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If the sum of first $n$ natural numbers is $\frac{1}{78}$ times the sum their cubes, then the value of $n$ is
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Verified Answer
The correct answer is:
12
Since, $\Sigma n=\frac{1}{78} \Sigma n^{3}$
$$
\begin{array}{ll}
\Rightarrow & \frac{n(n+1)}{2}=\frac{1}{78} \times \frac{n^{2}(n+1)^{2}}{4} \\
\Rightarrow & n^{2}+n-156=0 \\
\Rightarrow & (n+13)(n-12)=0 \\
\Rightarrow & n=12 \quad \quad[n \neq-13]
\end{array}
$$
$$
\begin{array}{ll}
\Rightarrow & \frac{n(n+1)}{2}=\frac{1}{78} \times \frac{n^{2}(n+1)^{2}}{4} \\
\Rightarrow & n^{2}+n-156=0 \\
\Rightarrow & (n+13)(n-12)=0 \\
\Rightarrow & n=12 \quad \quad[n \neq-13]
\end{array}
$$
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