数列求和问题，急迫!谢谢

n(3n+1) =3n^2+n

1*4+2*7+3*10+......+n(3n+1)
=(3*1^2+1)+(3*2^2+2)+~~+(3n^2+n)
=(3*1^2+3*2^2+~~+3n^2)+(1+2+3+~+n)
若楼n^2求和公式没错则
原式=3*[n(2n-1)(2n+1)]/6+(1+n)*n/2
=[n(2n-1)(2n+1)/2+(1+n)*n]/2
=[n(2n-1)(2n+1)+(1+n)*n]/2
={n*[(2n-1)(2n+1)+(1+n)]}/2
={n*[(4n^2-1)+(n+1)]}/2
={n*[(4n^2-1)+(n+1)]}/2
=n*(4n^2+n)/2
=(4n^3+n^2)/2
自己检验一下，应该没算错～

an=n(3n+1)=3n平方+n
3n平方求和=3*n(n+1)(2n+1)/6
n求和=n(n+1)/2

不懂可以问，

T=1*4+2*(4+3)+3*(4+6)+......n*(4+(3n-3))
=(1*4+2*4+3*4+......n*4)+1+2*3+3*6+4*9+....n*(3n-3)
=2n^2+2n+1+3(1*2+2*3+3*4+...n*(n-1)
设S=(1*2+2*3+3*4+...n*(n-1))
错位相减 An=Sn-Sn_1=1*2+2*2+3*2+...(n-1)*2=n^2-n
Sn=A1+A2+.....An=(1+2^2+3^2+4^2+.....n^2)-(1+2+3+4+...+n)=根据公式可以算出