Question

Sum of all the priducts of n positive interger taken two at a time

Answer 1

1+2+3+...+n)2=12+22+32+42+...+n2+2S.

But (1+2+3+...+n)2=[n(n+1)2]2.

12+22+32+42+...+n2=n(n+1)(2n+1)6.

So

S=n(n+1)(n−1)(3n+2)24