数列求和题 设数列前n项和为sn 点(n,Sn/n)在直线x-y+1=0上 ,求数列an的通项公式第二问,我主要是问第二问~bn=ana(n+2) 求证1/12<1/b1+1/b2+1/b3+``````+1/bn<3/16
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![数列求和题 设数列前n项和为sn 点(n,Sn/n)在直线x-y+1=0上 ,求数列an的通项公式第二问,我主要是问第二问~bn=ana(n+2) 求证1/12<1/b1+1/b2+1/b3+``````+1/bn<3/16](/uploads/image/z/1645751-47-1.jpg?t=%E6%95%B0%E5%88%97%E6%B1%82%E5%92%8C%E9%A2%98+%E8%AE%BE%E6%95%B0%E5%88%97%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BAsn+%E7%82%B9%28n%2CSn%2Fn%29%E5%9C%A8%E7%9B%B4%E7%BA%BFx-y%2B1%3D0%E4%B8%8A+%2C%E6%B1%82%E6%95%B0%E5%88%97an%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%E7%AC%AC%E4%BA%8C%E9%97%AE%2C%E6%88%91%E4%B8%BB%E8%A6%81%E6%98%AF%E9%97%AE%E7%AC%AC%E4%BA%8C%E9%97%AE%7Ebn%3Dana%28n%2B2%29+%E6%B1%82%E8%AF%811%2F12%EF%BC%9C1%2Fb1%2B1%2Fb2%2B1%2Fb3%2B%60%60%60%60%60%60%2B1%2Fbn%EF%BC%9C3%2F16)
数列求和题 设数列前n项和为sn 点(n,Sn/n)在直线x-y+1=0上 ,求数列an的通项公式第二问,我主要是问第二问~bn=ana(n+2) 求证1/12<1/b1+1/b2+1/b3+``````+1/bn<3/16
数列求和题 设数列前n项和为sn 点(n,Sn/n)在直线x-y+1=0上 ,求数列an的通项公式
第二问,我主要是问第二问~
bn=ana(n+2) 求证1/12<1/b1+1/b2+1/b3+``````+1/bn<3/16
数列求和题 设数列前n项和为sn 点(n,Sn/n)在直线x-y+1=0上 ,求数列an的通项公式第二问,我主要是问第二问~bn=ana(n+2) 求证1/12<1/b1+1/b2+1/b3+``````+1/bn<3/16
an=2n
bn=2n·2(n+2)=4n(n+2)
1/bn=(1/8)[1/n - 1/(n+2)]
所以 1/b1+1/b2+1/b3+...+1/bn=
(1/8)[1/1 -1/3 +1/2 - 1/4 +1/3-1/5 +...+1/(n-1) -1/(n+1) +1/n -1/(n+2)]
=(1/8)[1+1/2 -1/(n+1) -1/(n+2)]
=3/16 -(2n+3)/[8(n+1)(n+2)] <3/16
另一方面,(1/8)[1+1/2 -1/(n+1) -1/(n+2)]>(1/8)(1+1/2 -1/2 -1/3)=(1/8)·(2/3)=1/12
从而 1/12<1/b1+1/b2+1/b3+``````+1/bn<3/16
点(n,Sn/n)在直线x-y+1=0上
Sn/n=n+1
Sn=n^2+n
a1=S1=2
n>=2:an=Sn-S[n-1]=2n+1-1=2n
a1=2*1=2,符合,故通项是an=2n
bn=ana(n+2)=2n*2(n+2)=4n(n+2)
1/bn=1/[4n(n+2)]=1/8*[1/n-1/(n+2)]
所以有1/b...
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点(n,Sn/n)在直线x-y+1=0上
Sn/n=n+1
Sn=n^2+n
a1=S1=2
n>=2:an=Sn-S[n-1]=2n+1-1=2n
a1=2*1=2,符合,故通项是an=2n
bn=ana(n+2)=2n*2(n+2)=4n(n+2)
1/bn=1/[4n(n+2)]=1/8*[1/n-1/(n+2)]
所以有1/b1+1/b2+...+1/bn=1/8[1-1/3+1/2-1/4+1/3-1/5+....+1/n-1/(n+2)]
=1/8[1+1/2-1/(n+1)-1/(n+2)]
<1/8*(1+1/2)
=3/16
同时有1/8[1+1/2-1/(n+1)-1/(n+2)]>1/8[1+1/2-1/2-1/3]=1/8*2/3=1/12
故得证!
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