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(2
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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
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(2](/uploads/image/z/1213931-11-1.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%3Dn%5E2%2B1%2C%E6%95%B0%E5%88%97%7Bbn%7D%E6%BB%A1%E8%B6%B3%EF%BC%9Abn%3D2%2F%28an%2B1%29%2C%E4%B8%94%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BATn%2C%E8%AE%BECn%3DT%282)
Sn=n^2+1
Sn-1=(n-1)^2+1
∴an=2n-1
bn=2/(2n-1+1)=1/n
Cn=bn+1+...+b2n+1=1/(n+1)+1/(n+2)..+1/(2n+1)
Cn+1=1/(n+2)+...+1/(2n+1)+1/(2n+2)+1/(2n+3)
Cn-Cn+1=1/(n+1)-1/(2n+2)-1/(2n+3)=1/(2n+2)-1/(2n+2)+1/(2n+2)-1/(2n+3)=1/(2n+2)(2n+3)>0
∴Cn为递减函数
C2=38/60>16/21>C3=319/420
∴k=3
Sn-1=(n-1)^2+1
∴an=2n-1
bn=2/(2n-1+1)=1/n
Cn=bn+1+...+b2n+1=1/(n+1)+1/(n+2)..+1/(2n+1)
Cn+1=1/(n+2)+...+1/(2n+1)+1/(2n+2)+1/(2n+3)
Cn-Cn+1=1/(n+1)-1/(2n+2)-1/(2n+3)=1/(2n+2)-1/(2n+2)+1/(2n+2)-1/(2n+3)=1/(2n+2)(2n+3)>0
∴Cn为递减函数
C2=38/60>16/21>C3=319/420
∴k=3
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