(1) Sn=23n-n平方
当n=1时,a1=S1=23*1-1^2=22
当n>1时,an=Sn-S(n-1)=(23n-n^2)-[23(n-1)-(n-1)^2]=24-2n (显然n=1代入an=22)
于是合二为一,｛an}的通项公式是an=24-2n
(2) bn=|an|=|24-2n|
当1≤n≤12时,bn=24-2n,{bn}的前n项和Tn=24n-n(n+1)=23n-n^2；
当n>12时,{bn}的前n项和Tn=T12+(Tn-T12)=23*12-12^2+(Tn-T12)=132+(Tn-T12)
=132+[(2*13-24)+(2*14-24)+...+(2n-24)]
=132+(n-12)(n+13)-24(n-12)
=132+(n-12)(n-11)
=n^2-23n+264