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BCECE Engineering BCECE Engineering Solved Paper-2005

  • question_answer
    The series \[(1+3){{\log }_{e}}3+\frac{1+{{3}^{2}}}{2!}{{({{\log }_{e}}3)}^{2}}\]\[+\frac{1+{{3}^{2}}}{3!}{{({{\log }_{e}}3)}^{2}}+...\]is equal to:

    A)  28         

    B)                         30                         

    C)         25                         

    D)         0

    Correct Answer: A

    Solution :

    Let \[S={{\log }_{e}}3+\frac{{{(lo{{g}_{e}}3)}^{2}}}{2!}+\frac{{{(lo{{g}_{e}}3)}^{3}}}{3!}+...\] \[+\,3\,{{\log }_{e}}3+\frac{{{(3{{\log }_{e}}3)}^{2}}}{2!}+...\] \[=({{e}^{{{\log }_{e}}3}}-1)+({{e}^{3{{\log }_{e}}3}}-1)\] \[=(3-1)+({{3}^{3}}-1)\] \[=2+26=28\]


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