Question

Value of $\underset{n\to \infty }{lim}\frac{1}{n^3}[1+3+6+\cdots +\frac{n(n+1)}{2}]$ is

• A. $\frac{1}{24}$
• B. $\frac{1}{18}$
• C. $\frac{1}{12}$
• D. $\frac{1}{6}$

Answer 1

The correct option is D

$\frac{1}{6}$

$1+3+6+\cdots +\frac{n(n+1)}{2}=\sum _{r=1}^n\frac{r(r+1)}{2}\therefore L=\underset{n\to \infty }{lim}\frac{\sum _{r=1}^n\frac{r(r+1)}{2}}{n^3}=\frac{1}{2}\underset{n\to \infty }{lim}[\frac{\sum _{r=1}^n{r}^2}{n^3}+\frac{\sum _{r=1}^{n}r}{n^3}]=\frac{1}{2}\underset{n\to \infty }{lim}[\frac{n(n+1)(2n+1)}{6n^3}+\frac{n(n+1)}{2n^3}]=\frac{1}{2}\{\frac{1}{3}+0\}=\frac{1}{6}$