2. Solved Papers
  3. CEE Kerala Engineering

CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2006

  • question_answer
    Let\[n=2006!.\]Then \[\frac{1}{{{\log }_{2}}n}+\frac{1}{{{\log }_{3}}n}+...+\frac{1}{{{\log }_{2006}}n}\]is equal to:

    A)  2006                                     

    B)  2005

    C)  2005!                   

    D)         1

    E)  0

    Correct Answer: D

    Solution :

    Given that, \[n=2006!\] \[\therefore \] \[\frac{1}{{{\log }_{2}}n}+\frac{1}{{{\log }_{3}}n}+.....+\frac{1}{{{\log }_{2006}}n}\] \[={{\log }_{n}}2+{{\log }_{n}}3+.....+{{\log }_{n}}2006\] \[={{\log }_{n}}(2.3.4......2006)\] \[={{\log }_{n}}(2006!)={{\log }_{n}}n=1\]


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