a(n+1)=2*an+3
即an=2*a(n-1)+3
an+3=2*a(n-1)+6=2[a(n-1)+3]
则{an+3}是公比q=2的等比数列
a1+3=4,
an+3=(a1+3)*q^(n-1)=4*2^(n-1)=2^(n+1)
an=2^(n+1)-3