A) 1/30
B) 0
C) 1/4
D) 1/5
Correct Answer: D
Solution :
\[\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n}\sum\limits_{r=1}^{n}{{{\left( \frac{r}{n} \right)}^{4}}}=\int_{0}^{1}{{{x}^{4}}dx}\] Now, \[\underset{n\to \infty }{\mathop{\lim }}\,=\frac{1+{{2}^{4}}+{{3}^{4}}+....+{{n}^{4}}}{{{n}^{5}}}\] \[-\underset{n\to \infty }{\mathop{\lim }}\,\frac{1+{{2}^{3}}+{{3}^{3}}+.....+{{n}^{3}}}{{{n}^{5}}}\] \[=\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n}\sum\limits_{r=1}^{n}{{{\left( \frac{r}{n} \right)}^{4}}}-\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n}\times \underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n}\sum\limits_{r=1}^{n}{{{\left( \frac{r}{n} \right)}^{3}}}\] \[=\int_{0}^{1}{{{x}^{4}}dx-}\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n}\times \int_{0}^{1}{{{x}^{3}}dx=}\left[ \frac{{{x}^{5}}}{5} \right]_{0}^{1}-0=\frac{1}{5}\]
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