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

(1)
(an-1)n^2＋n-Sn=0
an = (Sn -n)/n^2 +1
Sn - S(n-1) = (Sn -n)/n^2 +1
n^2Sn -n^2S(n-1) = Sn -n + n^2
(n^2-1)Sn -n^2S(n-1) =n(n-1)
[(n+1)/n]Sn - [n/(n-1)]S(n-1) =1
{[(n+1)/n]Sn} 是等差数列, d=1
[(n+1)/n]Sn - [2/1)]S1 =n-1
[(n+1)/n]Sn =n+3
Sn = n(n+3)/(n+1)
an = Sn - S(n-1)
=n(n+3)/(n+1) - (n-1)(n+2)/n
= [n^2(n+3)-(n-1)(n+1)(n+2)]/[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/(n+1)

n=1时,S1=a1=2n≥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...