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

a1=S1=2当n≥2时,an=Sn-S(n-1)=n²+1-(n-1)²-1=2n-1则b1=2/3当n≥2时,bn=2/(an+1)=2/(2n-1+1)=1/nTn=2/3+1/2+1/3+……+1/nCn=T(2n+1)-Tn=1/(n+1)+1/(n+2)+……+1/(2n+1)当n=k时,Ck=1/(k+1)+1/(k+2)+……+1/(2k+1)当n=k+1时,C(k+1)=1/(k+2)+1/(k+3)+……+1/(2k+1)+1/(2k+2)+1/(2k+3)C(k+1)-Ck=1/(k+2)+1/(k+3)+……+1/(2k+1)+1/(2k+2)+1/(2k+3)-[1/(k+1)+1/(k+2)+……+1/(2k+1)]=1/(2k+2)+1/(2k+3)-1/(k+1)=1/(2k+3)-1/(2k+2)