(Ⅰ)由S20=0,S20=20a1+ 20×19 2 d=0,2a1+19d=0,d=2,a1=19,an=a1+(n-1)d=19-2(n-1)=-2n+21,Sn=na1+ n(n?1) 2 d=19n-n(n-1)=-n2+20n,(Ⅱ)bn-an=3n-1,∴bn=3n-1+an=3n-1-2n+21,Tn=Sn+(1+3+32+33+…+3n-1)=-n2+20n+ 3n?1 2 .