The lowest integer which is greater
than ${\left( {1 + {1 \over {{{10}^{100}}}}} \right)^{{{10}^{100}}}}$ is ______________.
Solution
Let $P = {\left( {1 + {1 \over {{{10}^{100}}}}} \right)^{{{10}^{100}}}}$<br><br>Let $x = {10^{100}}$<br><br>$\Rightarrow P = {\left( {1 + {1 \over x}} \right)^x}$<br><br>$$ \Rightarrow P = 1 + (x)\left( {{1 \over x}} \right) + {{(x)(x - 1)} \over {\left| \!{\underline {\,
2 \,}} \right. }}.{1 \over {{x^2}}} + {{(x)(x - 1)(x - 2)} \over {\left| \!{\underline {\,
3 \,}} \right. }}.{1 \over {{x^3}}} + ....$$<br><br>(upto 10<sup>100</sup> + 1 terms)<br><br>$$ \Rightarrow P = 1 + 1 + \left( {{1 \over {\left| \!{\underline {\,
2 \,}} \right. }} - {1 \over {\left| \!{\underline {\,
2 \,}} \right. {x^2}}}} \right) + \left( {{1 \over {\left| \!{\underline {\,
3 \,}} \right. }} - ...} \right) + ...$$ so on<br><br>$$ \Rightarrow P = 2 + \left( {Positive\,value\,less\,than\,{1 \over {\left| \!{\underline {\,
2 \,}} \right. }} + {1 \over {\left| \!{\underline {\,
3 \,}} \right. }} + {1 \over {\left| \!{\underline {\,
4 \,}} \right. }} + ...} \right)$$<br><br>Also $$e = 1 + {1 \over {\left| \!{\underline {\,
1 \,}} \right. }} + {1 \over {\left| \!{\underline {\,
2 \,}} \right. }} + {1 \over {\left| \!{\underline {\,
3 \,}} \right. }} + {1 \over {\left| \!{\underline {\,
4 \,}} \right. }} + ...$$<br><br>$$ \Rightarrow {1 \over {\left| \!{\underline {\,
2 \,}} \right. }} + {1 \over {\left| \!{\underline {\,
3 \,}} \right. }} + {1 \over {\left| \!{\underline {\,
4 \,}} \right. }} + ... = e - 2$$<br><br>$\Rightarrow$ P = 2 + (positive value less than e $-$ 2)<br><br>$\Rightarrow$ P $\in$ (2, 3)<br><br>$\Rightarrow$ least integer value of P is 3
About this question
Subject: Mathematics · Chapter: Binomial Theorem · Topic: Applications of Binomial Theorem
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