已知数列﹛an﹜的前n项和为Sn,满足 Sn=2an-n1 求﹛an﹜的通项公式an 2设bn=(2n+1)(an+1)求数列bn的前n项和tn
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![已知数列﹛an﹜的前n项和为Sn,满足 Sn=2an-n1 求﹛an﹜的通项公式an 2设bn=(2n+1)(an+1)求数列bn的前n项和tn](/uploads/image/z/4087699-43-9.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%EF%B9%9Ban%EF%B9%9C%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E6%BB%A1%E8%B6%B3+Sn%3D2an-n1+%E6%B1%82%EF%B9%9Ban%EF%B9%9C%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8Fan+2%E8%AE%BEbn%3D%282n%2B1%29%28an%2B1%29%E6%B1%82%E6%95%B0%E5%88%97bn%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8Ctn)
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