Sequence And Series Question 67
Question: $ {\log_{e}}2+{\log_{e}}( 1+\frac{1}{2} )+{\log_{e}}( 1+\frac{1}{3} )+….+{\log_{e}}( 1+\frac{1}{n-1} ) $ is equal to
Options:
A) $ {\log_{e}}1 $
B) $ {\log_{e}}n $
C) $ {\log_{e}}(1+n) $
D) $ {\log_{e}}(1-n) $
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Answer:
Correct Answer: B
Solution:
The given series reduces to $ {\log_{e}}2+{\log_{e}}( \frac{3}{2} )+{\log_{e}}( \frac{4}{3} )+….+{\log_{e}}( \frac{n}{n-1} ) $ $ ={\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 $ .
BETA
