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 .


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

$L e t a_{1} = \frac{1}{2}, a_{k + 1} = a_{k}^{2} + a_{k} \forall k ⩾ 1 .$ $t h e n a_{101} i s g r e a t e r t h a n$ $a) 1$ $b) 2$ $c) 3$ $d) 4 .$
Commented by mrW2 last updated on 13/Mar/18

$a_{101} > 2 \times 10^{17} >> 4!$
Commented by rahul 19 last updated on 13/Mar/18

$s i r h o w u g e t a_{101} > 2 \times 10^{17} ?$
Commented by MJS last updated on 13/Mar/18

$a_{1} = \frac{1}{2}; a_{2} = \frac{3}{4}; a_{3} = \frac{21}{16}; a_{4} = \frac{777}{256};$ $a_{5} = \frac{802641}{65536} \approx 12 > 4;$ $a_{k + 1} > a_{k} \Rightarrow a_{k} > 4 \forall k ≧ 5$
	Answered by mrW2 last updated on 13/Mar/18

	$a_{k + 1} = a_{k}^{2} + a_{k} > a_{k} > 0$ $\frac{a_{k + 1}}{a_{k}} = 1 + a_{k}$ $\frac{a_{101}}{a_{100}} = 1 + a_{100}$ $\frac{a_{100}}{a_{99}} = 1 + a_{99}$ $. . .$ $\frac{a_{2}}{a_{1}} = 1 + a_{1}$ $\Rightarrow \frac{a_{101}}{a_{1}} = (1 + a_{1}) (1 + a_{2}) . . . (1 + a_{100}) > {(1 + a_{1})}^{100}$ $\Rightarrow a_{101} > a_{1} {(1 + a_{1})}^{100} = \frac{1}{2} \times {(\frac{3}{2})}^{100} > 2 \times 10^{17}$
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