数列{an}的前n项和为Sn=npan(n∈N*)且a1≠a2, (1)求常数p的值; (2)证明:数列{an}是等差数列.

问题描述:

数列{an}的前n项和为Sn=npan(n∈N*)且a1≠a2
(1)求常数p的值;
(2)证明:数列{an}是等差数列.

(1)当n=1时,a1=pa1,若p=1时,a1+a2=2pa2=2a2,∴a1=a2,与已知矛盾,故p≠1.则a1=0.当n=2时,a1+a2=2pa2,∴(2p-1)a2=0.∵a1≠a2,故p=12.(2)由已知Sn=12nan,a1=0.n≥2时,an=Sn-Sn-1=12nan-12(n-1)...