已知数列{an}的前n项和为Sn,且对任意的n∈N*有an+Sn=n. (1)设bn=an-1,求证:数列{bn}是等比数列; (2)设c1=a1且cn=an-an-1(n≥2),求{cn}的通项公式.

问题描述:

已知数列{an}的前n项和为Sn,且对任意的n∈N*有an+Sn=n.
(1)设bn=an-1,求证:数列{bn}是等比数列;
(2)设c1=a1且cn=an-an-1(n≥2),求{cn}的通项公式.

(1)由a1+S1=1及a1=S1得a1=12.又由an+Sn=n及an+1+Sn+1=n+1,得an+1-an+an+1=1,∴2an+1=an+1.∴2(an+1-1)=an-1,即2bn+1=bn.∴数列{bn}是以b1=a1-1=-12为首项,12为公比的等比数列.(2):由(1)知bn=-12•...