Question

What is the sum of the products of every pair of the first `n` natural numbers?

Answer 1

`[(n+1)n(n-1)(3n+2)]/24`

Explanation:

Let the sum be 'S'
`(1+2+3......+n)^2 = (1^2 + 2^2 +......+n^2) + 2S`
`[n(n+1)/2]^2 = [n(n+1)(2n+1)]/6 + 2S`
After solvinf we will get,
`S=[(n+1)n(n-1)(3n+2)]/24`