Question

The value of $\underset{n\to \infty }{lim}\frac{1}{n^3}(\sqrt{n^2+1}+2\sqrt{n^2+2^2}+\cdots +n\sqrt{n^2+n^2})$ is

• A. $\frac{\sqrt{2}-1}{5}$
• B. $\frac{\sqrt{2}-1}{3}$
• C. $\frac{2\sqrt{2}-1}{5}$
• D. $\frac{2\sqrt{2}-1}{3}$

Answer 1

The correct option is D $\frac{2\sqrt{2}-1}{3}$

Let $L=\underset{n\to \infty }{lim}\frac{1}{n^3}(\sqrt{n^2+1}+2\sqrt{n^2+2^2}+\cdots +n\sqrt{n^2+n^2})$
$\Rightarrow L=\underset{n\to \infty }{lim}\frac{1}{n^3}\left[\sum _{r=1}^{n}r\sqrt{n^2+r^2}\right]\Rightarrow L=\underset{n\to \infty }{lim}\frac{1}{n}\sum _{r=1}^n\sqrt{1+{\left(\frac{r}{n}\right)}^2}\cdot \frac{r}{n}\Rightarrow L=\underset{0}{\overset{1}{\int }}\sqrt{1+x^2}\cdot xdx$

Assuming $\sqrt{1+x^2}=t$
$\Rightarrow \frac{2x}{2\sqrt{1+x^2}}dx=dt\Rightarrow xdx=tdt$
Now, $L=\underset{1}{\overset{\sqrt{2}}{\int }}{t}^{2}dt$
$\therefore L=\frac{2\sqrt{2}-1}{3}$