for x
Question:

For $x \in \mathrm{R}$, let $[x]$ denote the greatest integer $\leq x$, then the sum of the series

$\left[-\frac{1}{3}\right]+\left[-\frac{1}{3}-\frac{1}{100}\right]+\left[-\frac{1}{3}-\frac{2}{100}\right]+\cdots+\left[-\frac{1}{3}-\frac{99}{100}\right]_{\mathrm{i}}$

1. (1) $-153$

2. (2) $-133$

3. (3) $-131$

4. (4) $-135$

Correct Option: 2,

Solution:

$\because[x]+\left[x+\frac{1}{n}\right]+\left[x+\frac{2}{n}\right] \ldots\left[x+\frac{n-1}{n}\right]=[n x]$

and $[x]+[-x]=-1(x \notin z)$

$\therefore\left[-\frac{1}{3}\right]+\left[-\frac{1}{3}-100\right]+\ldots+\left[-\frac{1}{3}-\frac{99}{100}\right]$

$=-100-\left\{\left[\frac{1}{3}\right]+\left[\frac{1}{3}+\frac{1}{100}\right]+\ldots\left[\frac{1}{3}+\frac{99}{100}\right]\right\}$

$=-100-\left[\frac{100}{3}\right]=-133$