Question

Find $\underset{n\to \infty }{lim}\frac{1.n+2(n-1)+3(n-2)+...+(n-2).3+(n-1)2+n.1}{n^3}$ equals to

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $\frac{1}{6}$
• D. $\frac{1}{4}$

Answer 1

The correct option is C $\frac{1}{6}$

$\underset{n\to \infty }{lim}\frac{1.n+2(n-1)+3(n-2)+...+(n-2).3+(n-1)2+n.1}{n^3}$

$=\underset{{}_{n\to \infty }}{lim}\frac{{\sum }_{r=1}^{n}r(n-r+1)}{n^3}$

$=\underset{{}_{n\to \infty }}{lim}\frac{{\sum }_{r=1}^n(n+1)r-r^2}{n^3}$

$=\underset{{}_{n\to \infty }}{lim}\frac{\frac{n{(n+1)}^2}{2}-\frac{n(n+1)(2n+1)}{6}}{n^3}$

$=\underset{{}_{n\to \infty }}{lim}\frac{\frac{n(n+1)(n+2)}{6}}{n^3}$

$=\underset{{}_{n\to \infty }}{lim}\frac{\frac{n^3(1+\frac{1}{n})(1+\frac{2}{n})}{6}}{n^3}$

$=\underset{{}_{n\to \infty }}{lim}\frac{(1+\frac{1}{n})(1+\frac{2}{n})}{6}$

$=\frac{1}{6}$ $[\because n\to \infty \Rightarrow \frac{1}{n}\to 0]$