Prove that `1^(2)+2^(2)+3^(2)+.....+n^(2)=(n(n+1)(2n+1))/6`

Explanation

`1^(2) = 1`
`2^(2) = 2 + 2`
`3^(2) = 3 + 3 + 3`
`:.` Using weighted means
`(1 + (2 + 2) + (3 + 3 + 3) + ….. + (n + n + ….n "times"))/(1 + 2 + 3 + …..+ n)`
`ge (1^(1). 2^(2)……. N^(n))^((1)/(1 + 2 + 3 + … + n))`
`implies (1 + 2^(2) + 3^(2) + .... + n^(2))/((n(n + 1))/(2)) ge (1^(1) 2^(2)...... n^(n))^((2)/(n(n + 1)))`
`implies (n(n + 1)(2n + 1)/(6))/((n(n + 1))/(2)) ge (1^(1). 2^(2).... n^(n))^(n(n + 1))`
`implies (2n + 1)/(3) ge (1^(1) 2^(2) .... n^(2))^((2)/(n(n + 1))`
`implies 1^(1). 2^(2). 3^(3) .... n^(n) le ((2n + 1)/(3))^((n(n + 1))/(2))`