已知非负等差数列{an}的公差d不为0,前n项和为Sn,设m,n,p∈N*,且m+n=2p (1)求证:1/Sn+1/S
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已知非负等差数列{an}的公差d不为0,前n项和为Sn,设m,n,p∈N*,且m+n=2p (1)求证:1/Sn+1/Sm≥2/Sp
由题意得a1>=0 a2>0 ...an>0 d>0
sn=na1+(n-1)d sm=ma1+(m-1)d sp=pa1+(p-1)d
由1/sn+1/sm>=2/sp得
sp(sn+sm) >=2sn*sm
spsn-snsm+spsm-snsm>=0
sn(sp-sm)+sm(sp-sn)>=0
sn(pa1+(p-1)d -ma1-(m-1)d)+sm(pa1+(p-1)d -na1-(n-1)d)>=0
sn((p-m)a1+(p-m)d) +sm((p-n)a1+(p-n)d)>=0
sn(p-m)(a1+d) +sm(p-n)(a1+d)>=0
sn(p-m)+sm(p-n)>=0 (a1+d>0)
p-m=(m+n)/2-m =(n-m)/2 p-n=(m+n)/2 -n =(m-n)/2
sn*(n-m)/2 +sm(m-n)/2>=0
(n-m) (sn-sm)>=0
若n>=m 则sn>=sm
所以上式成立
若n
sn=na1+(n-1)d sm=ma1+(m-1)d sp=pa1+(p-1)d
由1/sn+1/sm>=2/sp得
sp(sn+sm) >=2sn*sm
spsn-snsm+spsm-snsm>=0
sn(sp-sm)+sm(sp-sn)>=0
sn(pa1+(p-1)d -ma1-(m-1)d)+sm(pa1+(p-1)d -na1-(n-1)d)>=0
sn((p-m)a1+(p-m)d) +sm((p-n)a1+(p-n)d)>=0
sn(p-m)(a1+d) +sm(p-n)(a1+d)>=0
sn(p-m)+sm(p-n)>=0 (a1+d>0)
p-m=(m+n)/2-m =(n-m)/2 p-n=(m+n)/2 -n =(m-n)/2
sn*(n-m)/2 +sm(m-n)/2>=0
(n-m) (sn-sm)>=0
若n>=m 则sn>=sm
所以上式成立
若n
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