题目内容：

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

优质解答

Sn=n^2+1Sn-1=(n-1)^2+1∴an=2n-1bn=2/(2n-1+1)=1/nCn=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/(2...