For $x\in R,$let $\left[x\right]$ denote the greatest integer  $\le$x  , then the sum of the series $\left[-\frac{1}{3}\right]+\left[-\frac{1}{3}-\frac{1}{100}\right]+\left[-\frac{1}{3}-\frac{2}{100}\right]+\mathrm{.......}+\left[-\frac{1}{3}-\frac{99}{100}\right]is$

We know that

$\left[x\right]+\left[x+\frac{1}{n}\right]+\left[x+\frac{2}{n}\right]+\mathrm{........}+\left[x+\frac{n-1}{n}\right]=\left[nx\right]$

and $\left[-x\right]=-\left[x\right]-1$

$\begin{array}{l}=-1-\left[\frac{1}{3}\right]-1-\left[\frac{1}{.3}+\frac{1}{100}\right]-1-\left[\frac{1}{3}+\frac{2}{100}\right]+\mathrm{........}-1-\left[\frac{1}{3}+\frac{99}{100}\right]\\ =-100-\left[\frac{1}{3}100\right]\\ =-100-33=-133\end{array}$