A) \[-\frac{1}{6}\]
B) \[\frac{1}{6}\]
C) \[\frac{1}{3}\]
D) \[-\frac{1}{3}\]
Correct Answer: C
Solution :
\[\underset{n\to \infty }{\mathop{\lim }}\,\,\left[ \frac{n\,(n+1)\,(2n+1)}{6{{n}^{3}}} \right]=\underset{n\to \infty }{\mathop{\lim }}\,\,\frac{\left( 1+\frac{1}{n} \right)\,\left( 2+\frac{1}{n} \right)}{6}=\frac{1}{3}.\] Note : Students should remember that \[\underset{n\to \infty }{\mathop{\lim }}\,\,\frac{\sum n}{{{n}^{2}}}=\frac{1}{2},\,\,\,\underset{n\to \infty }{\mathop{\lim }}\,\,\frac{\sum {{n}^{2}}}{{{n}^{3}}}=\frac{1}{3}\] and \[\underset{n\to \infty }{\mathop{\lim }}\,\,\frac{\sum {{n}^{3}}}{{{n}^{4}}}=\frac{1}{4}.\]
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