Question

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

• A. $-135$
• B. $-153$
• C. $-131$
• D. $-133$

Answer 1

The correct option is D $-133$

$[-\frac{1}{3}]+[-\frac{1}{3}-\frac{1}{100}]+[-\frac{1}{3}-\frac{2}{100}]+....+[-\frac{1}{3}-\frac{99}{100}]$
where $x\in R,\left[x\right]$ denote the greatest integer $\le x,$
Since $[-\frac{1}{3}]=-1$ and
$-\frac{1}{3}-\frac{x}{100}=-\frac{100+3x}{300}$
therefore,
For case 1-
$100+3x<300$
$\Rightarrow x<66.67\therefore [-\frac{1}{3}-\frac{x}{100}]=-1$
For case 2-
$300\le 100+3x<600\Rightarrow \frac{200}{3}\le x<\frac{500}{3}\Rightarrow 67\le x<166$
$\therefore [-\frac{1}{3}-\frac{x}{100}]=-2$
So the sum of the series equals
$=(-1-1-1-\cdots 67\text{times})+(-2-2-2-\cdots 33\text{times})$
$-67-2\left(33\right)=-133$