Question

If the sum of first $n$ natural numbers is $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.$ times the sum of their squares, then the value of $n$ is:

• A. $5$
• B. $6$
• C. $7$
• D. $8$

Answer 1

The correct option is C

$7$

Explanation for the correct option.

Sum of first $n$ natural numbers $\mathbf{=}\frac{\mathbf{n}\left(n+1\right)}{\mathbf{2}}$

Sum of square of first $n$ natural numbers $\mathbf{=}\frac{\mathbf{n}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}\left(2n+1\right)}{\mathbf{6}}$

As per the given information, $\frac{n\left(n+1\right)}{2}=\frac{1}{5}\times \frac{n\left(n+1\right)\left(2n+1\right)}{6}$

$\begin{array}{rcl}& \Rightarrow & \frac{30n\left(n+1\right)}{2n\left(n+1\right)}=2n+1\\ \Rightarrow 15& =& 2n+1\\ \Rightarrow 2n& =& 14\\ \Rightarrow n& =& 7\end{array}$

Hence, option C is correct.