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) \[{{\log }_{e}}3+{{\log }_{e}}2\]
Correct Answer: B
Solution :
| [b] \[\frac{2}{1}.\frac{1}{3}+\frac{3}{2}.\frac{1}{9}+\frac{4}{3}.\frac{1}{27}+\frac{5}{4}.\frac{1}{81}+....\infty \] |
| \[=(1+1)\frac{1}{3}+\left( 1+\frac{1}{2} \right){{\left( \frac{1}{3} \right)}^{2}}+\left( 1+\frac{1}{3} \right){{\left( \frac{1}{3} \right)}^{3}}\] |
| \[+\left( 1+\frac{1}{4} \right){{\left( \frac{1}{3} \right)}^{4}}+....\infty \] |
| \[=\left\{ \frac{1}{3}+{{\left( \frac{1}{3} \right)}^{2}}+{{\left( \frac{1}{3} \right)}^{3}}+....\infty \right\}+\] |
| \[\left\{ \left( \frac{1}{3} \right)+\frac{1}{2}{{\left( \frac{1}{3} \right)}^{2}}+\frac{1}{3}{{\left( \frac{1}{3} \right)}^{3}}....\infty \right\}\] |
| \[=\frac{\frac{1}{3}}{1-\frac{1}{3}}{{\log }_{e}}\left( 1-\frac{1}{3} \right)=\frac{1}{2}-{{\log }_{e}}\left( \frac{2}{3} \right)\] |
| \[=\frac{1}{2}+{{\log }_{e}}3-{{\log }_{e}}2\] |
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