数列题一道,已知数列an中,a1=3,前n项和为Sn=1/2(n+1)(an+1)-1.(1)求证:数列an是等差数列

来源：学生作业帮编辑：作业帮分类：数学作业时间：2024/05/21 16:28:19

数列题一道,
已知数列an中,a1=3,前n项和为Sn=1/2(n+1)(an+1)-1.
(1)求证:数列an是等差数列
(2)求数列an的通项公式
(3)设数列2/an*a(n-1)的前n项和为Tn,问是否存在实数M,使得Tn

1
n=(n+1)(an+1)/2-1
S(n-1)=n*[a(n-1)+1]/2-1
an=Sn-S(n-1)=[nan+1+an+n-na(n-1)-n]/2
2an=(n+1)an-n(n-1)+1
(n-1)an=na(n-1)+1
an/n=a(n-1)/(n-1)+1/n*(n-1)=a(n-1)/(n-1)+1/(n-1)-1/n
an/n+1/n=a(n-1)/(n-1)+1/(n-1)
(an+1)/n=[a(n-1)+1]/(n-1)
所以{an+1/n}为常数列
(an+1)/n=(a1+1)/1=a1+1=4
an+1=4n
an=4n-1
所以an为等差数列
3
bn=2/an*a(n-1)=2/(4n-1)*(4n-5)=(1/2)*[1/(4n-5)-1/(4n-1)] (n>1)
Tn=(1/2)*[1/3-1/7+1/7-1/11+..+1/(4n-5)-1/(4n-1)]
=(1/2)*[1/3-1/(4n-1)]
=1/6-1/(8n-2)
n>=2 8n-2>=14 0

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