Question

$\underset{n\to \infty }{\lim }\frac{3}{n}\left\{4+{\left(2+\frac{1}{n}\right)}^2+{\left(2+\frac{2}{n}\right)}^2+\dots +{\left(3-\frac{1}{n}\right)}^2\right\}$ is equal to

• A. $\frac{19}{3}$
• B. 12
• C. 0
• D. 19

Answer 1

Answer: (D)

$\underset{n\to \infty }{\lim }\frac{3}{n}\sum _{r=0}^{n-1}{\left(2+\frac{r}{n}\right)}^2$
$=3\underset{0}{\overset{1}{\int }}(2+x{)}^{2}dx=27-8=19$