Sequence And Series Question 216
Question: If $ S=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{3.4}-\frac{1}{4.5}+….+\infty , $ then $ e^{S}= $
[MP PET 1999]
Options:
A) $ {\log_{e}}( \frac{4}{e} ) $
B) $ \frac{4}{e} $
C) $ {\log_{e}}( \frac{e}{4} ) $
D) $ \frac{e}{4} $
Show Answer
Answer:
Correct Answer: B
Solution:
$ S=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{3.4}-\frac{1}{4.5}+…..\infty $ $ =1-\frac{1}{2}-\frac{1}{2}+\frac{1}{3}+\frac{1}{3}-\frac{1}{4}-\frac{1}{4}+\frac{1}{5}+…..\infty $ $ =1+2[ -\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}+…..\infty ] $ $ =1+2[{\log_{e}}2-1]=1+2{\log_{e}}2-2 $ $ ={\log_{e}}4-1={\log_{e}}4-{\log_{e}}e $ $ ={\log_{e}}( \frac{4}{e} ) $ ;
$ \therefore e^{S}=\frac{4}{e} $ .
BETA
