Binomial Theorem And Its Simple Applications Question 211
Question: The sum of the series $ \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 $ is equal to
Options:
A) $ {\log_{e}}3-{\log_{e}}2 $
B) $ \frac{1}{2}+{\log_{e}}3-{\log_{e}}2 $
C) $ \frac{1}{2}+{\log_{e}}3+{\log_{e}}2 $
D) $ {\log_{e}}3+{\log_{e}}2 $
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Answer:
Correct Answer: B
Solution:
- [b] $ \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 $
$ =(1+1)\frac{1}{3}+( 1+\frac{1}{2} ){{( \frac{1}{3} )}^{2}}+( 1+\frac{1}{3} ){{( \frac{1}{3} )}^{3}} $
$ +( 1+\frac{1}{4} ){{( \frac{1}{3} )}^{4}}+….\infty $
$ ={ \frac{1}{3}+{{( \frac{1}{3} )}^{2}}+{{( \frac{1}{3} )}^{3}}+….\infty }+ $
$ { ( \frac{1}{3} )+\frac{1}{2}{{( \frac{1}{3} )}^{2}}+\frac{1}{3}{{( \frac{1}{3} )}^{3}}….\infty } $
$ =\frac{\frac{1}{3}}{1-\frac{1}{3}}{\log_{e}}( 1-\frac{1}{3} )=\frac{1}{2}-{\log_{e}}( \frac{2}{3} ) $
$ =\frac{1}{2}+{\log_{e}}3-{\log_{e}}2 $
BETA
