因为n(n+1)=[n(n+1)(n+2)-(n-1)n(n+1)]/3，所以
1*2+2*3+3*4+......+n(n+1)
=[1*2*3-0+2*3*4-1*2*3+3*4*5-2*3*4+……+n(n+1)(n+2)-(n-1)n(n+1)]/3
=[n(n+1)(n+2)]/3
n(n+1)=n^2+n,分组求和，利用1^2+2^2+。。。。+n^2=n(n+1)(2n+1)/6求解
=2*c(2,2)+2*c(3,2)+……+2*c(n+1,2)
=2*[c(2,2)+c(3,2)+……+c(n+1,2)]
=2*c(n+2,3)
=2*(n+2)(n+1)n/6
=n(n+1)(n+2)/3
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