已知a1+2a2+3a3+……+nan=n(n+1)(n+2) (n属于自然数)，求Sn

a1+2a2+3a3+……+nan=n(n+1)(n+2)......1
还可以推出a1+2a2+3a3+……+（n-1)an-1=(n-1)n(n+1).......2
将1式减2式得，nan=n(n+1)(n+2)-(n-1)n(n+1)=n(n+1)(n+2-n+1)=3n(n+1),an=3(n+1)
sn=3*2+3*3+.....+3*(n+1)=3*[2+3+4....+(n+1)]=3*(2+n+1)*n/2=(3n*n+9n)/2

有两种方法：一是an=n^3+3n^2+2n,再分别求和
二是an=n(n+1)(n+2)=[n(n+1)(n+2)(n+3)-(n-1)n(n+1)(n+2)]/4
再用裂项消项法求和就得Sn=n(n+1)(n+2)(n+3)/4