For x∈R, let [x] denote the greatest integer ≤x. Then the sum of the series
\[\left[-\frac13\right]+\left[-\frac13-\frac{1}{100}\right]+\left[-\frac13-\frac{2}{100}\right]+\cdots+\left[-\frac13-\frac{99}{100}\right]\]
is:
Solutions · 1
Official solution
Using [x]+[−x]=−1 (for non-integer x) and the identity [x+n1]+[x+n2]+⋯=[nx]−…:
\[\sum_{k=0}^{99}\left[-\frac13-\frac{k}{100}\right] = -100-\sum_{k=0}^{99}\left(\frac13+\frac{k}{100}-\left[\ldots\right]\right)\cdots = -100-\frac{100}{3} = -133.\]