数列{an}的前n项和为Sn,已知Sn+an=-n(n∈N)恒成立.①求数列an的通项公式.②bn=ln(an+1),求{anbn}的前n项和
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![数列{an}的前n项和为Sn,已知Sn+an=-n(n∈N)恒成立.①求数列an的通项公式.②bn=ln(an+1),求{anbn}的前n项和](/uploads/image/z/13743168-24-8.jpg?t=%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E5%B7%B2%E7%9F%A5Sn%2Ban%3D-n%EF%BC%88n%E2%88%88N%EF%BC%89%E6%81%92%E6%88%90%E7%AB%8B.%E2%91%A0%E6%B1%82%E6%95%B0%E5%88%97an%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F.%E2%91%A1bn%3Dln%EF%BC%88an%2B1%EF%BC%89%2C%E6%B1%82%7Banbn%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C)
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