由a1=31
an=an-1-2
得an=33-2n (n≥2)
则bn=|an|=|33-2n|,分段讨论:
当 2≤n≤16时,bn=33-2n
当n≥17时,bn=2n-33
则Tn=16*(33+1)/2 +(n-16)(1+2n-33)/2
=272+(n-16)^2=n^2-32n+528