Question

If the value of ${lim}_{n\to \infty }\left(n^{-3/2}\right){\sum }_{j=1}^{6n}\sqrt{j}$ is equal to $\sqrt{N}$, then value of N/12 is___

Answer 1

${lim}_{n\to \infty }\frac{\sqrt{1}+\sqrt{2}+\sqrt{3}+....+\sqrt{6n}}{n\sqrt{n}}=\sqrt{N}={lim}_{n\to \infty }\frac{1}{n}{\sum }_{r=1}^{6n}\sqrt{\frac{r}{n}}={\int }_0^6\sqrt{x}dx=\sqrt{N}\Rightarrow {\left[\frac{2}{3}{x}^{\frac{3}{2}}\right]}_0^6=\sqrt{N}\Rightarrow \sqrt{N}=\frac{2}{3}.6\sqrt{6}\Rightarrow \sqrt{N}=\sqrt{96}\therefore N/12=\frac{96}{12}=8$