MCQ
$\lim _{n \rightarrow \infty} \frac{3}{n}\left\{4+\left(2+\frac{1}{n}\right)^2+\left(2+\frac{2}{n}\right)^2+\ldots+\left(3-\frac{1}{n}\right)^2\right\}$ is equal to
  • A
    $12$
  • B
    $\frac{19}{3}$
  • C
    $0$
  • $19$

Answer

Correct option: D.
$19$
d
$\lim _{n \rightarrow \infty} \frac{3}{n} \sum \limits_{r=0}^{n-1}\left(2+\frac{r}{n}\right)^2$

$=3 \int \limits_0^1(2+x)^2 d x=27-8=19$

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