`1+2+... +n=sum_{i=1}^{n}{:i:} =frac(n(n+1))(2) `

`(1+2+... +n)^2=(sum_{i=1}^{n}{:i:})^2 =frac(n^2(n+1)^2)(4) `

`1^2+2^2+... +n^2=sum_{i=1}^{n}{:i^2:} =frac(n(n+1)(2n+1))(6) `

`frac(n^2(n+1)^2)(4) -frac(n(n+1)(2n+1))(6) =frac(n(n^2-1)(3n+2))(12)`