Question

the mean of the squares of the numbers 0,1,2,3,........,n is?

Answer 1

mean = sum of observations/number of observations
mean = sum of squares till n/n. (n = number of observations)
mean = (n(n+1)(2n+1)/6)/n
mean of 1st n squares = (n+1) (2n+1) /6

Answer 2

Answer:

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

Step-by-step explanation:

Formula for mean:

$Mean=\frac{\sum x}{n}$

where, x is observations and n is number of observations.

The given numbers are 0,1,2,3,........,n.

We need to find the mean of the squares of the numbers 0,1,2,3,........,n.

Sum of squares of the numbers 0,1,2,3,........,n is

$\sum x^2=0^2+1^2+2^2+3^2+...+n^2$

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

The mean of the squares of the numbers 0,1,2,3,........,n.

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

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

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

Cancel out the common factors.

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

Therefore, the mean of the squares of the numbers 0,1,2,3,........,n is $\frac{(n+1)(2n+1)}{6}$.