s(n+1)=4a(n)+3,s(2)=a(1)+a(2)=4a(1)+3,a(2)=3a(1)+3=6.
s(n+2)=4a(n+1)+3,
a(n+2)=s(n+2)-s(n+1)=4a(n+1)-4a(n),
a(n+2)-2a(n+1)=2[a(n+1)-2a(n)],
{a(n+1)-2a(n)}是首项为a(2)-2a(1)=6-2=4,公比为2的等比数列.
a(n+1)-2a(n)=4*2^(n-1).
a(n+1)/2^(n+1)-a(n)/2^n = 1 = c(n+1)-c(n),
{c(n)}是首项为a(1)/2=1/2,公差为1的等差数列.
c(n)=1/2+(n-1)=n-1/2
Tn=(c1+cn)n/2=(1/2+n-1/2)n/2=n^2/2