已知正项数列{an}的前n项和为Sn,且满足Sn+an=1.
(I) 求数列{an}的通项公式;
(Ⅱ)设数列{bn}满足bn=(n-2)an,且数列{bn}的前n项和为Tn,求证:数列{2nTn}为等差数列.