设第n个数为an，则a1=1a2-a1=2a3-a2=3a4-a3=4…an-an-1=n叠加可得，an-a1=2+3+4+…+n∴an=1+2+3+…+n=n(n+1)2=12(n2+n)∴Sn=a1+a2+…+an=12(12+1+22+2+…+n2+n)=12[(12+22+…+n2)+(1+2+3+…+n)]=12×16n(n+1)(2n+1)+...