Question

If $\left[x\right]$ stands for the greatest integer function, then $\left[\frac{1}{2}+\frac{1}{1000}\right]+[\frac{1}{2}+\frac{2}{1000}]+\dots .+[\frac{1}{2}+\frac{999}{1000}]=$

• A. $498$
• B. $499$
• C. $500$
• D. $501$

Answer 1

Answer: (C)

$500$

$\left[\frac{1}{2}+\frac{1}{1000}\right]+[\frac{1}{2}+\frac{2}{1000}]+\dots .+[\frac{1}{2}+\frac{999}{1000}]$
$=\underset{–}{\left[\frac{1}{2}+\frac{1}{1000}\right]+[\frac{1}{2}+\frac{2}{1000}]+....+[\frac{1}{2}+\frac{499}{1000}]}+\underset{–}{[\frac{1}{2}+\frac{500}{1000}]+[\frac{1}{2}+\frac{501}{1000}].........+[\frac{1}{2}+\frac{999}{1000}]}$
$499\text{terms}$ $500\text{terms}$
First $499$ terms will vanish individually since they lie between $0$ and $1$
And next $500$ terms will be $1$ individually since they lie between $1$ and $2$
Hence required summation is $500$.