The correct option is C 1/3 nLt→∞n(n+1)(2n+1)6· #160; #10;n^3 #10; ∵ #160; 1+2^2+3^2+....... #160; #160; #10;n^2=n(n+1)(1n+1)6 #10; ...

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Definite Integral as Limit of Sum

nLt→∞[1+4+9+…...

Question

$n L t \to \infty \frac{[1 + 4 + 9 + \dots + n^{2}]}{n^{3}} =$

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Solution

The correct option is C 1/3
$n L t \to \infty \frac{n (n + 1) (2 n + 1)}{6 \cdot n^{3}}$
$∵ 1 + 2^{2} + 3^{2} + . . . . . . . n^{2} = \frac{n (n + 1) (1 n + 1)}{6}$
$n L t \to \infty \frac{n (n + 1) (1 n + 1)}{6 n^{3}}$
$= n L t \to \infty [\frac{1}{6} (1 + \frac{1}{n}) (2 + \frac{1}{n})]$
$= \frac{2}{6} = \frac{1}{3}$

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