:{for} \\ $$$${a}_{{n}+\mathrm{1}} ={a}_{{n}} ^{\mathrm{2}} −\mathrm{1}. \\ $$

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