已知数列{an}的前n项和为Sn,且Sn+an=n^2+2n-1对于所有自然数n恒成立1、求证:an-(2n-1)=[a(n-1)-(2n-3)]/2,求数列{an}的通项公式2、若数列{bn}满足:bn=1/a(n+1)an,求b1+b2+…+bn的和

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