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Question

$$n\overset{Lt}{\rightarrow}\infty \displaystyle \frac{[1+4+9+\ldots+n^{2}]}{n^{3}}=$$


A
2
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B
3
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C
1/3
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D
1
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Solution

The correct option is C 1/3

$$n\overset{Lt}{\rightarrow}\infty\dfrac{n(n+1)(2n+1)}{6\cdot  n^3}$$
$$\because  1+2^2+3^2+.......   n^2=\dfrac{n(n+1)(1n+1)}{6}$$
$$n\overset{Lt}{\rightarrow}\infty\dfrac{n(n+1)(1n+1)}{6n^3}$$
$$=n\overset{Lt}{\rightarrow}\infty\left [ \dfrac{1}{6} \left (1+ \dfrac{1}{n}\right )\left ( 2+ \dfrac{1}{n}\right ) \right ]$$
$$=\dfrac{2}{6}=\dfrac{1}{3}$$


Mathematics

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