Find the sum of all possible products of the first `n` natural numbers taken two by two.

Explanation

We find that
`S=1*2+1*3+1*4+"...."+2*3+2*4+"......."+3*4+3*5+"....."(n-1)*n"......(i)"`
`:.[1+2+3+"...."+(n-1)+n]^(2)=1^(2)+2^(2)+3^(2)+"...."+(n-1)^(2)+n^(2) +2[1*2+1*3+1*4+"...."+2*3+2*4+"...."+3*4+3*5+"......."+(n-1)*n]`
`(sumn)^(2)=sumn^(2)+25" " [" from Eq. (i) "]`
` implies S=((sumn)^(2)=sumn^(2))/(2)`
`({(n(n+1))/(2)}^(2)-(n(n+1)(2n+1))/(6))/(2)`
`((n^(2)(n+1)^(2))/(4)-(n(n+1)(2n+1))/(6))/(2)`
,`(n(n+1))/(24)[3n(n+1)-2(2n+1)]`
`(n(n+1)(3n^(2)-n+2))/(24)`
Hence, `S=((n-1)n(n+1)(3n+2))/(24)`