\[\lim_{n \to \infty} \left[ \frac{1^2 + 2^2 + . . . + n^2}{n^3} \right]\]

\[= \lim_{n \to \infty} \left[ \frac{1^2 + 2^2 + . . . + n^2}{n^3} \right]\]\[ = \lim_{n \to \infty} \left[ \frac{n\left( n + 1 \right) \left( 2n + 1 \right)}{6 n^3} \right]\]\[ = \lim_{n \to \infty} \left[ \frac{\left( n + 1 \right) \left( 2n + 1 \right)}{6 n^2} \right]\]\[ = \lim_{n \to \infty} \left[ \left( \frac{n + 1}{n} \right) \left( \frac{2n + 1}{n} \right) \times \frac{1}{6} \right] \]\[ = \lim_{n \to \infty} \left[ \left( 1 + \frac{1}{n} \right) \left( 2 + \frac{1}{n} \right) \times \frac{1}{6} \right]\]\[ = \frac{2}{6}\]\[ = \frac{1}{3}\]

lim n → ∞ [ 1 2 + 2 2 + . . . + n 2 n 3 ] - Mathematics

Question

$lim_{n \to \infty} [\frac{1^{2} + 2^{2} + . . . + n^{2}}{n^{3}}]$

Solution

$= lim_{n \to \infty} [\frac{1^{2} + 2^{2} + . . . + n^{2}}{n^{3}}]$
$= lim_{n \to \infty} [\frac{n (n + 1) (2 n + 1)}{6 n^{3}}]$
$= lim_{n \to \infty} [\frac{(n + 1) (2 n + 1)}{6 n^{2}}]$
$= lim_{n \to \infty} [(\frac{n + 1}{n}) (\frac{2 n + 1}{n}) \times \frac{1}{6}]$
$= lim_{n \to \infty} [(1 + \frac{1}{n}) (2 + \frac{1}{n}) \times \frac{1}{6}]$
$= \frac{2}{6}$
$= \frac{1}{3}$

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Concept of Limits - Algebra of Limits

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Q 14 Q 13 Q 15

Chapter 29: Limits - Exercise 29.6 [Page 39]

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RD Sharma Mathematics [English] Class 11

Chapter 29 Limits
Exercise 29.6 | Q 14 | Page 39

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lim n → ∞ [ 1 2 + 2 2 + . . . + n 2 n 3 ] - Mathematics

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