1.
当n=2时,
a2^2=S2+S1=2S1+a2=a2+2
a2^2-a2-2=0
a2=-1或a2=2
又an^2=Sn-S(n-1)≥0
an≥a(n-1)≥a1=1,
所以a2=2
an^2=Sn+S(n-1)
=Sn+[Sn-an]
=2Sn-an
2Sn=an^2+an
2S(n-1)=a(n-1)^2+a(n-1)
相减：
2an=2[Sn-S(n-1)]=an^2+an-a(n-1)^2-a(n-1)
an^2-a(n-1)^2=an+a(n-1)
[an+a(n-1)][an-a(n-1)-1]=0
又an≥a(n-1)≥a1=1
所以an+a(n-1)≠0
所以an-a(n-1)-1=0
所以an=a1+(n-1)=n
2.
bn=(1-an)^2-a(1-an)
=(1-n)^2-a(1-n)
b(n+1)=n^2-a*n
b(n+1)-bn=n^2-a*n-(1-n)^2+a(1-n)
=2(1-a)n+a-1
=2(1-a)(2n-1)＞0
1-a＞0
a＜1