Let \({\rm{f}}\left( {\rm{n}} \right) = \left[ {\frac{1}{4} + \frac{{\rm{n}}}{{1000}}} \right]\), where [x] denote the integral part of x. Then the value of \(\mathop \sum \limits_{{\rm{n}} = 1}^{1000} {\rm{f}}\left( {\rm{n}} \right)\) is

Let \({\rm{f}}\left( {\rm{n}} \right) = \left[ {\frac{1}{4} + \frac{{\

1.	Let \({\rm{f}}\left( {\rm{n}} \right) = \left[ {\frac{1}{4} + \frac{{\rm{n}}}{{1000}}} \right]\), where [x] denote the integral part of x. Then the value of \(\mathop \sum \limits_{{\rm{n}} = 1}^{1000} {\rm{f}}\left( {\rm{n}} \right)\) is
A.	251
B.	250
C.	1
D.	0
Answer» B. 250

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Let \({\rm{f}}\left( {\rm{n}} \right) = \left[ {\frac{1}{4} + \frac{{\rm{n}}}{{1000}}} \right]\), where [x] denote the integral part of x. Then the value of \(\mathop \sum \limits_{{\rm{n}} = 1}^{1000} {\rm{f}}\left( {\rm{n}} \right)\) is

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