What is the greatest value of the positive integer n satisfying the condition $1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + . . . . . . + \frac{1}{2^{n - 1}} < 2 - \frac{1}{1000}$ ?

8

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Solution

The correct option is C $10$
Given : $1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + . . . . . . + \frac{1}{2^{n - 1}} < 2 - \frac{1}{1000}$
Left side forms a sum of a finite geometric series with first term $1$ and common ratio $\frac{1}{2}$ with $n$ terms.
Sum $= \frac{a (1 - r^{n})}{(1 - r)} = \frac{1 (1 - (0.5)^{n})}{0.5} = 2 - 2^{1 - n}$
So, $2 - 2^{1 - n} < 2 - \frac{1}{1000}$
We have,
$2^{n - 1} < 1000$ we get the max value of $n = 10$ .
Hence, C is correct.

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