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InterviewSolution
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For a real number `x, [x]` denotes greatest integer function, then find value of `[1/2]+[1/2+1/100]+[1/2+2/100]+....+[1/2+99/100]`A. `-153`B. `-133`C. `-131`D. `-135` |
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Answer» Correct Answer - B Given series is `[-(1)/(3)] + [-(1)/(3) - (1)/(100)] + [-(1)/(3) - (2)/(100)] + ... ... +[-(1)/(3) - (99)/(100)]` [where, [x] denotes the greatest integer `le x`] Now, `[-(1)/(3)], [-(1)/(3) - (1)/(100)] , [-(1)/(3)-(2)/(100)],... + [-(1)/(3) - (66)/(100)]` all the term have value `-1` and `[-(1)/(3) - (67)/(100)], [-(1)/(3) -(68)/(100)], ..., [-(1)/(3) - (99)/(100)]` all the term have value `-2` So, `[-(1)/(3)] + [-(1)/(3) - (1)/(100)] + [-(1)/(3) - (2)/(100)] + ... + [-(1)/(3) - (66)/(100)] ` `= -1 -1 -1-1...67` times. `= (-1) xx 67 = -67` and `[-(1)/(3) - (67)/(100)] + [-(1)/(3) - (68)/(100)]+..+ [-(1)/(3) - (99)/(100)]` `= -2 -2-2 -2`...33 times `= (-2) xx 33 = -66` `:. [-(1)/(3)] + [-(1)/(3) - (1)/(100)] + [-(1)/(3) - (2)/(100)] + ....+ [-(1)/(3) - (99)/(100)]` `= (-67) + (-66) = - 133` Alternate Solution `:. [-x] = -[x] -1, " if " x !in` Integer, and `[x] + [x + (1)/(n)] + [x + (2)/(n)] + ...+ [x + (n-1)/(n)] = [nx], n in N` So given series `[-(1)/(3)] + [-(1)/(3) - (1)/(100)] + [-(1)/(3) - (2)/(100)] + [-(1)/(3) - (2)/(100)] + ....+ [(-1)/(3) - (99)/(100)]` `= (-[(1)/(3)] -1) + (-[(1)/(3) + (1)/(100)] -1) + (-[(1)/(3) + (2)/(100)] -1) + ...+ (-[(1)/(3) + (99)/(100)] -1)` `= (-1) xx 100 - [(1)/(3) xx 100] = - 100 - 33 = - 133` |
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