\[{{\log }_{e}}2+{{\log }_{e}}\left( 1+\frac{1}{2} \right)+{{\log }_{e}}\left( 1+\frac{1}{3} \right)+....+{{\log }_{e}}\left( 1+\frac{1}{n-1} \right)\] is equal to

\[{{\log }_{e}}2+{{\log }_{e}}\left( 1+\frac{1}{2} \right)+{{\log }_{e

1.	\[{{\log }_{e}}2+{{\log }_{e}}\left( 1+\frac{1}{2} \right)+{{\log }_{e}}\left( 1+\frac{1}{3} \right)+....+{{\log }_{e}}\left( 1+\frac{1}{n-1} \right)\] is equal to
A.	\[{{\log }_{e}}1\]
B.	\[{{\log }_{e}}n\]
C.	\[{{\log }_{e}}(1+n)\]
D.	\[{{\log }_{e}}(1-n)\]
Answer» C. \[{{\log }_{e}}(1+n)\]

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\[{{\log }_{e}}2+{{\log }_{e}}\left( 1+\frac{1}{2} \right)+{{\log }_{e}}\left( 1+\frac{1}{3} \right)+....+{{\log }_{e}}\left( 1+\frac{1}{n-1} \right)\] is equal to

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