A) 8
B) 9
C) 10
D) 11
Correct Answer: C
Solution :
[c] \[1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+..........+\frac{1}{{{2}^{n-1}}}<2-\frac{1}{1000}\] LHS of given inequality is in G.P. \[\therefore \frac{1-\frac{1}{{{2}^{n}}}}{1-\frac{1}{2}}<2-\frac{1}{1000}\] \[\Rightarrow 2-\frac{1}{{{2}^{n-1}}}<2-\frac{1}{100}\] \[\Rightarrow {{2}^{n-1}}<1000\] Now, \[{{(2)}^{9}}=512\,\,\And \,\,{{(2)}^{10}}=1024\] \[\therefore n-1=9\Rightarrow n=10\].
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