A) \[\frac{1}{3}{{\log }_{e}}3\]
B) \[\frac{1}{3}{{\log }_{e}}2\]
C) \[\frac{1}{3}{{\log }_{e}}\frac{1}{3}\]
D) None of these
Correct Answer: B
Solution :
Let \[S=\underset{n\to \infty }{\mathop{\lim }}\,\,\frac{1}{{{1}^{3}}+{{n}^{3}}}+\frac{4}{{{2}^{3}}+{{n}^{3}}}+...+\frac{1}{2n}\] \[\underset{x\to a}{\mathop{\lim }}\,\,\,\frac{1-\cos \,(a{{x}^{2}}+bx+c)}{{{(x-\alpha )}^{2}}}\,\] \[\therefore \,\,\,S=\underset{n\to \infty }{\mathop{\lim }}\,\,\,\sum\limits_{r=1}^{n}{\,\,\frac{{{r}^{2}}}{{{r}^{3}}+{{n}^{3}}}}\]\[=\underset{n\to \infty }{\mathop{\lim }}\,\,\,\,\sum\limits_{r=1}^{n}{{}}\frac{{{r}^{2}}}{{{n}^{3}}\,\left( \frac{{{r}^{3}}}{{{n}^{3}}}+1 \right)}\] \[=\underset{n\to \infty }{\mathop{\lim }}\,\,\,\,\sum\limits_{r=1}^{n}{{}}\frac{1}{n}.\frac{{{\left( \frac{r}{n} \right)}^{2}}}{\left[ 1+\,{{\left( \frac{r}{n} \right)}^{3}} \right]}\] Applying the formula, we get \[A=\int_{0}^{1}{{}}\frac{{{x}^{2}}}{1+{{x}^{3}}}dx\] \[A=\int_{0}^{1}{{}}\frac{{{x}^{2}}}{1+{{x}^{3}}}dx=\frac{1}{3}\int_{0}^{1}{{}}\frac{3{{x}^{2}}}{1+{{x}^{3}}}dx\]\[=\frac{1}{3}[{{\log }_{e}}(1+{{x}^{3}})]_{0}^{1}=\frac{1}{3}{{\log }_{e}}2.\]
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