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