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Let f(n) = [1/4 + n/1000], where [x] de...

Let `f(n) = [1/4 + n/1000]`, where `[x]` denote the integral part of x. Then the value of `sum_(n=1)^(1000) f(n)` is

A

`251`

B

`250`

C

`1`

D

`0`

Text Solution

Verified by Experts

The correct Answer is:
A
`f(n)= [1/4 + n/1000]`
`sum_(n=1)overset(1000)f(n) = [1/4+1/100] + [1/4+ 2/1000] +"……"+[1/4+ 1000/1000]`
`= [0.25+ 0.001] + [0.25+0.002] +"….."+[0.25+1]`
We get '0' for all values of n from 1 to 750.
From ` n = 75`, we get all the values as 1.
So, `overset(1000)underset(n=1)(sum) f(n) = 0 + 0 + 0 +"......" +[1/4 + 750/1000] + [1/4 + 751/1000] + [1.25]`
`= 1 + 1 + 1 + "......"(251 "times")`
`= 251`.
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