If then the least value of n is

JEE MainMathematicsToughSequences and Series

If
$S_{n} = 1 + \frac{1}{2} + \frac{1}{4} + \ldots\frac{1}{2^{n - 1}}\& 2 - S_{n} < \frac{1}{100}$
then the least value of n is

Show answer and solution

S_{n} = 2 - \frac{1}{2^{n - 1}}

2 - S_{n} = \frac{2}{2^{n - 1}} < \frac{1}{100}

n > 7

n_{least} = 8

Source: Grass-G / M-16 Mathematics Paper 25 Nov First.docx

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