题目

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. 答案：(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.

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