已知数列｛an｝的首项为a1=2,前n项和为.Sn,且满足（an-1）n∧2＋n-Sn=0.(1

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已知数列｛an｝的首项为a1=2,前n项和为.Sn,且满足（an-1）n∧2＋n-Sn=0.(1
已知数列｛an｝的首项为a1=2,前n项和为.Sn,且满足（an-1）n∧2＋n-Sn=0.
(1)证明数列｛((n+1)/n )×Sn}是等差数列,并求数列｛an｝的通项公式.
(2)设bn=an/(n∧2+n+2),记数列bn的前n项和为Tn,证明Tn＜1

n=1时,S1=a1=2
n≥2时,
(an -1)n^2+n-Sn=0
[Sn-S(n-1)-1]n^2+n-Sn=0
(n^2-1)Sn- n^2S(n-1)=n^2-n
(n+1)(n-1)Sn -n^2S(n-1)=n(n-1)
等式两边同除以n(n-1)
[(n+1)/n]Sn -[n/(n-1)]S(n-1)=1,为定值
(2/1)S1=2S1=2×2=4,数列{[(n+1)/n]Sn}是以4为首项,1为公差的等差数列.
[(n+1)/n]Sn=4+1×(n-1)=n+3
Sn=n(n+3)/(n+1)
=n(n+1+2)/(n+1)
=[n(n+1)+2n]/(n+1)
=[n(n+1)+2n+2-2]/(n+1)
=[n(n+1)+2(n+1)-2]/(n+1)
=n+2 -2/(n+1)
n≥2时,an=Sn-S(n-1)=n+2 -2/(n+1)-[(n-1)+2-2/n]=2/n -2/(n+1) +1
n=1时,a1=2/1 -2/2 +1=2,同样满足通项公式
数列{an}的通项公式为an=2/n -2/(n+1) +1
an=2/n -2/(n+1) +1
=[2(n+1)-2n+n(n+1)]/[n(n+1)]
=(2n+2-2n+n^2+n)/[n(n+1)]
=(n^2+n+2)/[n(n+1)]
bn=an/(n^2+n+2)=1/[n(n+1)]=1/n -1/(n+1)
Tn=b1+b2+...+bn
=1/1-1/2+1/2-1/3+...+1/n -1/(n+1)
=1- 1/(n+1)
1/(n+1)>0 1-1/(n+1)

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