# Please prove this formula?

**Question**

Prove

1*2+2*3+3*4+4*5+…+n*(n+1)

=1/3*n*(n+1)*(n+2)

Don’t use induction.

I need to know how someone wrote that formula.

**Answer**

This is “deduction”

1*2+2*3+3*4+4*5+…+n*(n+1)=?

Let S = 1*2+2*3+3*4+4*5+…+n*(n+1) … ( 1)

Use.

1 + 2 + … + n = n*( n + 1 ) / 2

and

1^2 + 2^2 + … + n^2 = [ n*(n+1)*(2n+1)/6 ]

So

S = Sum [ k =1 to k =n for k *( k + 1 ) ]

= Sum [ k =1 to k =n for k^2 ] + Sum [ k = 1 to k = n for k ]

= [1^2 + 2^2 + … + n^2 ] + [ 1 + 2 + … + n ]

= [ n*(n+1)*(2n+1)/6 ] + [ n*( n + 1 ) / 2 ] by employing the “formulas above ]

= [ n*(n + 1/2 ) ] * [ (2n + 1) / 3 + 1 ]

= [ n*(n + 1) /2 ] * [ (2n + 4) / 3 ]

= [ n*(n + 1) ] * [ (n + 2)*2 / 6 ]

= [ n*(n + 1 )*(n + 2) ] / 3