#### MathematicalPhysicist

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**finding the sum of....**

1*2+2*3+3*4+...+n(n+1)

i need to find to what it's equal?

i tried doing various ways:

2*(1+3)+4*(3+5)+6(5+7)+...

which means an even number times the sum of two odd numbers

so let's say:

a=2m

b=2m-1

a(b+3)+a(b+5)+a(b+7)...+n(n+1)

ab*n+a(3+5+7...+2n-1)*(n-1)=

a{[bn+(n-1)(3+5+7+...2n-1)]}

when 3+5+7+...(2n-1)=(3+s_n)n/2

when s_n=3+n(2n-1-3)=2n^2-4n+3

so (3+s_n)n/2=(2n^2-4n+6)n/2=n^3-2n^2+3n

here i pretty much given up, can someone help me.