关于1SnSn1的百科

• 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...

• 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=-1+Sn+1-Sn,即an=-1+an+1(n≥2),则an+1-an=1(n≥2),又a...
• 2020-06-0132738