3.(2014·兴化安丰中学检测)已知数列{an}中,a1=3,前n和Sn=(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)设数列的前n项和为Tn,是否存在实数M,使得Tn≤M对一切正整数n都成立?若存在,求M的最小值,若不存在,试说明理由.
[解] (1)Sn=(n+1)(an+1)-1,
Sn+1=(n+2)(an+1+1)-1.
an+1=Sn+1-Sn
=[(n+2)(an+1+1)-(n+1)(an+1)].
整理得,nan+1=(n+1)an-1.
(n+1)an+2=(n+2)an+1-1.
(n+1)an+2-nan+1=(n+2)an+1-(n+1)an.
2(n+1)an+1=(n+1)(an+2+an).
2an+1=an+2+an.
数列{an}为等差数列.
(2)a1=3,nan+1=(n+1)an-1,a2=2a1-1=5.
a2-a1=2,即公差为2.
an=a1+(n-1)d=3+(n-1)·2=2n+1.
(3)==,
Tn=×
=.
又当nN*时,Tn<,
要使得Tn≤M对一切正整数n恒成立,只要M≥,
存在实数M使得Tn≤M对一切正整数n都成立,M的最小值为.