21.D1、D3、E1、M3[2012·重庆卷] 设数列{an}的前n项和Sn满足Sn+1=a2Sn+a1,其中a2≠0.
(1)求证:{an}是首项为1的等比数列;
(2)若a2>-1,求证:Sn≤(a1+an),并给出等号成立的充要条件.
21.解:(1)证法一:由S2=a2S1+a1得a1+a2=a2a1+a1,即a2=a2a1.
因a2≠0,故a1=1,得=a2.
又由题设条件知
Sn+2=a2Sn+1+a1,Sn+1=a2Sn+a1,
两式相减得Sn+2-Sn+1=a2(Sn+1-Sn),
即an+2=a2an+1,
由a2≠0,知an+1≠0,因此=a2.