S<1>=1，S<2>=1+2+3+4，S<3>=1+2+3+…+9， 故S=1+2+…+n²=n²(1+n²)/2 a=S-S=[n²(1+n²)/2]-(n-1)²[1+(n-1)²]/2 =2n³-3n²+3n-1
根据规律,可以得出,第n项共有2n-1项. 所以前面共有1+3+5+…+（2n-3)=(n-1)^2=n^2-2n+1 所以an=(n^2-2n+2)+(n^2-2n+3)+(n^2-2n+4)+……+(n^2-2n+2n) =(2n-1)(n^2-2n+2+n^2)/2=2n^3-3n^2+3n-1.
