Question

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

• A. $-100$
• B. $-123$
• C. $-135$
• D. $-153$

Answer 1

The correct option is C $-135$

For $0\le r\le 66,0\le \frac{r}{100}<\frac{2}{3}$

$\Rightarrow -\frac{2}{3}<-\frac{r}{100}\le 0$

$\Rightarrow -\frac{1}{3}-\frac{2}{3}<-\frac{1}{3}-\frac{r}{100}\le \frac{1}{3}$

$\therefore [-\frac{1}{3}-\frac{r}{100}]=-1$ or $0\le r\le 66$

Also, for $67\le r\le 100,\frac{67}{100}\le \frac{r}{100}\le 1$

$\Rightarrow -1\le -\frac{r}{100}\le -\frac{67}{100}$

$\Rightarrow -\frac{1}{3}-1\le -\frac{1}{3}-\frac{r}{100}\le -\frac{1}{3}-\frac{67}{100}$

$\therefore [-\frac{1}{3}-\frac{r}{100}]=-2$ for $67\le r\le 100$

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