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(C) We know that the sum of the squares of the first $n$ natural numbers is given by $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$.Substituting this i

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$\frac{1}{3}$

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(C) We know that the sum of the squares of the first $n$ natural numbers is given by $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$.Substituting this into the limit expression:$\mathop {\lim }\limits_{n 	o \infty } \left[ {\frac{{n(n + 1)(2n + 1)}}{{6{n^3}}}} ight]$$= \mathop {\lim }\limits_{n 	o \infty } \frac{n}{n} \cdot \frac{(n + 1)}{n} \cdot \frac{(2n + 1)}{n} \cdot \frac{1}{6}$$= \mathop {\lim }\limits_{n 	o \infty } \left( 1 ight) \left( 1 + \frac{1}{n} ight) \left( 2 + \frac{1}{n} ight) \cdot \frac{1}{6}$$= 1 \cdot (1 + 0) \cdot (2 + 0) \cdot \frac{1}{6} = \frac{2}{6} = \frac{1}{3}$.Thus,the correct option is $C$.

$\mathop {\lim }\limits_{n \to \infty } \left[ {\frac{{\sum_{k=1}^{n} {k^2}}}{{{n^3}}}} \right] = $

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