解(1)an=sn-s(n-1)=(n²+n)-((n-1)²+(n-1))=2n(n>=2),
当n=1时,a1=s1=2,也满足上式,所以an=2n
(2)Bn=(1/2)^(2n)+n
Tn=(1/2)^2+(1/2)^4+(1/2)^6+……+(1/2)^(2n)+
(1+2+3+…..+n)
=[1/4*(1-(1/4)^n)]/(1-1/4)+n(1+n)/2
=1/3-1/3*(1/4)^n+n(1+n)/2