Q. 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
Explanation
(b)
Since, $\Sigma n = \frac{1}{78}\Sigma n^{3}$
$\Rightarrow \frac{n(n + 1)}{2} = \frac{1}{78} \times \frac{n^{2}(n + 1)^{2}}{4}$
$\Rightarrow n^{2} + n - 156 = 0$
$\Rightarrow (n + 13)(n - 12) = 0$
$\Rightarrow n = 12\ \ \ \ \ \ \ \ \ \lbrack\because n \neq - 13\rbrack$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.