Sum of square of `n` natural numbers

The mean of square of 1st n natural number is

The difference between the sum of squares of first n natural numbers and the sum of first n natural numbers is always divisible by 2.

The sum of squares of first n natural numbers is given by 1/6 n (n+1)(2n+1) or 1/6 (2n^3 + 3n^2 + n) . Find the sum of squares of the first 10 natural numbers.

Statement 1: The variance of first n even natural numbers is (n^2-1)/4 Statement 2: The sum of first n natural numbers is (n(n+1)/2 and the sum of squares of first n natural numbers is (n(n+1)(2n+1)/6 (1) Statement1 is true, Statement2 is true, Statement2 is a correct explanation for statement1 (2) Statement1 is true, Statement2 is true; Statement2 is not a correct explanation for statement1. (3) Statement1 is true, statement2 is false. (4) Statement1 is false, Statement2 is true

The sum of the squares of first n natural numbers is -

Prove by mathematical induction that the sum of the squares of first n natural numbers is 1/6n(n+1)(2n+1) .

Statement - I : The variance of first n even natural numbers is (n^(2) - 1)/(4) Statement - II : The sum of first n natural numbers is (n(n+1))/(2) and the sum of the squares of first n natural numbers is (n(n+1)(2n+1))/(6)

Statement - I : The variance of first n even natural numbers is (n^(2) - 1)/(4) Statement - II : The sum of first n natural numbers is (n(n+1))/(2) and the sum of the squares of first n natural numbers is (n(n+1)(2n+1))/(6)

Find the sum of squares of first 9 natural numbers