Question

The sum of first $n$ natural numbers is $\dfrac{1}{78}$ times the sum of their cube, then the value of $n$ is:
1. 11
2. 12
3. 13
4. 14

Answer 1

Hint: To solve this question, we need to know the basic formula related to sequence and series. As we know the formula of sum of first $n$ natural numbers are square of $\dfrac{n\left( n+1 \right)}{2}$, we will directly use this as per the question statement, that is the sum of the first $n$ natural numbers is $\dfrac{1}{78}$ times the sum of their cube and then find the value of $n$.

Complete step-by-step solution:
We know that,
The sum of the first $n$ natural numbers = $\dfrac{n\left( n+1 \right)}{2}$.
Also the sum of the cube of $n$ natural numbers = ${{\left[ \dfrac{n\left( n+1 \right)}{2} \right]}^{2}}$.
According to the question,
The sum of first $n$ natural numbers = $\dfrac{1}{78}\times $ (sum of cube $n$ natural numbers).
So, it can be written as,
$\begin{align}
  & \dfrac{n\left( n+1 \right)}{2}=\dfrac{1}{78}\times {{\left[ \dfrac{n\left( n+1 \right)}{2} \right]}^{2}} \\
 & \Rightarrow \dfrac{n\left( n+1 \right)}{2}\times \dfrac{1}{78}=1 \\
\end{align}$
And if we solve this for getting the values of $n$, then,
$\begin{align}
  & \Rightarrow n\left( n+1 \right)=78\times 2 \\
 & \Rightarrow n\left( n+1 \right)=156 \\
 & \Rightarrow {{n}^{2}}+n=156 \\
 & \Rightarrow {{n}^{2}}+n-156=0 \\
\end{align}$
By solving this we will get the value of $n=-13$ and $12$.
So, the value of $n$ will be 12.
Hence the correct answer is option 2.

Note: The sum of cube refers to the sum of the cube of the numbers. It is basically the addition of cubed numbers. The cube term could be 2 term, 3 term or n term or first n even terms or odd terms, set of natural numbers or consecutive numbers etc.