A) \[0\]
B) \[2\]
C) \[\frac{1}{3}\]
D) \[\infty \]
E) \[\frac{1}{4}\]
Correct Answer: E
Solution :
\[\underset{k\to \infty }{\mathop{\lim }}\,\left( \frac{{{1}^{3}}+{{2}^{3}}+{{3}^{3}}+...+{{k}^{3}}}{{{k}^{4}}} \right)\] \[=\underset{k\to \infty }{\mathop{\lim }}\,\left( \frac{{{k}^{2}}{{(k+1)}^{2}}}{4}\times \frac{1}{{{k}^{4}}} \right)\] \[\left\{ \because {{1}^{3}}+{{2}^{3}}+....+{{k}^{3}}={{\left[ \frac{k(k+1)}{2} \right]}^{2}} \right\}\] \[=\underset{x\to \infty }{\mathop{\lim }}\,\left( \frac{{{k}^{4}}{{(1+1/k)}^{2}}}{4}\times \frac{1}{{{k}^{4}}} \right)\] \[=\frac{{{(1+0)}^{2}}}{4}=\frac{1}{4}\]
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