The least positive integer n such that 1 - fr | vedclass

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(C) The given expression is $1 - \sum_{k=1}^{n-1} \frac{2}{3^k} < \frac{1}{100}$.This can be written as $1 - 2 \left( \frac{1}{3} + \frac{1}{3^2} + \d

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(C) The given expression is $1 - \sum_{k=1}^{n-1} \frac{2}{3^k} This can be written as $1 - 2 \left( \frac{1}{3} + \frac{1}{3^2} + \dots + \frac{1}{3^{n-1}} \right) The sum inside the bracket is a geometric progression with $a = \frac{1}{3}$,$r = \frac{1}{3}$,and $n-1$ terms.The sum $S_{n-1} = \frac{a(1-r^{n-1})}{1-r} = \frac{\frac{1}{3}(1 - (\frac{1}{3})^{n-1})}{1 - \frac{1}{3}} = \frac{\frac{1}{3}(1 - \frac{1}{3^{n-1}})}{\frac{2}{3}} = \frac{1}{2} (1 - \frac{1}{3^{n-1}}) = \frac{1}{2} - \frac{1}{2 \cdot 3^{n-1}}$.Substituting this back: $1 - 2 \left( \frac{1}{2} - \frac{1}{2 \cdot 3^{n-1}} \right) $1 - 1 + \frac{1}{3^{n-1}} $\frac{1}{3^{n-1}}  100$.For $n=5$,$3^{5-1} = 3^4 = 81$ (which is not $> 100$).For $n=6$,$3^{6-1} = 3^5 = 243$ (which is $> 100$).Thus,the least positive integer $n$ is $6$.

The least positive integer $n$ such that $1 - \frac{2}{3} - \frac{2}{3^2} - \dots - \frac{2}{3^{n-1}} < \frac{1}{100}$ is

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