设Sn是正项数列{An}的前n项和,且Sn=1／2An^2+1／2An-1,n属于N*

a(1)=s(1)=(1/2)[a(1)]^2 + (1/2)a(1) - 1,
0 = [a(1)]^2 - a(1) - 2 = [a(1)-2][a(1)+1],因a(1)>0,所以,只能 a(1)=2.
s(n)=(1/2)[a(n)]^2 + a(n)/2 - 1,
s(n+1) = (1/2)[a(n+1)]^2 + a(n+1)/2 - 1,
a(n+1)=s(n+1)-s(n)=(1/2)[a(n+1)]^2 + a(n+1)/2 - (1/2)[a(n)]^2 - a(n)/2,
0 = [a(n+1)]^2 - [a(n)]^2 - a(n+1) - a(n) = [a(n+1)+a(n)][a(n+1)-a(n)-1],
因a(n+1)+a(n)>0.
只能,0 = a(n+1)-a(n)-1,
a(n+1)=a(n)+1,
{a(n)}是首项为a(1)=2,公差为1的等差数列.
a(n)=2+(n-1)=n+1.
b(n)=a^2a(n)-3=(n+1)a^2-3=na^2 + a^2-3,
t(n)=(1/2)n(n+1)a^2 + n(a^2-3)