已知数列{an}的前n项和为Sn ,点(n,Sn)均在函数f(x)=-x^2+3x+2的图象上 1求an通项公式 2若数列{bn-an}的首项是1,公比为q（q≠0）的等比数列,求数列{bn}的前n项和Tn

an=Sn - Sn-1 = -n^2+3n+2 -[-(n-1)^2+3(n-1)+2]
=(n-1)^2-n^2+3n-3n-3+2-2
=(n-1+n)(n-1-n)-3
=2n-1-3
=2n-4
设bn-an=q^(n-1),则bn=q^(n-1)-an=q^(n-1)+2n-4
则Tn=q第二问an=Sn - Sn-1 = -n^2+3n+2 -[-(n-1)^2+3(n-1)+2]=(n-1)^2-n^2+3n-3n+3+2-2=(n-1+n)(n-1-n)+3=-2n+1+3=-2n+4设bn-an=q^(n-1),则bn=q^(n-1)-an=q^(n-1)-2n+4则Tn=(1-q^n)/(1-q)-x^2+3x+2