已知数列{an}的前n项和为sn,a1=1,数列｛an+sn｝是公差为2的等差数列

1 数列｛an+sn｝是公差为2的等差数列,数列｛an+sn｝首项a1+s1=2a1=2,数列｛an+sn｝的通项=2+2(n-1)=2n,a2+s2=a2+a2+a1=4,a2=3/2,a3+s3=a3+a3+a2+a1=6,a3=7/4；2.an+sn=2n,a(n-1)+s(n-1)=2n-2,前式减后式得：an-a(n-1...