UPSC NDA 2 2017 Mathematics Solved Paper

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Question : 87 of 120

Marks: +1, -0

Let f(n)=[‌‌

1

4

+‌‌

n

1000

] where [x] denotes the integral part of x. then the value of

1000

∑

n=1

f(n) is

[NDA II 2017]

251
250
1
0

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