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a2-a1=1,故{an}是公差为1的等差数列.

(2)解由(1)知数列{an}的公差为1.

由Sn=0,Sn+1=4,得an+1=4,即a1+n=4,

由Sn=0,得na1+=0,即a1+=0,

联立解得n=7.

知识点：数列

题型：解答题