What is the average of the cubes of the first nine natural numbers?

The correct answer is Option (4) → 225

The sum of the cubes of the first \(n\) natural numbers is \(\displaystyle \left(\frac{n(n+1)}{2}\right)^{2}\), the square of their ordinary sum. The average is this sum divided by \(n\).

• Sum of the first 9 natural numbers \(\displaystyle =\frac{9\times10}{2}=45\)

• Sum of their cubes \(=45^{2}=2025\)

• Average \(\displaystyle =\frac{2025}{9}=225\)

The average of the cubes is 225.