Q115 / 180 · COMEDK 2014 · Mathematics

If the sum of first $n$ natural numbers is $\frac{1}{78}$ times the sum their cubes, then the value of $n$ is

  • A
    11
  • Correct answer
    B
    12
  • C
    13
  • D
    14

Solution

Since, $\Sigma n=\frac{1}{78} \Sigma n^{3}$ $$ \begin{array}{ll} \Rightarrow & \frac{n(n+1)}{2}=\frac{1}{78} \times \frac{n^{2}(n+1)^{2}}{4} \\ \Rightarrow & n^{2}+n-156=0 \\ \Rightarrow & (n+13)(n-12)=0 \\ \Rightarrow & n=12 \quad \quad[n \neq-13] \end{array} $$

Hence, the correct answer is option: 'B'.

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