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

Options

A. `(1)/(24)n(n+1)(n-1)(3n+2)`
B. `(n(n+1)(2n+1))/(6)`
C. `(n(n+1)(n-1)(2n+3))/(24)`
D. none of these

Answer

A

Explanation

We have,
`(a_(1)+a_(2)+ . . . . .+a_(n))^(2)=underset(i=1)overset(n)suma_(i)^(2)+2underset(iltj)suma_(i)a_(j)`
Let S be the required sum. Putting `a_(1)=1,a_(2)=2, . . . . ,a_(n)=n`, we get
`{(n(n+1))/(2)}^(2)=(n(n+1)(2n+1))/(6)+2S`
`rArr" "S=(1)/(24)n(n+1)(n-1)(3n+2)`