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