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[(4)/(5)] + [(4)/(5)+(1)/(1000)] + [(4)/...

`[(4)/(5)] + [(4)/(5)+(1)/(1000)] + [(4)/(5)+(2)/(1000)] + ...+[(4)/(5) + (999)/(1000)] =`
where [.] denotes greatest integer function

A

998

B

980

C

800

D

801

Text Solution

Verified by Experts

The correct Answer is:
C

`underset(200" terms are zero")ubrace([(4)/(5)]+[(4)/(5)+(1)/(1000)]+....+[(4)/(5)+(199)/(1000)])+underset(800" terms are equal to 1")ul([(4)/(5)+(200)/(1000)]+[(4)/(5)+(201)/(1000)]+....+[(4)/(5)+(999)/(1000)])`
`rArr 800`
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