设数列{an}和{bn}满足a1=b1=6,a2=b2=4,a3=b3=3且数列{a(n+1)-an}是等差数列,数列{bn-2}是等比数列(1)分别求{an}{bn}的通项公式(2)是否存在k∈N*,使bk-ak∈(0,1/2)?若存在,求出k;若不存在,说明理由.
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![设数列{an}和{bn}满足a1=b1=6,a2=b2=4,a3=b3=3且数列{a(n+1)-an}是等差数列,数列{bn-2}是等比数列(1)分别求{an}{bn}的通项公式(2)是否存在k∈N*,使bk-ak∈(0,1/2)?若存在,求出k;若不存在,说明理由.](/uploads/image/z/2427975-63-5.jpg?t=%E8%AE%BE%E6%95%B0%E5%88%97%7Ban%7D%E5%92%8C%7Bbn%7D%E6%BB%A1%E8%B6%B3a1%3Db1%3D6%2Ca2%3Db2%3D4%2Ca3%3Db3%3D3%E4%B8%94%E6%95%B0%E5%88%97%7Ba%28n%2B1%29-an%7D%E6%98%AF%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%2C%E6%95%B0%E5%88%97%7Bbn-2%7D%E6%98%AF%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%281%29%E5%88%86%E5%88%AB%E6%B1%82%7Ban%7D%7Bbn%7D%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%282%29%E6%98%AF%E5%90%A6%E5%AD%98%E5%9C%A8k%E2%88%88N%2A%2C%E4%BD%BFbk-ak%E2%88%88%280%2C1%2F2%29%3F%E8%8B%A5%E5%AD%98%E5%9C%A8%2C%E6%B1%82%E5%87%BAk%EF%BC%9B%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.)
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