A) \[{{\log }_{e}}1\]
B) \[{{\log }_{e}}n\]
C) \[{{\log }_{e}}(1+n)\]
D) \[{{\log }_{e}}(1-n)\]
Correct Answer: B
Solution :
The given series reduces to \[{{\log }_{e}}2+{{\log }_{e}}\left( \frac{3}{2} \right)+{{\log }_{e}}\left( \frac{4}{3} \right)+....+{{\log }_{e}}\left( \frac{n}{n-1} \right)\] \[={{\log }_{e}}2+{{\log }_{e}}3-{{\log }_{e}}2+{{\log }_{e}}4-{{\log }_{e}}3+..\].... \[+{{\log }_{e}}(n)-{{\log }_{e}}(n-1)\]\[={{\log }_{e}}n\].
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