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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2009

  • question_answer
    If \[n=(2020)!\]then\[\frac{1}{{{\log }_{2}}n}+\frac{1}{{{\log }_{3}}n}+\frac{1}{{{\log }_{4}}n}+....+\frac{1}{{{\log }_{2020}}n}\]is equal to

    A)  \[2020\]                             

    B)  \[1\]

    C)  \[(2020)!\]                        

    D)  \[0\]

    Correct Answer: B

    Solution :

    \[\frac{1}{{{\log }_{2}}n}+\frac{1}{{{\log }_{3}}n}+\frac{1}{{{\log }_{4}}n}+....+\frac{1}{\log {{ & }_{2020}}n}\] \[={{\log }_{n}}2+{{\log }_{n}}3+{{\log }_{n}}4+...+{{\log }_{n}}2020\] \[={{\log }_{n}}(2\times 3\times 4\times .....\times 2020)!\]  \[(\because \,n=2020!\,given)\]                 \[=1\]


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