设数列{An}的前n项和Sn等于三分之四An减去2的n+1次方的三分之一再加上三分之二.求首项A1与通项An.

a1=S1=(4/3)a1-(1/3)2^2+2/3,a1=2.a(n)=S(n)-S(n-1)==[(4/3)a(n)-(1/3)2^(n+1)+2/3]-[(4/3)a(n-1)-(1/3)2^n+2/3]=(4/3)a(n)-(4/3)a(n-1)-(1/3)2^n.a(n)=4a(n-1)+2^n,a(n)+2^n=4[a(n-1)+2^(n-1)],a(n)+2^n是等比数列,...