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Lim N → ∞ 1 2 + 2 2 + 3 2 + . . . + N 2 N 3 - Mathematics

MCQ

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

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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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 29 Limits
Q 1 | Page 77
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