(1)求常数p的值;
(2)证明数列{an}是等差数列.
(1)求常数p的值;
(2)证明数列{an}是等差数列.
(1)解
:当n=1时,a1=pa1,若p=1,则a1+a2=2a2,∴a1=a2.与已知矛盾,∴p≠1,则a1=0.
当n=2时,a1+a2=2pa2,∴(2p-1)a2=0.
∵a1≠a2,故p=
.
(2)证明：
由(1)知Sn=
nan,a1=0.当n≥2时,an=Sn-Sn-1=
nan-
(n-1)an-1.
∴
=
.
则
=
,…,
=
.
∴
=n-1.
∴an=(n-1)a2.
故{an}是以0为首项以a2为公差的等差数列.