$lim_{n \to \infty} \frac{1^{2} + 2^{2} + 3^{2} + . . . . . n^{2}}{n^{3}}$ किसके बराबर है?

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Question

This question was previously asked in

UP TGT Mathematics 2021 Official Paper

Answer 1

संकल्पना:

$1^{2} + 2^{2} + 3^{2} + . . . . . n^{2} = \frac{n (n + 1) (2 n + 1)}{6}$

गणना:

दिया गया:

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

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

= $lim_{n \to \infty} \frac{(n + 1) (2 n + 1)}{6 n^{2}}$

= $lim_{n \to \infty} \frac{n^{2} (1 + \frac{1}{n}) (2 + \frac{1}{n})}{6 n^{2}}$

= $(1 + \frac{1}{\infty}) (2 + \frac{1}{\infty}) \frac{1}{6}$

= $\frac{1}{3}$

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