设数列{an}的前n项和Sn=4/3an-{(1/3)*2^n+1}+2/3

由Sn=4/3an-{(1/3)*2^n+1}+2/3知S(n-1)=4/3a(n-1)-{(1/3)*2^(n-1)+1}+2/3两式相减,得到Sn-S(n-1)=4/3(an-a(n-1))-(1/3)*2^n+(1/3)*2^(n-1)即an=4/3(an-a(n-1))-(1/3)*2^(n-1),an-(1/2)*2^n=4a(n-1),改写成an+(1/2)*2...