已知数列{an}中,a1=1,a(n+1)=2Sn+1(n属于N*),数列{bn}中,b1=1,b4=7,且满足b(n+
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已知数列{an}中,a1=1,a(n+1)=2Sn+1(n属于N*),数列{bn}中,b1=1,b4=7,且满足b(n+2)+bn=2b(n+1),求数列{an}与{bn}的通项公式.设Sn是数列{an·bn}的前n项和,求Sn
![已知数列{an}中,a1=1,a(n+1)=2Sn+1(n属于N*),数列{bn}中,b1=1,b4=7,且满足b(n+](/uploads/image/z/17953554-66-4.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%EF%BD%9Ban%EF%BD%9D%E4%B8%AD%2Ca1%3D1%2Ca%28n%2B1%29%3D2Sn%2B1%28n%E5%B1%9E%E4%BA%8EN%2A%29%2C%E6%95%B0%E5%88%97%EF%BD%9Bbn%EF%BD%9D%E4%B8%AD%2Cb1%3D1%2Cb4%3D7%2C%E4%B8%94%E6%BB%A1%E8%B6%B3b%28n%2B)
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