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Q. $\displaystyle\lim _{n \rightarrow \infty} \frac{3}{n}\left\{4+\left(2+\frac{1}{n}\right)^2+\left(2+\frac{2}{n}\right)^2+\ldots+\left(3-\frac{1}{n}\right)^2\right\}$ is equal to

JEE MainJEE Main 2023Limits and Derivatives

Solution:

$ \displaystyle\lim _{n \rightarrow \infty} \frac{3}{n} \displaystyle\sum_{r=0}^{n-1}\left(2+\frac{r}{n}\right)^2 $
$ =3 \int\limits_0^1(2+x)^2 d x=27-8=19$