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BCECE Engineering BCECE Engineering Solved Paper-2006

  • question_answer
    If the sum of first \[n\]natural numbers is \[\frac{1}{78}\] times the sum of their cubes, then the value of \[n\] is:

    A)  11                         

    B)         12                         

    C)         13                         

    D)         14

    Correct Answer: B

    Solution :

    Given, the sum of first n natural numbers \[=\frac{1}{78}\] (the sum of their cubes)                 \[\Rightarrow \]               \[\frac{n(n+1)}{2}=\frac{1}{78}\times \frac{{{n}^{2}}{{(n+1)}^{2}}}{4}\] \[\Rightarrow \]               \[156=n(n+1)\] \[\Rightarrow \]               \[{{n}^{2}}+n-156=0\] \[\Rightarrow \]               \[(n+13)(n-12)=0\] \[\Rightarrow \]               \[n=12\]                               \[(\because \,n\ne -13)\]


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