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  3. CEE Kerala Engineering

CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2004

  • question_answer
    The sum of\[{{15}^{2}}+{{16}^{2}}+{{17}^{2}}+.....+{{30}^{2}}\]is equal to:

    A)  8840                                     

    B)  8440

    C)  8540                     

    D)         8450

    E)  8000

    Correct Answer: B

    Solution :

    \[{{15}^{2}}+{{16}^{2}}+{{17}^{2}}......+{{30}^{2}}\] \[={{1}^{2}}+{{2}^{2}}+...+{{30}^{2}}-({{1}^{2}}+{{2}^{2}}+....+{{14}^{2}})\] \[=\frac{30(31)(61)}{6}-\frac{14(15)(29)}{6}\] \[=\frac{56730-6090}{6}=8440\]


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