Question

Let $f\left(n\right)=[\frac{1}{2}+\frac{n}{100}]$ where $\left[x\right]$ denotes the integral part of $x$. Then the value of ${\sum }_{n=1}^{100}f\left(n\right)$ is

• A. $50$
• B. $51$
• C. $1$
• D. none of these

Answer 1

Answer: (B)

$51$

For, $1\le n<50$

$\Rightarrow \frac{51}{100}\le \frac{1}{2}+\frac{n}{100}<1$
$\Rightarrow [\frac{1}{2}+\frac{n}{100}]=0$ ...(1)

For, $50\le n\le 100$
$\Rightarrow 1\le \frac{1}{2}+\frac{n}{100}\le \frac{3}{2}$
$\Rightarrow [\frac{1}{2}+\frac{n}{100}]=1$ ...(2)

Now, ${\sum }_{n=1}^{100}f\left(n\right)={\sum }_{n=1}^{49}[\frac{1}{2}+\frac{n}{100}]+{\sum }_{n=50}^{100}[\frac{1}{2}+\frac{n}{100}]$
$\Rightarrow {\sum }_{n=1}^{100}f\left(n\right)=0+{\sum }_{n=50}^{100}1=51$ ....[ From (1) & (2) ]