(理) 设数列{an}为正项数列,其前n项和为Sn,且有an,sn,a2n
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(理) 设数列{an}为正项数列,其前n项和为Sn,且有an,sn,
a | 2 n |
(1)∵an,sn,
a2n成等差数列
∴2Sn=an+
a2n,
∴n≥2时,2Sn-1=an-1+
a2n−1,
两式相减得:2an=an2+an-
a2n−1-an-1,
∴(an+an-1)(an-an-1-1)=0
∵数列{an}为正项数列,∴an-an-1=1
即{an}是公差为1的等差数列
又2a1=a12+a1,∴a1=1
∴an=1+(n-1)×1=n;
(2)由(1)知,Sn=
n(n+1)
2,
∴f(n)=
Sn
(n+50)Sn+1=
n
n2+52n+100=
1
n+
100
n+52≤
1
72
当且仅当n=10时,f(n)有最大值
1
72.
a2n成等差数列
∴2Sn=an+
a2n,
∴n≥2时,2Sn-1=an-1+
a2n−1,
两式相减得:2an=an2+an-
a2n−1-an-1,
∴(an+an-1)(an-an-1-1)=0
∵数列{an}为正项数列,∴an-an-1=1
即{an}是公差为1的等差数列
又2a1=a12+a1,∴a1=1
∴an=1+(n-1)×1=n;
(2)由(1)知,Sn=
n(n+1)
2,
∴f(n)=
Sn
(n+50)Sn+1=
n
n2+52n+100=
1
n+
100
n+52≤
1
72
当且仅当n=10时,f(n)有最大值
1
72.
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