lim n 3 n { 4 + ( 2 + 1 n ) 2 + ( 2 + 2 n ) 2 + ? + ( 3 - 1 n ) 2 } is equal to [2023]

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Solution

Q.
$\lim_{n \to \infty} \frac{3}{n} {4 + {(2 + \frac{1}{n})}^{2} + {(2 + \frac{2}{n})}^{2} + \dots + {(3 - \frac{1}{n})}^{2}}$ is equal to [2023]

1 $\frac{19}{3}$

2 12

3 0

4 19

Ans.
(4)

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

$= \lim_{n \to \infty} \frac{3}{n} {{(2 + \frac{0}{n})}^{2} + {(2 + \frac{1}{n})}^{2} + \dots + (2 + {(\frac{n - 1}{n}))}^{2}}$

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

Let $2 + x = t d x = d t$ $\Rightarrow 3 \int_{2}^{3} t^{2} d t = 3 \times {[\frac{t^{3}}{2}]}_{2}^{3} = \frac{3}{3} [27 - 8] = 19$

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