MCQ
The least positive integer $n$ such that $1 - \frac{2}{3} - \frac{2}{{{3^2}}} - .... - \frac{2}{{{3^{n - 1}}}} < \frac{1}{{100}},$ is
- A$4$
- ✓$5$
- C$6$
- D$7$
✓
Answer
Correct option: B.
$5$
b
$1 - \frac{2}{3} - \frac{2}{{{3^2}}}...\frac{2}{{{3^{n - 1}}}} < \frac{1}{{100}}$
$1 - \frac{2}{3} - \frac{2}{{{3^2}}}...\frac{2}{{{3^{n - 1}}}} < \frac{1}{{100}}$
$ \Rightarrow 1 - \frac{2}{3}\left[ {\frac{1}{3} + \frac{1}{{{3^2}}} + \frac{1}{{{3^3}}} + ....\frac{1}{{{3^{n - 1}}}}} \right] < \frac{1}{{100}}$
$ \Rightarrow \frac{{1 - 2\left[ {\frac{1}{3}\left( {\frac{1}{{{3^n}}} - 1} \right)} \right]}}{{\frac{1}{3} - 1}} < \frac{1}{{100}}$
$ \Rightarrow 1 - 2\left[ {\frac{{{3^n} - 1}}{{{{2.3}^n}}}} \right] < \frac{1}{{100}}$
$ \Rightarrow 1 - \left[ {\frac{{{3^n} - 1}}{{{3^n}}}} \right] < \frac{1}{{100}}$
$ \Rightarrow 1 - 1 + \frac{1}{{{3^n}}} < \frac{1}{{100}}$
$ \Rightarrow 100 < {3^n}$
Thus, least value of $n$ is $5$
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