Question

Let $f\left(n\right)=[\frac{1}{4}+\frac{n}{1000}]$, where $\left[x\right]$ denotes the integral part of $x$. Then the value of $\sum _{n=1}^{1000}f\left(n\right)$ is

• A. $251$
• B. $250$
• C. $1$
• D. $0$

Answer 1

Answer: (A)

$251$

For all $n<750$, the value of $f\left(n\right)$ would be 0 as $\frac{1}{4}+\frac{n}{1000}<1$.

But, for $750\le n\le 1000$, the value of $f\left(n\right)$ would be 1 as $\frac{1}{4}+\frac{n}{1000}>1$ and hence, the greatest integer function would round it off to $1$.

So, a total of $251$ values give 1 and hence, their sum is $251$.