Question

Sum of the squares of first $n$ natural numbers exceeds their sum by $330$, then $n$ equals

• A. $8$
• B. $10$
• C. $15$
• D. $20$

Answer 1

The correct option is B

$10$

Explanation for the correct option

Finding the value of $n$

Given,

We know that the sum of the first $n$ natural numbers is $\frac{n\left(n+1\right)}{2}$ and the sum of the squares of the first $n$ natural numbers is$\frac{n\left(n+1\right)\left(2n+1\right)}{6}$.

It is given that the sum of the first $n$ natural numbers exceeds their sum by $330$. So, we must get :

$\frac{n\left(n+1\right)\left(2n+1\right)}{6}-\frac{n\left(n+1\right)}{2}=330$

Solving the equation $\frac{n\left(n+1\right)\left(2n+1\right)}{6}-\frac{n\left(n+1\right)}{2}=330$ to find the value of $n$

We have,

$$\begin{gathered} \frac{n\left(n+1\right)\left(2n+1\right)}{6}-\frac{n\left(n+1\right)}{2}=330 \\ \Rightarrow \frac{n\left(n+1\right)\left(2n+1\right)-3n\left(n+1\right)}{6}=300 \\ \Rightarrow \frac{n\left(n+1\right)\left[\left(2n+1\right)-3\right]}{6}=330 \\ \Rightarrow \frac{n\left(n+1\right)\left(2n-2\right)}{6}=330 \\ \Rightarrow \frac{2n\left(n+1\right)\left(n-1\right)}{6}=330 \\ \Rightarrow \frac{n\left(n+1\right)\left(n-1\right)}{3}=330 \\ \Rightarrow n\left(n+1\right)\left(n-1\right)=330\times 3 \\ \Rightarrow n\left(n+1\right)\left(n-1\right)=\left(3\times 10\times 11\right)\times 3 \\ \Rightarrow n\left(n+1\right)\left(n-1\right)=9\times 10\times 11 \\ \Rightarrow n\left(n+1\right)\left(n-1\right)=10\left(10+1\right)\left(10-1\right) \end{gathered}$$

So, on comparing both sides, we get $n=10$.

Hence, the correct answer is Option (B).