A) \[1\]
B) 2/3
C) 1/3
D) \[0\]
Correct Answer: C
Solution :
Given limit \[=\underset{n\to \infty }{\mathop{\lim }}\,\,\,\frac{{{1}^{2}}+{{2}^{2}}+{{3}^{2}}+.....+{{n}^{2}}}{1+{{n}^{3}}}=\underset{n\to \infty }{\mathop{\lim }}\,\,\,\frac{\Sigma {{n}^{2}}}{1+{{n}^{3}}}\] \[=\underset{n\to \infty }{\mathop{\lim }}\,\,\,\frac{1}{6}\frac{n\,(n+1)\,(2n+1)}{1+{{n}^{3}}}\]\[=\underset{n\to \infty }{\mathop{\lim }}\,\,\frac{1}{6}\frac{\left( 1+\frac{1}{n} \right)\,\left( 2+\frac{1}{n} \right)}{\left( \frac{1}{{{n}^{3}}}+1 \right)}\] \[=\frac{1}{6}\,.\,1\,.\,\frac{2}{(1)}=\left( \frac{1}{3} \right).\]
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