Question

The mean of thr squares of the first n natural numbers is:​

Answer 1

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Answer 2

Answer:-

$\sf{⟹\frac{(n+1)(2n+1)}{6}✓}$

Explanation:-

➢ The first n natural numbers are 1,2,3,...,n

Sum of squares of n natural numbers is

$1^2+2^2 +3^2 +...+n^2$

We know that sum of squares of first n natural numbers is given by

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

Mean is the average of the values of the data set.

Therefore, mean of the squares of first n natural numbers is

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

$\sf{⟹\frac{n(n+1)(2n+1)}{6}}$

$\sf{⟹\frac{(n+1)(2n+1)}{6}✓}$