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BCECE Engineering BCECE Engineering Solved Paper-2006

  • question_answer
    If \[n=1000!,\]then the value of sum \[\frac{1}{{{\log }_{2}}n}+\frac{1}{{{\log }_{3}}n}+...+\frac{1}{{{\log }_{1000}}n}\]is:

    A)  0                            

    B)         1                            

    C)  10                         

    D)        \[{{10}^{3}}\]

    Correct Answer: B

    Solution :

    Given that \[n=1000!\] Now, \[\frac{1}{{{\log }_{2}}n}+\frac{1}{{{\log }_{3}}n}+...+\frac{1}{{{\log }_{1000}}n}\] \[={{\log }_{n}}2+{{\log }_{n}}3+....+{{\log }_{n}}1000\] \[={{\log }_{n}}2.3.4....1000\] \[={{\log }_{n}}(1000!)=lo{{g}_{n}}n=1\]


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