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]`

Answer

B

Explanation

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`