S(n+1)=2Sn+n+1,故
Sn=2S(n-1)+(n-1)+1,两式相减,得
a(n+1)=2an+1,两边同时加1,得
a(n+1)+1=2(an+1),即
(a(n+1))/(an+1)=2,又a1+1=a+1,故
an+1是以a+1为首项,2为公差的等比数列,故
an+1=(a+1)*2^(n-1),
an=(a+1)*2^(n-1)-1