a(1)=s(1)=(3-1)/2=1,
a(n+1)=s(n+1)-s(n)=[3(n+1)^2-(n+1)]/2 - [3n^2-n]/2 = (3/2)(2n+1) - 1/2 = 3n + 1,
a(n) = 3(n-1)+1=3n-2.
b(n) = a[2^n] = 3*2^n - 2,
t(n) = b(1)+b(2)+...+b(n) = 3[2+2^2+...+2^n] - 2n = 6[1+2+...+2^(n-1)] -2n = 6[2^n-1]/(2-1) -2n
=6(2^n - 1) - 2n
=3*2^(n+1) - 6 -2n