因为S(n+1)=2S(n)+a(1)所以有a(n+1)=S(n)+a(1)当然有a(n+2)=S(n+1)+a(1)两式子相减可得a(n+2)-a(n+1)=a(n+1)既有a(n+2)=2a(n+1)且有a(2)=2a(1),故a(n)的通项公式为a(n)=a(1)*2^(n-1)所以有S（n）/a(n)=[2^(n-1)]/[(2^n)-1]
