1*2+2*3+3*4+4*5+5*6+6*7+…+n(n+1)=n(n+1)(n+2)/3

左边=2(2C2+3C2+……+(n+1）C2）
=2（3C3+3C2+4C2……+(n+1）C2)
=2(4C3+4C2……+(n+1）C2)
=2((n+1)C3+(n+1)C2)
=2 (n+2)C3
=n(n+1)(n+2)/3=右边

n(n+1)=1/3[n(n+1)(n+2)-(n-1)n(n+1)]言尽于此.