Sequence And Series Question 27
Question: $ \frac{2}{1},.,\frac{1}{3}+\frac{3}{2}.\frac{1}{9}+\frac{4}{3}.\frac{1}{27}+\frac{5}{4}.\frac{1}{81}+……\infty = $
Options:
A) $ \frac{1}{2}-{\log_{e}}\frac{2}{3} $
B) $ -{\log_{e}}\frac{2}{3} $
C) $ \frac{1}{2}+{\log_{e}}( \frac{2}{3} ) $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ S=\frac{2}{1}.\frac{1}{3}+\frac{3}{2}.\frac{1}{9}+\frac{4}{3}.\frac{1}{27}+……..+\frac{n+1}{n}.\frac{1}{3^{n}}+……..\infty $ , where $ T_{n}=\frac{n+1}{n}.\frac{1}{3^{n}}=( 1+\frac{1}{n} )\frac{1}{3^{n}}=\frac{1}{3^{n}}+\frac{1}{n{{.3}^{n}}} $
$ \Rightarrow S=\Sigma T_{n}=\Sigma \frac{1}{3^{n}}+\Sigma \frac{1}{n\ .\ 3}n $ $ =\frac{\frac{1}{3}}{1-\frac{1}{3}}+{ -{\log_{e}}( 1-\frac{1}{3} ) }=\frac{1}{2}-{\log_{e}}( \frac{2}{3} ) $ . Trick: As the sum of the series upto 3 or 4 terms is approximately 0.9. Obviously gives the value nearer to 0.9.
BETA
