11th Class Mathematics Other Series Question Bank Critical Thinking

  • question_answer \[\frac{1}{3}+\frac{1}{2\,.\,{{3}^{2}}}+\frac{1}{3\,.\,{{3}^{3}}}+\frac{1}{4\,.\,{{3}^{4}}}+.....\infty =\]  [MNR 1975]

    A) \[{{\log }_{e}}2-{{\log }_{e}}3\]

    B) \[{{\log }_{e}}3-{{\log }_{e}}2\]

    C) \[{{\log }_{e}}6\]

    D) None of these

    Correct Answer: B

    Solution :

    \[\frac{1}{3}+\frac{1}{{{2.3}^{2}}}+\frac{1}{{{3.3}^{3}}}+\frac{1}{{{4.3}^{4}}}+......\infty \] \[=\left( \frac{1}{3} \right)+\frac{{{(1/3)}^{2}}}{2}+\frac{{{(1/3)}^{3}}}{3}+\frac{{{(1/3)}^{4}}}{4}+......\infty \] \[=-{{\log }_{e}}\left( 1-\frac{1}{3} \right)=-{{\log }_{e}}\left( \frac{2}{3} \right)={{\log }_{e}}\left( \frac{3}{2} \right)\] \[={{\log }_{e}}3-{{\log }_{e}}2\].


You need to login to perform this action.
You will be redirected in 3 sec spinner