Question Number 31726 by rahul 19 last updated on 13/Mar/18
$${Let}\:{a}_{\mathrm{1}} =\:\frac{\mathrm{1}}{\mathrm{2}}\:,\:{a}_{{k}+\mathrm{1}} ={a}_{{k}} ^{\mathrm{2}} +{a}_{{k}} \forall\:{k}\geqslant\:\mathrm{1}. \\ $$$${then}\:{a}_{\mathrm{101}} \:\:{is}\:{greater}\:{than} \\ $$$$\left.{a}\right)\:\mathrm{1}\: \\ $$$$\left.{b}\right)\:\mathrm{2} \\ $$$$\left.{c}\right)\:\mathrm{3} \\ $$$$\left.{d}\right)\:\mathrm{4}\:. \\ $$
Commented by mrW2 last updated on 13/Mar/18
$${a}_{\mathrm{101}} >\mathrm{2}×\mathrm{10}^{\mathrm{17}} >>\mathrm{4}! \\ $$
Commented by rahul 19 last updated on 13/Mar/18
$${sir}\:{how}\:{u}\:{get}\:{a}_{\mathrm{101}} >\mathrm{2}×\mathrm{10}^{\mathrm{17}} ? \\ $$
Commented by MJS last updated on 13/Mar/18
$${a}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}};\:{a}_{\mathrm{2}} =\frac{\mathrm{3}}{\mathrm{4}};\:{a}_{\mathrm{3}} =\frac{\mathrm{21}}{\mathrm{16}};\:{a}_{\mathrm{4}} =\frac{\mathrm{777}}{\mathrm{256}}; \\ $$$$\mathrm{a}_{\mathrm{5}} =\frac{\mathrm{802641}}{\mathrm{65536}}\approx\mathrm{12}>\mathrm{4}; \\ $$$${a}_{{k}+\mathrm{1}} >{a}_{{k}} \:\Rightarrow\:{a}_{{k}} >\mathrm{4}\forall{k}\geqq\mathrm{5} \\ $$
Answered by mrW2 last updated on 13/Mar/18
$${a}_{{k}+\mathrm{1}} ={a}_{{k}} ^{\mathrm{2}} +{a}_{{k}} >{a}_{{k}} >\mathrm{0} \\ $$$$\frac{{a}_{{k}+\mathrm{1}} }{{a}_{{k}} }=\mathrm{1}+{a}_{{k}} \\ $$$$\frac{{a}_{\mathrm{101}} }{{a}_{\mathrm{100}} }=\mathrm{1}+{a}_{\mathrm{100}} \\ $$$$\frac{{a}_{\mathrm{100}} }{{a}_{\mathrm{99}} }=\mathrm{1}+{a}_{\mathrm{99}} \\ $$$$… \\ $$$$\frac{{a}_{\mathrm{2}} }{{a}_{\mathrm{1}} }=\mathrm{1}+{a}_{\mathrm{1}} \\ $$$$\Rightarrow\frac{{a}_{\mathrm{101}} }{{a}_{\mathrm{1}} }=\left(\mathrm{1}+{a}_{\mathrm{1}} \right)\left(\mathrm{1}+{a}_{\mathrm{2}} \right)…\left(\mathrm{1}+{a}_{\mathrm{100}} \right)>\left(\mathrm{1}+{a}_{\mathrm{1}} \right)^{\mathrm{100}} \\ $$$$\Rightarrow{a}_{\mathrm{101}} >{a}_{\mathrm{1}} \left(\mathrm{1}+{a}_{\mathrm{1}} \right)^{\mathrm{100}} =\frac{\mathrm{1}}{\mathrm{2}}×\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{100}} >\mathrm{2}×\mathrm{10}^{\mathrm{17}} \\ $$