Question

$\underset{n\to \infty }{\text{lim}}\frac{1}{n^4}\sum _{r=1}^{n}r(r+2)(r+4)=$

• A. $\frac{3}{4}$
• B. $0$
• C. $\frac{1}{8}$
• D. $\frac{1}{4}$

Answer 1

The correct option is D $\frac{1}{4}$

Given $\underset{n\to \infty }{\text{lim}}\frac{1}{n^4}\sum _{r=1}^{n}r(r+2)(r+4)$
$=\underset{n\to \infty }{\text{lim}}\frac{1}{n^4}\left[\sum r^3+6\sum r^2+8\sum r\right]$
$=\underset{n\to \infty }{\text{lim}}\frac{1}{n^4}[\frac{n^2(n+1{)}^2}{4}+6\cdot \frac{n(n+1)(2n+1)}{6}+8\frac{n(n+1)}{2}]$
Take $n^4$ common from the numerator and denominator and then applying the limits, we get
$=\frac{1}{4}$