Question

Let $f\left(n\right)=[\frac{1}{2}+\frac{n}{100}]$ where [.] denotes the greatest integer function, then the value of ${\sum }_{n=1}^{151}f\left(n\right)$ is

• A. $101$
• B. $102$
• C. $104$
• D. $103$

Answer 1

The correct option is C $104$

Clearly till $n=49$ $f\left(n\right)=0$

At $n=50$ $f\left(n\right)=\left[1\right]=1$

At $n=150$ $f\left(n\right)=\left[\frac{200}{100}\right]=2$ Also $f\left(151\right)=2$

Till $n=149$ $f\left(n\right)=1$

Hence sum $=100\times 1+2\times 2=104$