Indian Institute of Foreign Trade 2010 Solved Paper

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Question : 13 of 135

Marks: +1, -0

If [x] is the greater integer less than or equal to ‘x’, then find the value of the following series

4389
4839
3498
3489

Solution:

[√{1}]+[√{2}]+[√{3}]+[√{4}]+. . .+[√{360}]=1+1+1+2+2+2+2+2+3+3+3. . .7

times

+4+4+......9

times

+.........+18+18+......37

times

=1×3+2×5+3×7+4×9+............+18×37

=∑^18_1n(2n+1)

=∑^18_{1}2n^2+n

={2n×(n+1)×(2n+1)}/6+{n×(n+1)}/2

={2×18×(19)×(37)}/6+{18×19}/2

=4389

Therefore our answer is option 'A'

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