1*4,2*7,3*30,……,n(3n+1)的前n项和

∑n(3n+1)=∑(3n^2+n)=∑3n^2+∑n=3∑n^2+n(n+1)/2
=3[n(n+1)(2n+1)/6]+n(n+1)/2
=n(n+1)(2n+1)/2+n(n+1)/2
=n(n+1)(2n+2)/2
=n(n+1)(n+1)
=n(n+1)^2

通项是3n^2+n,看成两个数列分别求和，n^2的前n项和为n（n+1)(2n+1))6,另一个是n（1+n）/2,所以这个数列的前n项和为n（n+1）（2n+1)/2+n（1+n）/2= n(n+1)^2

分开求和，对3n^2和n分别求和，再加起来

An=n(3n+1)=3n^2+n
Sn=n(n+1)(2n+1)/2 +n(n+1)=n^3+2n^2+n