数列{an}的前n项和记为Sn,已知a1=1,an+1=Sn(n=1,2,3,…),*:(1)数列是等比数列...
问题详情:数列{an}的前n项和记为Sn,已知a1=1,an+1=Sn(n=1,2,3,…),*:(1)数列是等比数列;(2)Sn+1=4an.【回答】*:(1)∵an+1=Sn+1-Sn,an+1=Sn(n=1,2,3,…),∴(n+2)Sn=n(Sn+1-Sn),整理得nSn+1=2(n+1)Sn,∴,即=2,∴数列是等比数列.(2)由(1)知:=4·(n≥2),于是Sn+1=4·(n+1...