Question

The mean of the following date $1^2,2^2,3^2,....n^2$ is

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

Answer 1

The correct option is A $\frac{(n+1)(2n+1)}{6}$

mean of $1^2,2^2,3^2,....n^2$

$=\frac{1^2+2^2+3^2+........+n^2}{n}$

$=\frac{\sum n^2}{n}$

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

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