已知数列{an}的前n项和Sn=-an-1/2^n-1+2(n为整数2) (1)令bn=2^n×an,求证数列{bn}是等差数列,并求数并求数列{an}的通项公式2)令Cn=(n+1)|n*an,求Tn=c1+c2+...+cn

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