若{an}为正项数列.Sn为其前N项和,且an,Sn,an^2成等差数列’ 1.求{an},2,设f(n)=Sn/(n+
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若{an}为正项数列.Sn为其前N项和,且an,Sn,an^2成等差数列’ 1.求{an},2,设f(n)=Sn/(n+50)Sn+1,
求f(n)的最大值.注:除号下面的那个Sn+1的n+1是下标
求f(n)的最大值.注:除号下面的那个Sn+1的n+1是下标
an,Sn,an^2成等差数列
2Sn=an^2+an
2Sn-1=an-1^2+an-1 相减
2an=an^2+an-(an-1^2+an-1)
(an+an-1)=(an-an-1)(an+an-1) 若{an}为正项数列
an-an-1=1
an=an-1+1
{an}为 首项是 a1=1 d=1的等差数列
an=a1+(n-1)d=n
Sn=n(n+1)/2
2.f(n)=(n(n+1)/2)/(n+50)*(n+1)(n+2)/2
=n/(n+50)(n+2)
=n/[n^2+52n+100]
=1/[n+100/n+52]
n+100/n+52>=20+52 均值定理
n+100/n+52最小值=72
f(n)的最大值=1/72
再问: 可以不用均值定理来做么,拜托我们还没学过,.
再答: a+b>=2√ab n+100/n>=2√(n*100/n)=20
2Sn=an^2+an
2Sn-1=an-1^2+an-1 相减
2an=an^2+an-(an-1^2+an-1)
(an+an-1)=(an-an-1)(an+an-1) 若{an}为正项数列
an-an-1=1
an=an-1+1
{an}为 首项是 a1=1 d=1的等差数列
an=a1+(n-1)d=n
Sn=n(n+1)/2
2.f(n)=(n(n+1)/2)/(n+50)*(n+1)(n+2)/2
=n/(n+50)(n+2)
=n/[n^2+52n+100]
=1/[n+100/n+52]
n+100/n+52>=20+52 均值定理
n+100/n+52最小值=72
f(n)的最大值=1/72
再问: 可以不用均值定理来做么,拜托我们还没学过,.
再答: a+b>=2√ab n+100/n>=2√(n*100/n)=20
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