已知等差数列{an}的前n项和为Sn,公差d≠0,a1=1,且a1,a2,a7成等比数列(1)求数列{an}的前n项和Sn(2)设bn=2Sn/(2n-1),数列{bn}的前n项和为Tn,求证:2Tn-9b(n-1)+18>64bn/(n+9)b(n+1)注:(n+1)、(n-1)为下标
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![已知等差数列{an}的前n项和为Sn,公差d≠0,a1=1,且a1,a2,a7成等比数列(1)求数列{an}的前n项和Sn(2)设bn=2Sn/(2n-1),数列{bn}的前n项和为Tn,求证:2Tn-9b(n-1)+18>64bn/(n+9)b(n+1)注:(n+1)、(n-1)为下标](/uploads/image/z/1998964-28-4.jpg?t=%E5%B7%B2%E7%9F%A5%E7%AD%89%E5%B7%AE%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%85%AC%E5%B7%AEd%E2%89%A00%2Ca1%3D1%2C%E4%B8%94a1%2Ca2%2Ca7%E6%88%90%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%EF%BC%881%EF%BC%89%E6%B1%82%E6%95%B0%E5%88%97%EF%BD%9Ban%EF%BD%9D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8CSn%EF%BC%882%EF%BC%89%E8%AE%BEbn%3D2Sn%2F%282n-1%29%2C%E6%95%B0%E5%88%97%EF%BD%9Bbn%EF%BD%9D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BATn%2C%E6%B1%82%E8%AF%81%EF%BC%9A2Tn-9b%28n-1%29%2B18%3E64bn%2F%28n%2B9%29b%28n%2B1%29%E6%B3%A8%EF%BC%9A%EF%BC%88n%2B1%29%E3%80%81%28n-1%29%E4%B8%BA%E4%B8%8B%E6%A0%87)
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已知等差数列{an}的前n项和为Sn,公差d≠0,a1=1,且a1,a2,a7成等比数列(1)求数列{an}的前n项和Sn(2)设bn=2Sn/(2n-1),数列{bn}的前n项和为Tn,求证:2Tn-9b(n-1)+18>64bn/(n+9)b(n+1)注:(n+1)、(n-1)为下标
已知等差数列{an}的前n项和为Sn,公差d≠0,a1=1,且a1,a2,a7成等比数列
(1)求数列{an}的前n项和Sn
(2)设bn=2Sn/(2n-1),数列{bn}的前n项和为Tn,求证:2Tn-9b(n-1)+18>64bn/(n+9)b(n+1)
注:(n+1)、(n-1)为下标
a(1) a(2) a(7)
OK?
请严肃对待,谢谢合作
已知等差数列{an}的前n项和为Sn,公差d≠0,a1=1,且a1,a2,a7成等比数列(1)求数列{an}的前n项和Sn(2)设bn=2Sn/(2n-1),数列{bn}的前n项和为Tn,求证:2Tn-9b(n-1)+18>64bn/(n+9)b(n+1)注:(n+1)、(n-1)为下标
d=q-1
q^2=1+6d
所以q^2=6q-5
(q-3)^2=1
q-3=1
q=4
所以d=3
所以a1=1,a2=4,a3=7.a7=16
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