CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

The value of $$\underset{n\rightarrow \infty}{\lim} \begin{Bmatrix}\dfrac{\sqrt{n+1}+\sqrt{n+2}+.....+\sqrt{2n-1}}{n^{\tfrac{3}{2}}}\end{Bmatrix}$$ is


A
23(221)
loader
B
23(21)
loader
C
23(2+1)
loader
D
23(22+1)
loader

Solution

The correct option is A $$\dfrac{2}{3}(2\sqrt{2}-1)$$
$$\underset{n\rightarrow \infty}{\lim} \begin{Bmatrix}\dfrac{\sqrt{n+1}+\sqrt{n+2}+.....+\sqrt{2n-1}}{n^{\tfrac{3}{2}}}\end{Bmatrix}$$

$$=\underset{n\rightarrow \infty}{\lim} \begin{Bmatrix}\dfrac{\sqrt{n+1}+\sqrt{n+2}+.....+\sqrt{2n-1}}{\sqrt n}\end{Bmatrix}\dfrac{1}{n}$$

$$=\underset{n\rightarrow \infty}{\lim} \begin{Bmatrix}{\sqrt{1+\dfrac{1}{n}}+\sqrt{1+\dfrac{2}{n}}+.....+\sqrt{1+\dfrac{n-1}{n}}}{}\end{Bmatrix}\dfrac{1}{n}$$
Now using limit as an integrals 
$$\displaystyle =\int_0^1\sqrt{1+x}dx=\dfrac{2}{3}[(1+x)\sqrt{1+x}]_0^1=\dfrac{2}{3}(2\sqrt2-1)$$

Mathematics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image