已知数列an中,sn=n^2-6n,若设Tn=|a1|+|a2|+...+|an|.1.求an的通项公式2.Tn

n=1,a1=s1=1-6=-5
n>=2,an=sn-s(n-1下标）=n^2-6n-[(n-1)^2-6(n-1)]=2n-7
综上an=2n-7
n>=4时,an>0,否则小于0
n=1,Tn=5
n=2,tn=8
n=3,tn=9
n>=4,tn=9+1+3+...+2n-7=9+(n-3)^2