作业帮已知数列{an}与{bn}满足关系:a1=2,a(n+1)=(an^2+1)/2an,bn=(an+1)/(an-1).(n∈N*).(1)求证：数列{lg bn}是等比数列;(2)求证：(an-1)/[a(n+1)-1]=3^[2^(n-1)]+1;(3)设Sn是数列{an}的前n项和,当n≥2时,求证：Sn
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![已知数列{an}与{bn}满足关系:a1=2,a(n+1)=(an^2+1)/2an,bn=(an+1)/(an-1).(n∈N*).(1)求证：数列{lg bn}是等比数列;(2)求证：(an-1)/[a(n+1)-1]=3^[2^(n-1)]+1;(3)设Sn是数列{an}的前n项和,当n≥2时,求证：Sn](/uploads/image/z/10420391-47-1.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%E4%B8%8E%7Bbn%7D%E6%BB%A1%E8%B6%B3%E5%85%B3%E7%B3%BB%3Aa1%3D2%2Ca%28n%2B1%29%3D%28an%5E2%2B1%29%2F2an%2Cbn%3D%28an%2B1%29%2F%28an-1%29.%28n%E2%88%88N%2A%29.%281%29%E6%B1%82%E8%AF%81%EF%BC%9A%E6%95%B0%E5%88%97%7Blg+bn%7D%E6%98%AF%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%3B%282%29%E6%B1%82%E8%AF%81%EF%BC%9A%28an-1%29%2F%5Ba%28n%2B1%29-1%5D%3D3%5E%5B2%5E%28n-1%29%5D%2B1%3B%283%29%E8%AE%BESn%E6%98%AF%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%2C%E5%BD%93n%E2%89%A52%E6%97%B6%2C%E6%B1%82%E8%AF%81%EF%BC%9ASn)
已知数列{an}与{bn}满足关系:a1=2,a(n+1)=(an^2+1)/2an,bn=(an+1)/(an-1).(n∈N*).(1)求证：数列{lg bn}是等比数列;(2)求证：(an-1)/[a(n+1)-1]=3^[2^(n-1)]+1;(3)设Sn是数列{an}的前n项和,当n≥2时,求证：Sn
已知数列{an}与{bn}满足关系:a1=2,a(n+1)=(an^2+1)/2an,bn=(an+1)/(an-1).(n∈N*).
(1)求证：数列{lg bn}是等比数列;
(2)求证：(an-1)/[a(n+1)-1]=3^[2^(n-1)]+1;
(3)设Sn是数列{an}的前n项和,当n≥2时,求证：Sn
已知数列{an}与{bn}满足关系:a1=2,a(n+1)=(an^2+1)/2an,bn=(an+1)/(an-1).(n∈N*).(1)求证：数列{lg bn}是等比数列;(2)求证：(an-1)/[a(n+1)-1]=3^[2^(n-1)]+1;(3)设Sn是数列{an}的前n项和,当n≥2时,求证：Sn
(1) 由bn=(an+1)/(an-1) （1）
得 b(n+1)=[a(n+1)+1]/[a(n+1)-1] （2）
再将a(n+1)=(an^2+1)/2an 代入（2）
化简得 b(n+1)=(an+1)^2/(an-1)^2
故 b(n+1)=bn^2 再对两边取对数 得lgb(n+1)=2lgbn
故数列{lg bn}是首项为lgb1=lg3 公比为2 的等比数列
(2) 由(1)的结论得 bn=3^[2^(n-1)]
(an-1)/[a(n+1)-1]=(an-1)/[(an^2+1)/2an-1]=2+2/(an-1)
而bn=(an+1)/(an-1) =1+2/(an-1)
故：(an-1)/[a(n+1)-1]=bn+1=3^[2^(n-1)]+1
(3) 由bn=(an+1)/(an-1) 得an=(bn+1)/(bn-1)=1+2/(bn-1)
则Sn=a1+a2+……+an=n+Tn(其中Tn是2/(bn-1)的n项和）
故要证Sn