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