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

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Question

(C) The given expression is $1 - \left( \frac{2}{3} + \frac{2}{3^2} + \dots + \frac{2}{3^{n-1}} \right) < \frac{1}{100}$.The term inside the parenthes

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(C) The given expression is $1 - \left( \frac{2}{3} + \frac{2}{3^2} + \dots + \frac{2}{3^{n-1}} \right) The term inside the parenthesis is a geometric progression with first term $a = \frac{2}{3}$ and common ratio $r = \frac{1}{3}$ having $(n-1)$ terms.The sum of the geometric progression is $S_{n-1} = a \frac{1 - r^{n-1}}{1 - r} = \frac{2}{3} \frac{1 - (1/3)^{n-1}}{1 - 1/3} = \frac{2}{3} \frac{1 - (1/3)^{n-1}}{2/3} = 1 - \frac{1}{3^{n-1}}$.Substituting this back into the inequality: $1 - (1 - \frac{1}{3^{n-1}}) $\Rightarrow \frac{1}{3^{n-1}} $\Rightarrow 3^{n-1} > 100$.We know that $3^4 = 81$ and $3^5 = 243$.Therefore,$n-1$ must be at least $5$,which means $n-1 \ge 5$,so $n \ge 6$.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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