数列{an}的前n项和为Sn=npan(n∈N*)且a1≠a2,
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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)求常数p的值;
(2)证明:数列{an}是等差数列.
![数列{an}的前n项和为Sn=npan(n∈N*)且a1≠a2,](/uploads/image/z/15641693-53-3.jpg?t=%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%3Dnpan%EF%BC%88n%E2%88%88N%2A%EF%BC%89%E4%B8%94a1%E2%89%A0a2%EF%BC%8C)
(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=
1
2.
(2)由已知Sn=
1
2nan,a1=0.
n≥2时,an=Sn-Sn-1=
1
2nan-
1
2(n-1)an-1.
∴
an
an−1=
n−1
n−2.则
an−1
an−2=
n−2
n−3,
a3
a2=
2
1.
∴
an
a2=n-1.∴an=(n-1)a2,an-an-1=a2.
故{an}是以a2为公差,以a1为首项的等差数列.
∴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=
1
2.
(2)由已知Sn=
1
2nan,a1=0.
n≥2时,an=Sn-Sn-1=
1
2nan-
1
2(n-1)an-1.
∴
an
an−1=
n−1
n−2.则
an−1
an−2=
n−2
n−3,
a3
a2=
2
1.
∴
an
a2=n-1.∴an=(n-1)a2,an-an-1=a2.
故{an}是以a2为公差,以a1为首项的等差数列.
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