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The lowest integer which is greater than $${\left( {1 + {1 \over {{{10}^{100}}}}} \right)^{{{10}^{100}}}}$$ is ______________.

JEE · Math · previous-year question

  1. A.3correct
  2. B.4
  3. C.2
  4. D.1

Answer

A. 3

Explanation

Let $$P = {\left( {1 + {1 \over {{{10}^{100}}}}} \right)^{{{10}^{100}}}}$$ Let $$x = {10^{100}}$$ $$ \Rightarrow P = {\left( {1 + {1 \over x}} \right)^x}$$ $$ \Rightarrow P = 1 + (x)\left( {{1 \over x}} \right) + {{(x)(x - 1)} \over {\left| \!{\underline {\, 2 \,}} \right. }}.{1 \over {{x^2}}} + {{(x)(x - 1)(x - 2)} \over {\left| \!{\underline {\, 3 \,}} \right. }}.{1 \over {{x^3}}} + ....$$ (upto 10100 + 1 terms) $$ \Rightarrow P = 1 + 1 + \left( {{1 \over {\left| \!{\underline {\, 2 \,}} \right. }} - {1 \over {\left| \!{\underline {\, 2 \,}} \right. {x^2}}}} \right) + \left( {{1 \over {\left| \!{\underline {\, 3 \,}} \right. }} - ...} \right) + ...$$ so on $$ \Rightarrow P = 2 + \left( {Positive\,value\,less\,than\,{1 \over {\left| \!{\underline {\, 2 \,}} \right. }} + {1 \over {\left| \!{\underline {\, 3 \,}} \right. }} + {1 \over {\left| \!{\underline {\, 4 \,}} \right. }} + ...} \right)$$ Also $$e = 1 + {1 \over {\left| \!{\underline {\, 1 \,}} \right. }} + {1 \over {\left| \!{\underline {\, 2 \,}} \right. }} + {1 \over {\left| \!{\underline {\, 3 \,}} \right. }} + {1 \over {\left| \!{\underline {\, 4 \,}} \right. }} + ...$$ $$ \Rightarrow {1 \over {\left| \!{\underline {\, 2 \,}} \right. }} + {1 \over {\left| \!{\underline {\, 3 \,}} \right. }} + {1 \over {\left| \!{\underline {\, 4 \,}} \right. }} + ... = e - 2$$ $$\Rightarrow$$ P = 2 + (positive value less than e $$-$$ 2) $$\Rightarrow$$ P $$\in$$ (2, 3) $$\Rightarrow$$ least integer value of P is 3

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