Question

已知数列{an}的前n项和为Sn,满足an+Sn=2n.(1)证明：数列{an,-2}为等比数列,并求出an；（2）设b

Answer 1

an+Sn=2n①

a(n+1)+S(n+1)=2(n+1)②

②-①,得2a(n+1)-an=2

即a(n+1)-2=(an-2)/2

所以{an-2}是以1/2为公比的等比数列

由①有a1=1,a1-2=-1

所以an-2=(a1-2)*(1/2)^(n-1)

an=2-1/2^(n-1)