设数列｛an｝的前n项和Sn=4/3an-1/3*2n+1+2/3，求通项公式

当n=1时，a1=S1=(4/3)a1-(1/3)*2^(1+1)+2/3=(4/3)a1-2/3，解得:a1=2；
当n>1时：
Sn=(4/3)an-(1/3)*2^(n+1)+2/3=(4/3)an-2*(1/3)*2^n+2/3
S(n-1)=(4/3)a(n-1)-(1/3)*2^n+2/3=(4/3)a(n-1)-1*(1/3)*2^n+2/3
an
=Sn-S(n-1)
=[(4/3)an-2*(1/3)*2^n+2/3]-[(4/3)a(n-1)-1*(1/3)*2^n+2/3]
=(4/3)an-(4/3)a(n-1)-(1/3)*2^n
∴(1/3)an=(4/3)a(n-1)+(1/3)*2^n
即 an=4*a(n-1)+2^n
4*a(n-1)=4^2*a(n-2)+4*2^(n-1)
……
4^(n-2)*a2=4^(n-1)*a1+4^(n-2)*2^2
上述式子相加，得：
an=4^(n-1)*a1+2^n+4*2^(n-1)+…+4^(n-2)*2^2
=2^(2n-2)*2+2^n+2^2*2^(n-1)+…+2^(2n-4)*2^2
=2^(2n-1)+2^n+2^(n+1)+…+2^(2n-2)
=2^(2n-1)+2^n[2^0+2^1+…+2^(n-2)]
=2^(2n-1)+2^n*2^0*[1-2^(n-1)]/(1-2)
=2^(2n-1)+2^n*[2^(n-1)-1]
=2^(2n-1)+2^(2n-1)-2^n
=2^1*2^(2n-1)-2^n
=2^(2n)-2^n
∵a1=2=2^2-2^1，符合上式
∴数列{an}的通项公式是an=2^(2n)-2^n.