A) \[x/2\]
B) \[x/3\]
C) \[x/6\]
D) 0
Correct Answer: B
Solution :
| \[\because \]\[\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{{{n}^{3}}}\{[{{1}^{2}}x]+[{{2}^{2}}x]+[{{3}^{2}}x]+...+[{{n}^{2}}x]\}\] |
| \[=\underset{n\to \infty }{\mathop{\lim }}\,\left\{ \frac{\sum\limits_{r=1}^{n}{[{{r}^{2}}x]}}{{{n}^{3}}} \right\}=\underset{n\to \infty }{\mathop{\lim }}\,\left( \frac{\sum\limits_{r=1}^{n}{{{r}^{2}}x-\{{{r}^{2}}x\}}}{{{n}^{3}}} \right)\] |
| \[=\underset{n\to \infty }{\mathop{\lim }}\,\left( \frac{x\frac{n(n+1)(2n+1)}{6}}{{{n}^{3}}}-\sum\limits_{r=1}^{n}{\frac{\{{{r}^{2}}x\}}{{{n}^{3}}}} \right)\] |
| \[=x.\frac{(1)(1)(2)}{6}-0=\frac{x}{3}\] |
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