a(n+1)=Sn+2的（n+1）次方an=S(n-1)+2的n次方a(n+1)-an=sn-s(n-1)+2^na(n+1)-an=an+2^na(n+1)=2an+2^n （同除2^（n+1）得a（n+1）/2^（n+1）=an/2^n+1/2令cn=an/2^n b1=1/2则c(n+1)=cn+1/2故cn=c1+(n-1)d=1/2*n故a...