Question

Let $f\left(n\right)=\left[\frac{\left(n+50\right)}{100}\right]$ where $\left[n\right]$ denotes the integral part of $x.$ Then

• A. ${\sum _{n=1}}^{100}f\left(n\right)=51$
• B. ${\sum _{n=50}}^{100}f\left(n\right)=51$
• C. ${\sum _{n=1}}^{1000}f\left(n\right)=51$
• D. ${\sum _{n=1}}^{10}f\left(n\right)=51$

Answer 1

Answer: (A), (B)

${\sum _{n=50}}^{100}f\left(n\right)=51$

Explanation for the correct option:

Checking for Option (B):

We have, $f\left(n\right)=\left[\frac{\left(n+50\right)}{100}\right]$

For $1<n<50,f\left(n\right)=0.$

For $50\le n\le 100,f\left(n\right)=1$

Now,

$$\begin{gathered} {\sum _{n=1}}^{100}f\left(n\right)={\sum _{n=1}}^{49}f(n)+{\sum _{n=50}}^{100}f(n) \\ =0+1+1+(upto51terms) \\ {\sum _{n=50}}^{100}f\left(n\right)=51 \end{gathered}$$

Therefore, Option (B) is correct.

Checking for Option(A):

$f\left(n\right)=\left[\frac{\left(n+50\right)}{100}\right]$

For $1<n<50,f\left(n\right)=0$

$$\begin{gathered} {\sum _{n=1}}^{100}f\left(n\right)={\sum _{n=1}}^{49}f(n)+{\sum _{n=50}}^{100}f(n) \\ =0+1+1+(upto51terms) \\ {\sum _{n=0}}^{100}f\left(n\right)=51 \end{gathered}$$

Therefore, option A, is correct.

Explanation for incorrect options:

Checking for Option (C):

For $1\le n\le 49,f\left(n\right)=0$

For $50\le n\le 149,f\left(n\right)=1$

For $150\le n\le 249,f\left(n\right)=2$

$$\begin{gathered} \sum _{n=1}^{1000}f\left(n\right)=\sum _{n=0}^{49}f\left(n\right)+\sum _{n=50}^{149}f\left(n\right)+\sum _{n=150}^{249}f\left(n\right)+...+\sum _{n=950}^{1000}f\left(n\right) \\ =0+1\left(100\right)+2\left(100\right)+....+10\left(51\right) \\ =5010 \end{gathered}$$

Hence, option C is incorrect.

Checking for Option(D):

For $1\le n\le 10,f\left(n\right)=0$

$\underset{n=1}{\overset{10}{\sum f\left(n\right)}}=0$

Hence, option C is incorrect.

Therefore, the correct answer is option (A) and (B)