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天津高考2019数学_2019天津高考数学解析
tamoadmin 2024-05-15 人已围观
简介a1=0,a2=2,a3=4;a4=8,a5=12,a6=18(根据条件可得,因为k是自然数)a7=24,a8=32a1=0*2+0*2a2=0*2+0*2+1*2a3=0*2+0*2+1*2+1*2a4=0*2+0*2+1*2+1*2+2*2a5=0*2+0*2+1*2+1*2+2*2+2*2a6=0*2+0*2+1*2+1*2+2*2+2*2+3*2a(2*k-1)=0*2+0*2+1*2+1
a1=0,a2=2,a3=4;a4=8,a5=12,a6=18(根据条件可得,因为k是自然数)a7=24,a8=32
a1=0*2+0*2
a2=0*2+0*2+1*2
a3=0*2+0*2+1*2+1*2
a4=0*2+0*2+1*2+1*2+2*2
a5=0*2+0*2+1*2+1*2+2*2+2*2
a6=0*2+0*2+1*2+1*2+2*2+2*2+3*2
a(2*k-1)=0*2+0*2+1*2+1*2+2*2+2*2+3*2+.....+(k-1)*2+(k-1)*2
a(2*k)=0*2+0*2+1*2+1*2+2*2+2*2+3*2+.....+(k-1)*2+(k-1)*2+k*2
a(2*k-1)=2*k*(k-1)
a(2*k)=2*k^2
归纳法验证
3、Tn'=Tn,但是:取an=n^2/2,n为任意自然数。有Tn'<=Tn [这里用换元,取n=2*k]
于是:Tn'=2*n-2
同理:Tn''=Tn.但是an=(n^2-1)/2 取n=2*k-1
于是:Tn''=sum(n^2/(n^2-1)) n=2 3 4 5.....
Tn''=2*n-sum(1/(n^2-1))<2*n-sum(1/n^2)=2*n-e e是自然对数
由此求的不等式。当n=2时等号成立。
sum(1/n^2)=e