Question

The value of $\underset{n\to \infty }{lim}\frac{1.2+2.3+3.4+...+n.(n+1)}{n^3}$ is:

• A. $1$
• B. $-1$
• C. $\frac{1}{3}$
• D. $-\frac{1}{3}$

Answer 1

Answer: (C)

$\frac{1}{3}$

$\underset{n\to \infty }{lim}\frac{1.2+2.3+3.4+...+n.(n+1)}{n^3}$
$={lim}_{n\to \infty }\frac{{\sum }_{r=1}^{n}r(r+1)}{n^3}$
$=\underset{n\to \infty }{lim}\frac{{\sum }_{r=1}^n(r^2+r)}{n^3}$
$=\underset{n\to \infty }{lim}\frac{\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}}{n^3}$
$=\frac{1}{3}$