已知数列{an}的前n项和为Sn=n^2+1,数列{bn}满足:bn=2/(an+1),且前n项和为Tn,设Cn=T(2n+1)-Tn.若对n>=k时,总有Cn

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已知数列{an}的前n项和为Sn=n^2+1,数列{bn}满足:bn=2/(an+1),且前n项和为Tn,设Cn=T(2n+1)-Tn.若对n>=k时,总有Cn
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已知数列{an}的前n项和为Sn=n^2+1,数列{bn}满足:bn=2/(an+1),且前n项和为Tn,设Cn=T(2n+1)-Tn.若对n>=k时,总有Cn
已知数列{an}的前n项和为Sn=n^2+1,数列{bn}满足:bn=2/(an+1),且前n项和为Tn,设Cn=T(2n+1)-Tn.
若对n>=k时,总有Cn

已知数列{an}的前n项和为Sn=n^2+1,数列{bn}满足:bn=2/(an+1),且前n项和为Tn,设Cn=T(2n+1)-Tn.若对n>=k时,总有Cn
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