For x∈R, let [x] denote the greatest integer ≤x, then the sum of the series [−13]+[−13−1100]+[−13−2100]+....+[−13−99100] is :
A
−135
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B
−153
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C
−131
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D
−133
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Solution
The correct option is D−133 [−13]+[−13−1100]+[−13−2100]+....+[−13−99100]
where x∈R,[x] denote the greatest integer ≤x,
Since [−13]=−1 and −13−x100=−100+3x300
therefore,
For case 1- 100+3x<300 ⇒x<66.67∴[−13−x100]=−1
For case 2- 300≤100+3x<600⇒2003≤x<5003⇒67≤x<166 ∴[−13−x100]=−2
So the sum of the series equals =(−1−1−1−⋯67 times)+(−2−2−2−⋯33 times) −67−2(33)=−133