数列{an+Sn}是公差为2的等差数列∴an+Sn=a1+s1+2(n-1)=1+1+2n-2=2n∵当n=2时,a2+a2+a1=4∴a2=3/2(2)当n>=2时由an+Sn=2n得a(n-1)+S(n-1)=2(n-1)两式相减得an-a(n-1)+an=22an-4=a(n-1)-22(an-2)=a(n-1)-2∴{an-2}是首...