What is the average of the squares of the first ten natural numbers?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 38.5
The sum of the squares of the first \(n\) natural numbers is \(\displaystyle \frac{n(n+1)(2n+1)}{6}\); the average is this sum divided by \(n\).
\(\displaystyle 1^{2}+2^{2}+\cdots+10^{2}=\frac{10\times11\times21}{6}=385\)
\(\displaystyle \text{Average}=\frac{385}{10}=38.5\)
The average is 38.5.