Let f(n)= [(1)/(4)+ n 1000 ], where [ x] denote the integral part of x . Then the value of ∑_ n=1 ^ 1000 f(n) is

Correct answer: A. 251 aligned & f(n)=[(1)/(4)+(n)/(1000)] & ∑_ n=1 ^ 1000 f(n)=[(1)/(4)+(1)/(1000)]+[(1)/(4)+(2)/(1000)]+ .+[(1)/(4)+(1000)/(1000)] aligned =[0.25+0.001]+[0.25+0.002]+ .+[0.25+1] We get ' 0 ' for all values of n from 1 to 750 . From n=750 , we get all the values as 1 . So, array rl ∑_ n=1 ^ 1000 & f(n)=0+0+0+ .+[(1)/(4)+(750)/(1000)]+[(1)/(4)+(751)/(1000)]+ .[1.25] & =1+1+1+ .(251 times ) & =251 array

NDA 2017

Exam: NDA
Year: 2017
Subject: Mathematics
Chapter: Sequences and Series

NDA 2017 · Mathematics · Sequences and Series

Let f(n)= [(1)/(4)+ n 1000 ], where [ x] denote the integral part...

Let

f(\mathrm{n})= \left[\frac{1}{4}+\frac{\mathrm{n}}{1000}\right], where

[

\mathrm{x}] denote the integral part of

. Then the value of \sum_{n=1}^{1000} f(n)

Options

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

Correct answer

A. $251$

Step-by-step solution

\begin{aligned} & f(n)=\left[\frac{1}{4}+\frac{n}{1000}\right] \\ & \sum_{n=1}^{1000} f(n)=\left[\frac{1}{4}+\frac{1}{1000}\right]+\left[\frac{1}{4}+\frac{2}{1000}\right]+\ldots .+\left[\frac{1}{4}+\frac{1000}{1000}\right]\end{aligned}

=[0.25+0.001]+[0.25+0.002]+\ldots .+[0.25+1]

We get ' 0 ' for all values of

from 1 to 750 . From

\mathrm{n}=750

, we get all the values as 1 . So,

\begin{array}{rl}\sum_{n=1}^{1000} & f(n)=0+0+0+\ldots .+\left[\frac{1}{4}+\frac{750}{1000}\right]+\left[\frac{1}{4}+\frac{751}{1000}\right]+\ldots .[1.25] \\ & =1+1+1+\ldots .(251 \text{ times }) \\ & =251\end{array}

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