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Question
\[\lim_{n \to \infty} \frac{1^2 + 2^2 + 3^2 + . . . + n^2}{n^3}\]
Options
(a) 1
(b) 1/2
(c) 1/3
(d) 0
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Solution
(c) 1/3
\[\lim_{n \to \infty} \frac{1^2 + 2^2 + 3^2 . . . . . n^2}{n^3}\]
\[ = \lim_{n \to \infty} \frac{\Sigma n^2}{n^3}\]
\[ = \lim_{n \to \infty} \frac{n\left( n + 1 \right) \left( 2n + 1 \right)}{6 n^3}\]
\[ = \lim_{n \to 0} \frac{\left( n + 1 \right) \left( 2n + 1 \right)}{6 n^2}\]
\[ \text{Dividing the numerator and the denominator by n}^2 , we get:\]
\[ \lim_{n \to \infty} \frac{\frac{\left( n + 1 \right)}{n} \times \frac{\left( 2n + 1 \right)}{n}}{6}\]
\[ = \lim_{n \to \infty} \frac{\left( 1 + \frac{1}{n} \right) \left( 2 + \frac{1}{n} \right)}{6}\]
\[ \Rightarrow \frac{2}{6} = \frac{1}{3}\]
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