1SnSn1的精选

问题详情：已知数列{an}满足a2-a1=1,其前n项和为Sn,当n≥2时,Sn-1-1,Sn,Sn+1成等差数列.(1)求*:{an}为等差数列;(2)若Sn=0,Sn+1=4,求n.【回答】(1)*当n≥2时,由Sn-1-1,Sn,Sn+1成等差数列,得2Sn=Sn-1-1+Sn+1,即Sn-Sn-1=-...