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

  • question_answer
    \[(1+3){{\log }_{e}}3+\frac{1+{{3}^{2}}}{2\,!}{{({{\log }_{e}}3)}^{2}}+\frac{1+{{3}^{3}}}{3\,!}{{({{\log }_{e}}3)}^{3}}+.....\infty =\] [Roorkee 1989]

    A) 28

    B) 30

    C) 25

    D) 0

    Correct Answer: A

    Solution :

    \[S={{\log }_{e}}3+\frac{{{({{\log }_{e}}3)}^{2}}}{2!}+\frac{{{({{\log }_{e}}3)}^{3}}}{3!}+........+3{{\log }_{e}}3+\frac{{{(3{{\log }_{e}}3)}^{2}}}{2!}+\frac{{{(3{{\log }_{e}}3)}^{3}}}{3!}+....\] \[=({{e}^{{{\log }_{e}}3}}-1)+({{e}^{3{{\log }_{e}}3}}-1)=(3-1)+({{3}^{3}}-1)=28\].


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