A) \[\frac{{{({{\log }_{e}}2)}^{2}}}{4}\]
B) \[\frac{{{({{\log }_{e}}2)}^{2}}}{3}\]
C) \[\frac{{{({{\log }_{e}}2)}^{2}}}{2}\]
D) \[{{({{\log }_{e}}2)}^{2}}\]
Correct Answer: C
Solution :
[c] \[\underset{k=1}{\mathop{\overset{n}{\mathop{\Sigma }}\,}}\,\frac{1}{n\left( 1+\frac{k}{n} \right)}.{{\log }_{e}}\left( 1+\frac{k}{n} \right)=\int\limits_{0}^{1}{\frac{{{\log }_{e}}(1+x)}{(1+x)}}dx\] \[=\int\limits_{1}^{2}{\frac{{{\log }_{e}}t}{t}}dt=\left[ \frac{{{({{\log }_{e}}t)}^{2}}}{2} \right]_{1}^{2}=\frac{{{({{\log }_{e}}2)}^{2}}}{2}\]
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