Sequence And Series Question 551
Question: $ \frac{1}{3}+\frac{1}{2,.,3^{2}}+\frac{1}{3,.,3^{3}}+\frac{1}{4,.,3^{4}}+…..\infty = $
[MNR 1975]
Options:
A) $ {\log_{e}}2-{\log_{e}}3 $
B) $ {\log_{e}}3-{\log_{e}}2 $
C) $ {\log_{e}}6 $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ \frac{1}{3}+\frac{1}{{{2.3}^{2}}}+\frac{1}{{{3.3}^{3}}}+\frac{1}{{{4.3}^{4}}}+……\infty $ $ =( \frac{1}{3} )+\frac{{{(1/3)}^{2}}}{2}+\frac{{{(1/3)}^{3}}}{3}+\frac{{{(1/3)}^{4}}}{4}+……\infty $ $ =-{\log_{e}}( 1-\frac{1}{3} )=-{\log_{e}}( \frac{2}{3} )={\log_{e}}( \frac{3}{2} ) $ $ ={\log_{e}}3-{\log_{e}}2 $ .
BETA
