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JEE Main & Advanced Mathematics Sequence & Series Question Bank Logarithmic series

  • question_answer
    If \[b=a-\frac{{{a}^{2}}}{2}+\frac{{{a}^{3}}}{3}-\frac{{{a}^{4}}}{4}+..\]then \[b+\frac{{{b}^{2}}}{2\,!}+\frac{{{b}^{3}}}{3\,!}+\frac{{{b}^{4}}}{4\,!}+...\infty =\]

    A) \[{{\log }_{e}}a\]

    B) \[{{\log }_{e}}b\]

    C) \[a\]

    D) \[{{e}^{a}}\]

    Correct Answer: C

    Solution :

    Given \[b={{\log }_{e}}(1+a)\Rightarrow 1+a={{e}^{b}}\] \[\Rightarrow \,\,\,1+a=1+\frac{b}{1!}+\frac{{{b}^{2}}}{2!}+....\]Þ \[a=b+\frac{{{b}^{2}}}{2\,!}+.....\]


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