已知等差数列an的首项a1>0,公差d>0,前n项和为Sn
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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
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属于N*,且m+n=2p 求证(1)Sn+Sm大于等于2Sp(2)Sn*Sm≤(Sp)^2
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属于N*,且m+n=2p 求证(1)Sn+Sm大于等于2Sp(2)Sn*Sm≤(Sp)^2
Sn = d / 2 n^2 + (a1 - d / 2) n ,
(1).
Sn + Sm = d / 2 (n^2 + m^2) + (a1 - d / 2) (n + m)
>= d / 2 (n^2 + m^2 + 2 n m) / 2 + (a1 - d / 2) (n + m)
= d / 2 2 p^2 + (a1 - d / 2) 2 p
= 2 [d / 2 p^2 + (a1 - d / 2) p]
= 2 Sp
(2.)
Sn Sm = d^2 / 4 m^2 n^2 + (a1 - d / 2)^2 m n + d / 2 (a1 - d / 2) m n (m + n)
= d^2 / 4 [√(m n)]^4 + (a1 - d / 2)^2 [√(m n)]^2 + d / 2 (a1 - d / 2) [√(m n)]^2 (m + n)
(1).
Sn + Sm = d / 2 (n^2 + m^2) + (a1 - d / 2) (n + m)
>= d / 2 (n^2 + m^2 + 2 n m) / 2 + (a1 - d / 2) (n + m)
= d / 2 2 p^2 + (a1 - d / 2) 2 p
= 2 [d / 2 p^2 + (a1 - d / 2) p]
= 2 Sp
(2.)
Sn Sm = d^2 / 4 m^2 n^2 + (a1 - d / 2)^2 m n + d / 2 (a1 - d / 2) m n (m + n)
= d^2 / 4 [√(m n)]^4 + (a1 - d / 2)^2 [√(m n)]^2 + d / 2 (a1 - d / 2) [√(m n)]^2 (m + n)
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