Question

What is the mean of 1square, 2square, 3square........... N square?

Answer 1

The mean of $1^2,2^2,3^2,...,n^2$ is $\frac{(n+1)(2n+1)}{6}$.

Step-by-step explanation:

To find : The mean of $1^2,2^2,3^2,...,n^2$ ?

Solution :

The mean is defined as the sum of the observation divided by number of observation.

$\text{Mean}=\frac{\sum x_n}{n}$

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

We know that,

$\sum 1^2+2^2+3^2+....+n^2 =\frac{n(n+1)(2n+1)}{6}$

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

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

Therefore, The mean of $1^2,2^2,3^2,...,n^2$ is $\frac{(n+1)(2n+1)}{6}$.

#Learn more

Find the mean of first seven natural numbers divisible by 2,3,4​

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