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-

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}

\dots

\frac{a_{2}}{a_{1}} = 1 + a_{1}

\Rightarrow \frac{a_{101}}{a_{1}} = (1 + a_{1}) (1 + a_{2}) \dots (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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