已知等差数列{an}的首项a1>0,公差d>0,前n项和为Sn,设m,n,p∈N*,且m+n=2p (1)求证:Sn+Sm≥2Sp;(2)求证:Sn•Sm≤(Sp)2; (3)若S1005=1,求证:∑1/Sn≥ 2009 ∑上方为2009 下方为n=1
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![已知等差数列{an}的首项a1>0,公差d>0,前n项和为Sn,设m,n,p∈N*,且m+n=2p (1)求证:Sn+Sm≥2Sp;(2)求证:Sn•Sm≤(Sp)2; (3)若S1005=1,求证:∑1/Sn≥ 2009 ∑上方为2009 下方为n=1](/uploads/image/z/2430719-71-9.jpg?t=%E5%B7%B2%E7%9F%A5%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E9%A6%96%E9%A1%B9a1%EF%BC%9E0%2C%E5%85%AC%E5%B7%AEd%EF%BC%9E0%2C%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E8%AE%BEm%2Cn%2Cp%E2%88%88N%2A%2C%E4%B8%94m%2Bn%3D2p+%EF%BC%881%EF%BC%89%E6%B1%82%E8%AF%81%EF%BC%9ASn%2BSm%E2%89%A52Sp%EF%BC%9B%EF%BC%882%EF%BC%89%E6%B1%82%E8%AF%81%EF%BC%9ASn%26%238226%3BSm%E2%89%A4%EF%BC%88Sp%EF%BC%892%EF%BC%9B+%EF%BC%883%EF%BC%89%E8%8B%A5S1005%3D1%2C%E6%B1%82%E8%AF%81%EF%BC%9A%E2%88%911%2FSn%E2%89%A5+2009+%E2%88%91%E4%B8%8A%E6%96%B9%E4%B8%BA2009+%E4%B8%8B%E6%96%B9%E4%B8%BAn%3D1)
已知等差数列{an}的首项a1>0,公差d>0,前n项和为Sn,设m,n,p∈N*,且m+n=2p (1)求证:Sn+Sm≥2Sp;(2)求证:Sn•Sm≤(Sp)2; (3)若S1005=1,求证:∑1/Sn≥ 2009 ∑上方为2009 下方为n=1
已知等差数列{an}的首项a1>0,公差d>0,前n项和为Sn,设m,n,p∈N*,且m+n=2p (1)求证:Sn+Sm≥2Sp;
(2)求证:Sn•Sm≤(Sp)2;
(3)若S1005=1,求证:∑1/Sn≥ 2009 ∑上方为2009 下方为n=1
已知等差数列{an}的首项a1>0,公差d>0,前n项和为Sn,设m,n,p∈N*,且m+n=2p (1)求证:Sn+Sm≥2Sp;(2)求证:Sn•Sm≤(Sp)2; (3)若S1005=1,求证:∑1/Sn≥ 2009 ∑上方为2009 下方为n=1
(1)由等差数列前n项和公式可得Sn=d2
n2+(a1-d
2
)n,代入Sn+Sm,利用m+n=2p可证
(2)由等差数列前n项和公式可得Sn=d
2
n2+(a1-d
2
)n,代入SnSm,利用m+n=2p可证
(3)由(2)可得Sp
Sm
≥Sn
Sp
,从而有2009
n=1
1
Sn
=S1005
S1
+S1005
S2
+S1005
S2009
=S2009
S1005
+S2008
S1005
+S1
S1005
,再利用(1)的结论可证.证明:(1)由等差数列前n项和公式可得Sn=d 2 n2+(a1-d 2 )n,∴Sn+Sm=d 2 n2+(a1-d 2 )n+ d 2 m2+(a1-d 2 )m=d 2 (n2+m2)+( m+n)a1-d 2 (m+n)≥2Sp
(2)SnSm=[d 2 n2+(a1-d 2 )n][d 2 m2+(a1-d 2 )m]≤d2 4 p2+d 2 (a1-d 2 )p3+(a1-d 2 )2p2,∴SnSm≤(Sp)2
(3)2009 n=1 1 Sn =S1005 S1 +S1005 S2 +S1005 S2009 =S2009 S1005 +S2008 S1005 +S1 S1005 ≥2009
点评:本题主要考查等差数列的前n项和公式及不等式的基本性质,考查等价转化思想,属于中档题.
http://www.jyeoo.com/math2/ques/detail/e2227bb2-ce74-468f-9998-9553dafe9750