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Question Number 31726 by rahul 19 last updated on 13/Mar/18

Let a_1 = (1/2) , a_(k+1) =a_k ^2 +a_k ∀ k≥ 1. then a_(101) is greater than a) 1 b) 2 c) 3 d) 4 .

$${Let}\:{a}_{\mathrm{1}} =\:\frac{\mathrm{1}}{\mathrm{2}}\:,\:{a}_{{k}+\mathrm{1}} ={a}_{{k}} ^{\mathrm{2}} +{a}_{{k}} \forall\:{k}\geqslant\:\mathrm{1}. \\ $$$${then}\:{a}_{\mathrm{101}} \:\:{is}\:{greater}\:{than} \\ $$$$\left.{a}\right)\:\mathrm{1}\: \\ $$$$\left.{b}\right)\:\mathrm{2} \\ $$$$\left.{c}\right)\:\mathrm{3} \\ $$$$\left.{d}\right)\:\mathrm{4}\:. \\ $$

Commented by mrW2 last updated on 13/Mar/18

a_(101) >2×10^(17) >>4!

$${a}_{\mathrm{101}} >\mathrm{2}×\mathrm{10}^{\mathrm{17}} >>\mathrm{4}! \\ $$

Commented by rahul 19 last updated on 13/Mar/18

sir how u get a_(101) >2×10^(17) ?

$${sir}\:{how}\:{u}\:{get}\:{a}_{\mathrm{101}} >\mathrm{2}×\mathrm{10}^{\mathrm{17}} ? \\ $$

Commented by MJS last updated on 13/Mar/18

a_1 =(1/2); a_2 =(3/4); a_3 =((21)/(16)); a_4 =((777)/(256)); a_5 =((802641)/(65536))≈12>4; a_(k+1) >a_k ⇒ a_k >4∀k≧5

$${a}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}};\:{a}_{\mathrm{2}} =\frac{\mathrm{3}}{\mathrm{4}};\:{a}_{\mathrm{3}} =\frac{\mathrm{21}}{\mathrm{16}};\:{a}_{\mathrm{4}} =\frac{\mathrm{777}}{\mathrm{256}}; \\ $$$$\mathrm{a}_{\mathrm{5}} =\frac{\mathrm{802641}}{\mathrm{65536}}\approx\mathrm{12}>\mathrm{4}; \\ $$$${a}_{{k}+\mathrm{1}} >{a}_{{k}} \:\Rightarrow\:{a}_{{k}} >\mathrm{4}\forall{k}\geqq\mathrm{5} \\ $$

Answered by mrW2 last updated on 13/Mar/18

a_(k+1) =a_k ^2 +a_k >a_k >0 (a_(k+1) /a_k )=1+a_k (a_(101) /a_(100) )=1+a_(100) (a_(100) /a_(99) )=1+a_(99) ... (a_2 /a_1 )=1+a_1 ⇒(a_(101) /a_1 )=(1+a_1 )(1+a_2 )...(1+a_(100) )>(1+a_1 )^(100) ⇒a_(101) >a_1 (1+a_1 )^(100) =(1/2)×((3/2))^(100) >2×10^(17)

$${a}_{{k}+\mathrm{1}} ={a}_{{k}} ^{\mathrm{2}} +{a}_{{k}} >{a}_{{k}} >\mathrm{0} \\ $$$$\frac{{a}_{{k}+\mathrm{1}} }{{a}_{{k}} }=\mathrm{1}+{a}_{{k}} \\ $$$$\frac{{a}_{\mathrm{101}} }{{a}_{\mathrm{100}} }=\mathrm{1}+{a}_{\mathrm{100}} \\ $$$$\frac{{a}_{\mathrm{100}} }{{a}_{\mathrm{99}} }=\mathrm{1}+{a}_{\mathrm{99}} \\ $$$$... \\ $$$$\frac{{a}_{\mathrm{2}} }{{a}_{\mathrm{1}} }=\mathrm{1}+{a}_{\mathrm{1}} \\ $$$$\Rightarrow\frac{{a}_{\mathrm{101}} }{{a}_{\mathrm{1}} }=\left(\mathrm{1}+{a}_{\mathrm{1}} \right)\left(\mathrm{1}+{a}_{\mathrm{2}} \right)...\left(\mathrm{1}+{a}_{\mathrm{100}} \right)>\left(\mathrm{1}+{a}_{\mathrm{1}} \right)^{\mathrm{100}} \\ $$$$\Rightarrow{a}_{\mathrm{101}} >{a}_{\mathrm{1}} \left(\mathrm{1}+{a}_{\mathrm{1}} \right)^{\mathrm{100}} =\frac{\mathrm{1}}{\mathrm{2}}×\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{100}} >\mathrm{2}×\mathrm{10}^{\mathrm{17}} \\ $$

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