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设Cn=T(2n+_数学解答

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
设Cn=T(2n+1)-Tn.求bn通项以及Cn增减性

人气：458 ℃　时间：2019-08-19 08:29:32

解答

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/n
Tn=2/3+1/2+1/3+……+1/n
Cn=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)