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

JEE Main · 2023 · 30 Jan 2023 · Shift 2 · Definite Integrals

lim_{n \to \infty} \frac{3}{n} {4 + (2 + \frac{1}{n})^{2} + (2 + \frac{2}{n})^{2} + \dots + (3 - \frac{1}{n})^{2}}

is equal to

Answer: (D)

The bracket is

r = 0 \sum n - 1 (2 + \frac{r}{n})^{2}

, so the expression is the Riemann sum

\frac{3}{n} r = 0 \sum n - 1 (2 + \frac{r}{n})^{2} \to 3 \int_{0}^{1} (2 + x)^{2} d x

. Evaluating,

3 \int_{0}^{1} (2 + x)^{2} d x = [(2 + x)^{3}]_{0}^{1} = 3^{3} - 2^{3} = 19

JEE Main · 2023 · 30 Jan 2023 · Shift 2 · Single correct option

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