Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • The value of <img alt="\mathrm{\lim _{n \rightarrow \infty} \frac{[r]+[2 r]+\ldots .+[n r]}{n^2}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Clim%20_%7Bn%20%5Crightarrow%20

The value of \mathrm{\lim _{n \rightarrow \infty} \frac{[r]+[2 r]+\ldots .+[n r]}{n^2}}, where \mathrm{r} is a non-zero real number and \mathrm{[r]} denotes the greatest integer less than or equal to \mathrm{r}, is equal to \mathrm{\frac{a r}{8}}, then value of \mathrm{{a }} is

Option: 1

4


Option: 2

1


Option: 3

2


Option: 4

5


Answers (1)

best_answer

\mathrm{ \text { We have, } r-1<[r] \leq r, 2 r-1<[2 r] \leq 2 r \text {, ......, } }

\mathrm{ n r-1<[n r] \leq n r \\ }

\mathrm{ \Rightarrow \frac{n(n+1) r}{2}-n<[r]+[2 r]+\ldots \ldots+[n r] \leq \frac{n(n+1) r}{2} }

\mathrm{ \Rightarrow \frac{n(n+1) r}{2 n^2}-\frac{1}{n}<\frac{[r]+[2 r]+\ldots \ldots+[n r]}{n^2} \leq \frac{n(n+1) r}{2 n^2} }

\mathrm{ \Rightarrow \lim _{n \rightarrow \infty}\left(\frac{n(n+1) r}{2 n^2}-\frac{1}{n}\right) }

\mathrm{ <\lim _{n \rightarrow \infty} \frac{[r]+[2 r]+[3 r]+\ldots+[n r]}{n^2} \leq \lim _{n \rightarrow \infty} \frac{n(n+1)}{2 n^2} r }

\mathrm{ \Rightarrow \frac{r}{2}<\lim _{n \rightarrow \infty} \frac{[r]+[2 r]+[3 r]+\ldots+[n r]}{n^2} \leq \frac{r}{2} }

\mathrm{ \text { Thus, } \lim _{n \rightarrow \infty} \frac{[r]+[2 r]+\ldots \ldots+[n r]}{n^2}=\frac{r}{2}=\frac{a r}{8}, a=4 }




 

Posted by

Divya Prakash Singh

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JAC Chandigarh Counselling 2025: Round 1 seat allotment result out for BTech admissions
anam-khan

JoSAA round 5 seat allotment result 2025 out on josaa.nic.in
anam-khan

BTech aspirant from north-east or UT? Apply for CSAB NEUT counselling 2025 today

Latest Question