Question

$li{m}_{\pi \to \infty }\frac{1}{n}{\sum }_{r=1}^{2n}\frac{r}{\sqrt{n^2+r^2}}equals$ [IIT 1997 Re-exam]

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

Answer 1

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

$L=li{m}_{\pi \to \infty }{\sum }_{r=1}^{2n}\frac{1}{n}.\frac{\frac{r}{n}}{\sqrt{1+{\left(\frac{r}{n}\right)}^2}}={\int }_0^2\frac{x}{\sqrt{1+x^2}}dx=\sqrt{5}-1$