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If S denotes the sum to infinity and S_n, the sum of n terms of the series $1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots, such that S - S_{n} < \frac{1}{1000}$ then the least value of n is

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11

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detailed solution

Correct option is D

We have
$\begin{array}{l} S = \frac{1}{1 - \frac{1}{2}} = 2 \\ S_{n} = \frac{(1 - 1 / 2^{n})}{(1 - 1 / 2)} = 2 (1 - \frac{1}{2^{n}}) = 2 - \frac{1}{2^{n - 1}} \\ ∴ S = S_{n} < \frac{1}{1000} \Rightarrow \frac{1}{2^{n - 1}} < \frac{1}{1000} \\ \Rightarrow 2^{n - 1} > 1000 \\ \Rightarrow n - 1 \geq 10 \\ \Rightarrow n \geq 11 \end{array}$
Hence, the least value of n is 11.

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