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

Question

设Cn=T(2n+1)-Tn.求bn通项以及Cn增减性

Answer 1

解：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)<0
∴Cn严格单调递减

Answer 2

当n≥2且n∈N+时：an=Sn-S（n-1）=n²+1-（n-1）²-1=2n-1
当n=1时：a1=S1=1²+1=2不满足上式
2 ，n=1
∴an={2n-1，n≥2且n∈N+

∴ 2 /2+1=2/3 ，n=1
bn={2/(2n-1+1)=1/n ，n≥2且n∈N+

C（n+1）-C(n)=T(2n+3)-T(n+1)-T(2n+1)+Tn
=T(2n+3)-T(2n+1)-[T(n+1)-Tn]
=b(2n+3)+b(2n+2)-b(n+1)
=1/(2n+3)+1/(2n+2)-1/(n+1)
=(-1)/[2(2n+3)(n+1)]
∵n∈N+
∴C（n+1）-C(n)＜0
∴C（n+1）＜C(n)
∴{Cn}单调递减