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
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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设Cn=T(2n+1)-Tn.求bn通项以及Cn增减性](/uploads/image/z/745645-13-5.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%E8%AE%BECn%3DT%282n%2B1%29-Tn.%E6%B1%82bn%E9%80%9A%E9%A1%B9%E4%BB%A5%E5%8F%8ACn%E5%A2%9E%E5%87%8F%E6%80%A7)
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
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
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
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)
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)...
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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
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