A) \[\frac{1}{2}-{{\log }_{e}}\frac{2}{3}\]
B) \[-{{\log }_{e}}\frac{2}{3}\]
C) \[\frac{1}{2}+{{\log }_{e}}\left( \frac{2}{3} \right)\]
D) None of these
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 \[{{T}_{n}}=\frac{n+1}{n}.\frac{1}{{{3}^{n}}}=\left( 1+\frac{1}{n} \right)\frac{1}{{{3}^{n}}}=\frac{1}{{{3}^{n}}}+\frac{1}{n{{.3}^{n}}}\] \[\Rightarrow S=\Sigma {{T}_{n}}=\Sigma \frac{1}{{{3}^{n}}}+\Sigma \frac{1}{n\ .\ 3}n\] \[=\frac{\frac{1}{3}}{1-\frac{1}{3}}+\left\{ -{{\log }_{e}}\left( 1-\frac{1}{3} \right) \right\}=\frac{1}{2}-{{\log }_{e}}\left( \frac{2}{3} \right)\]. Trick: As the sum of the series upto 3 or 4 terms is approximately 0.9. Obviously gives the value nearer to 0.9.
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