已知等比数列{an}共有n+1项,其首项a1=1,末项a(n+1)=2002,公比q>0 (1) 记Tn=a1a2a3...an,求Tn(2) 记bn=t^log(2002)Tn(t>0且t≠1),求数列{bn}的前n项之和Sn(3) 在(2)的条件下,试比较Sn/bn与S(n+1)/b(n+1)的大小
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![已知等比数列{an}共有n+1项,其首项a1=1,末项a(n+1)=2002,公比q>0 (1) 记Tn=a1a2a3...an,求Tn(2) 记bn=t^log(2002)Tn(t>0且t≠1),求数列{bn}的前n项之和Sn(3) 在(2)的条件下,试比较Sn/bn与S(n+1)/b(n+1)的大小](/uploads/image/z/14792398-70-8.jpg?t=%E5%B7%B2%E7%9F%A5%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%7Ban%7D%E5%85%B1%E6%9C%89n%2B1%E9%A1%B9%2C%E5%85%B6%E9%A6%96%E9%A1%B9a1%3D1%2C%E6%9C%AB%E9%A1%B9a%28n%2B1%29%3D2002%2C%E5%85%AC%E6%AF%94q%EF%BC%9E0+%281%29+%E8%AE%B0Tn%3Da1a2a3...an%2C%E6%B1%82Tn%282%29+%E8%AE%B0bn%3Dt%5Elog%282002%29Tn%28t%EF%BC%9E0%E4%B8%94t%E2%89%A01%29%2C%E6%B1%82%E6%95%B0%E5%88%97%7Bbn%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E4%B9%8B%E5%92%8CSn%283%29+%E5%9C%A8%282%29%E7%9A%84%E6%9D%A1%E4%BB%B6%E4%B8%8B%2C%E8%AF%95%E6%AF%94%E8%BE%83Sn%2Fbn%E4%B8%8ES%28n%2B1%29%2Fb%28n%2B1%29%E7%9A%84%E5%A4%A7%E5%B0%8F)
已知等比数列{an}共有n+1项,其首项a1=1,末项a(n+1)=2002,公比q>0 (1) 记Tn=a1a2a3...an,求Tn(2) 记bn=t^log(2002)Tn(t>0且t≠1),求数列{bn}的前n项之和Sn(3) 在(2)的条件下,试比较Sn/bn与S(n+1)/b(n+1)的大小
已知等比数列{an}共有n+1项,其首项a1=1,末项a(n+1)=2002,公比q>0 (1) 记Tn=a1a2a3...an,求Tn
(2) 记bn=t^log(2002)Tn(t>0且t≠1),求数列{bn}的前n项之和Sn
(3) 在(2)的条件下,试比较Sn/bn与S(n+1)/b(n+1)的大小
已知等比数列{an}共有n+1项,其首项a1=1,末项a(n+1)=2002,公比q>0 (1) 记Tn=a1a2a3...an,求Tn(2) 记bn=t^log(2002)Tn(t>0且t≠1),求数列{bn}的前n项之和Sn(3) 在(2)的条件下,试比较Sn/bn与S(n+1)/b(n+1)的大小
a1=1,a2=q ,an=q^(n-1) a2*an=a(n+1)=2002
Tn=a1a2a3...an=2002^[(n-1)/2]
bn=t^log(2002)Tn=t*(n-1)/2=n*(t/2)-t/2 ,为首项0,公差为t/2的等差数列
Sn=t*(n-1)*n/4
Sn/bn=[t*(n-1)*n/4]/[t*(n-1)/2]=n/2
S(n+1)/b(n+1)=(n+1)/2
Sn/bn<S(n+1)/b(n+1)
。。。。。。
Tn=a1a2a3...an=2002^[(n-1)/2]
bn=t^log(2002)Tn=t*(n-1)/2=n*(t/2)-t/2 ,为首项0,公差为t/2的等差数列
Sn=t*(n-1)*n/4
Sn/bn=[t*(n-1)*n/4]/[t*(n-1)/2]=n/2
S(n+1)/b(n+1)=(n+1)/2
Sn/bn<S(n+1)/b(n+1)
1+q