已知数列an的前n项和Sn满足Sn=2an-1求数列{an}的通项公式an,我算到Sn-S(n-1)=(2an-1)-(2a(n-1) - 1)这一我算到Sn-S(n-1)=(2an-1)-(2a(n-1) - 1)这一步就不会了
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![已知数列an的前n项和Sn满足Sn=2an-1求数列{an}的通项公式an,我算到Sn-S(n-1)=(2an-1)-(2a(n-1) - 1)这一我算到Sn-S(n-1)=(2an-1)-(2a(n-1) - 1)这一步就不会了](/uploads/image/z/12781777-49-7.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97an%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8CSn%E6%BB%A1%E8%B6%B3Sn%3D2an-1%E6%B1%82%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8Fan%2C%E6%88%91%E7%AE%97%E5%88%B0Sn-S%28n-1%29%3D%282an-1%29-%282a%28n-1%29+-+1%29%E8%BF%99%E4%B8%80%E6%88%91%E7%AE%97%E5%88%B0Sn-S%28n-1%29%3D%282an-1%29-%282a%28n-1%29+-+1%29%E8%BF%99%E4%B8%80%E6%AD%A5%E5%B0%B1%E4%B8%8D%E4%BC%9A%E4%BA%86)
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