1SnSn1的精選

問題詳情：已知數列{an}滿足a2-a1=1,其前n項和爲Sn,當n≥2時,Sn-1-1,Sn,Sn+1成等差數列.(1)求*:{an}爲等差數列;(2)若Sn=0,Sn+1=4,求n.【回答】(1)*當n≥2時,由Sn-1-1,Sn,Sn+1成等差數列,得2Sn=Sn-1-1+Sn+1,即Sn-Sn-1=-...