Question

${lim}_{\pi \to \infty }\frac{1}{n}{\sum }_{r=1}^{2n}\frac{r}{\sqrt{n^2+r^2}}=$

• A.
• B.
  
• C.
  
• D.
  

Answer 1

The correct option is A. $-1+\sqrt{5}$

$L={lim}_{\pi \to \infty }{\sum }_{r=1}^{2n}\frac{1}{n}.\frac{\frac{r}{n}}{\sqrt{1+{\left(\frac{r}{n}\right)}^2}}={\int }_0^2\frac{x}{\sqrt{1+x^2}}dx$

Put $1+x^2=t^2$

$\Rightarrow xdx=tdt$

$x=0$ then $t=1$

$x=2$ then $t=\sqrt{5}$

Now,

${\int }_0^2\frac{x}{\sqrt{1+x^2}}dx$

$={\int }_1^{\sqrt{5}}\frac{t}{t}dt$

$={\int }_1^{\sqrt{5}}1dt$

$=[t{]}_1^{\sqrt{5}}$

$=\sqrt{5}-1$

Hence, Option A is correct.