Question

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

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

Answer 1

Answer: (D)

$\frac{1}{4}$

$\underset{n\to \infty }{lim}\frac{1}{n^4}\sum _{r=1}^{n}r(r+2)(r+4)$
$=\underset{n\to \infty }{lim}\frac{1}{n^4}\sum _{r=1}^n(r^3+4r^2+2r^2+8r)$ (use summation formula)
$=\underset{n\to \infty }{lim}\frac{1}{n^4}({\left(\frac{n(n+1)}{2}\right)}^2+\frac{6n(n+1)(2x+1)}{6}+\frac{8\left(n\right)(n+1)}{2})$
$=\underset{n\to \infty }{lim}\frac{n^4}{n^4}(\frac{(1+\frac{1}{n}{)}^2}{4}+\frac{6}{n}\frac{(1+\frac{1}{n})(1+\frac{1}{n})}{6}+\frac{8}{n^2}\frac{(1+\frac{1}{n})}{2})$
$=\frac{1}{4}$