设数列an前n项和为sn,对任意正整数nh,都有an=5sn+1,记bn=(4+an)/(1-an),(1)求an与bn的通项公式;(2)设bn前n项和为Rn,是否存在正数k,使得Rn>=4k成立?若存在,找出一个正整数k,若不存在,说明理由;(3)记cn=b(2n)
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![设数列an前n项和为sn,对任意正整数nh,都有an=5sn+1,记bn=(4+an)/(1-an),(1)求an与bn的通项公式;(2)设bn前n项和为Rn,是否存在正数k,使得Rn>=4k成立?若存在,找出一个正整数k,若不存在,说明理由;(3)记cn=b(2n)](/uploads/image/z/9637513-25-3.jpg?t=%E8%AE%BE%E6%95%B0%E5%88%97an%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BAsn%2C%E5%AF%B9%E4%BB%BB%E6%84%8F%E6%AD%A3%E6%95%B4%E6%95%B0nh%2C%E9%83%BD%E6%9C%89an%3D5sn%2B1%2C%E8%AE%B0bn%3D%284%2Ban%29%2F%281-an%29%2C%281%29%E6%B1%82an%E4%B8%8Ebn%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%3B%282%29%E8%AE%BEbn%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BARn%2C%E6%98%AF%E5%90%A6%E5%AD%98%E5%9C%A8%E6%AD%A3%E6%95%B0k%2C%E4%BD%BF%E5%BE%97Rn%3E%3D4k%E6%88%90%E7%AB%8B%3F%E8%8B%A5%E5%AD%98%E5%9C%A8%2C%E6%89%BE%E5%87%BA%E4%B8%80%E4%B8%AA%E6%AD%A3%E6%95%B4%E6%95%B0k%2C%E8%8B%A5%E4%B8%8D%E5%AD%98%E5%9C%A8%2C%E8%AF%B4%E6%98%8E%E7%90%86%E7%94%B1%EF%BC%9B%283%29%E8%AE%B0cn%3Db%282n%29)
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