For $\mathrm{x \in R}$, let $\mathrm{\left [ x \right ]}$  denote the greatest integer $\mathrm{\leq x}$, then value of  $\mathrm{\left[-\frac{1}{3}\right]+\left[-\frac{1}{3}-\frac{1}{100}\right]+\left[-\frac{1}{3}-\frac{2}{100}\right]+\ldots+\left[-\frac{1}{3}-\frac{99}{100}\right]}$ isOption: 1 -100Option: 2 -123Option: 3 -135Option: 4 -153

For $\mathrm{0 \leq r \leq 66,0 \leq \frac{r}{100}<\frac{2}{3} \Rightarrow-\frac{2}{3}<-\frac{r}{100} \leq 0}$
$\mathrm{\Rightarrow \quad-\frac{1}{3}-\frac{2}{3}<-\frac{1}{3}-\frac{r}{100} \leq-\frac{1}{3}}$
$\mathrm{\therefore \quad\left[-\frac{1}{3}-\frac{r}{100}\right]=-1 \quad\, for \, 0 \leq r \leq 66}$

Also, for $\mathrm{67 \leq r \leq 100, \frac{67}{100} \leq \frac{r}{100} \leq 1 \Rightarrow-1 \leq-\frac{r}{100} \leq-\frac{67}{100}}$

$\mathrm{\Rightarrow \quad-\frac{1}{3}-1 \leq-\frac{1}{3}-\frac{r}{100} \leq-\frac{1}{3}-\frac{67}{100}}$
$\mathrm{\therefore \quad\left[-\frac{1}{3}-\frac{r}{100}\right]=-2\, for \, 67 \leq r \leq 100}$

Hence , $\mathrm{\sum_{r=0}^{100}\left[-\frac{1}{3}-\frac{r}{100}\right]=67(-1)+2(-34)=-135}$.