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Find [x]+\sum_{r=1}^{2000} \frac {\{x+r\}}{2000} , where [x] = GIF and {x} = FP.
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3 Answers
1note that as r is a positive integer so {x+r} = {x}
=> [x] + {x} *2000/2000 = x.
481After simplifying,we get.....
(2000[x]+x+1-[x+1]+x+2-[x+2]........+x+2000-[x+2000])/2000
[x+n]=[x]+n....where n is an integer
Applying this in d above eq. we get.....2000x/2000=x
Thus d value of above expression is x.
