已知数列{an}的前n项和Sn=n²-8n,求数列{|an|}的前n项和Tn

已知数列{an}的前n项和Sn=n²-8n,求数列{|an|}的前n项和Tn
已知数列{an}的前n项和Sn=n²-8n,求数列{|an|}的前n项和Tn

已知数列{an}的前n项和Sn=n²-8n,求数列{|an|}的前n项和Tn
an =Sn -S(n-1)
= 2n-9
an >0
2n-9 >0
n >9/2
for n 5
Tn = -(a1+..+a4) +(a5+a6+..+an)
= 16 + (n-4)(n-4)
= n^2-8n+32

an=Sn-Sn-1=n²-8n-[(n-1)²-8(n-1)]=n²-8n-[n²-2n+1-8n+8]=n²-8n-[n²-10n+9]=2n-9,首项为a1=S1=-7，所以满足此式，当n>=5时，an>0，所以Tn=|S4|+|Sn-S4|=16+n²-8n+16=n²-8n+32(n>4),T1=7,T2=12,T3=15,T4=16