1.

Find the sum of all possible products of the first `n`natural numbers taken two by two.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» Correct Answer - A
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)`


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