logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
0 votes

$\lim\limits_{n\to \infty}\large\frac{1}{n}$$\sum\limits_{r=1}^{2n}\large\frac{r}{\sqrt{n^2+r^2}}$ equals

$(a)\;1+\sqrt 5\qquad(b)\;-1+\sqrt 5\qquad(c)\;-1+\sqrt 2\qquad(d)\;1+\sqrt 2$

Can you answer this question?
 
 

1 Answer

0 votes
We have =$\lim\limits_{n\to \infty}\large\frac{1}{n}\sum\limits_{r=1}^{2n}\large\frac{r}{\sqrt{n^2+r^2}}$
$\Rightarrow \lim\limits_{n\to \infty}\large\frac{1}{n}\sum\limits_{r=1}^{2n}\large\frac{r}{n\sqrt{1+(r/n)^2}}$
$\Rightarrow \int_0^2\large\frac{x}{\sqrt{1+x^2}}$$dx$
$\lim\limits_{n\to \infty}\large\frac{1}{r}\sum_{r=0}^{a_n}f(\Large\frac{r}{n})=$$\int_0^a f(x) dx$
$\bigg[\sqrt{1+x^2}\bigg]_0^2=\sqrt 5-1$
Hence (b) is the correct answer.
answered Jan 3, 2014 by sreemathi.v
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...