Question

The arithmetic mean of the squares of the first $n$ natural numbers is

• A. $\frac{n(n+1)(2n+1)}{6}$
• B. $\frac{n(n+1)(2n+1)}{2}$
• C. $\frac{(n+1)(2n+1)}{6}$
• D. $\frac{(n+1)(2n+1)}{3}$

Answer 1

Answer: (C)

$\frac{(n+1)(2n+1)}{6}$

Sum of squares of the first $n$ natural numbers $=\frac{n(n+1)(2n+1)}{6}$

Their arithmetic mean is $\frac{Sum}{n}=\frac{n(n+1)(2n+1)}{6n}=\frac{(n+1)(2n+1)}{6}$

Hence, C is correct.