已知an是正数等差数列,lga1、lga2、lga4成等差数列,又bn=1/a2n,n=1,2,3,...求证bn为等比数列大概是数学课代表抄错题了
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已知an是正数等差数列,lga1、lga2、lga4成等差数列,又bn=1/a2n,n=1,2,3,...求证bn为等比数列大概是数学课代表抄错题了
已知an是正数等差数列,lga1、lga2、lga4成等差数列,又bn=1/a2n,n=1,2,3,...求证bn为等比数列
大概是数学课代表抄错题了
已知an是正数等差数列,lga1、lga2、lga4成等差数列,又bn=1/a2n,n=1,2,3,...求证bn为等比数列大概是数学课代表抄错题了
证：
由lga1,lga2,lga4成等差数列,得
2lga2=lga1+lga4
lg(a2)²=lg(a1a4)
a2²=a1a4
(a1+d)²=a1(a1+3d)
整理,得
d(a1-d)=0
d=0或a1=d
d=0时,an=a1
bn=1/a(2n)=1/a1,为定值,数列{bn}是以1/a1为首项,1为公比的等比数列.
a1=d时,a2n=a1+(2n-1)d=a1+2na1-a1=2na1
bn=1/a(2n)=1/2na1
bn-1=1/[2(n-1)a1]
bn/b(n-1)=n/(n-1),不是定值.数列不是等比数列,题目有问题.
因为 lga1 lga2 lga4 成等差数列
故 2lga2= lga1+lga4
即 a2^2 =a1*a4
又 an 是不同正数的等差数列
故 a4 = a1+3d a2 =a1+d (d>0,a1>0)
(a1+d)^2 = a1*(a1+3d)
a1d = d*d 即 a1 = d
an = nd...
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因为 lga1 lga2 lga4 成等差数列
故 2lga2= lga1+lga4
即 a2^2 =a1*a4
又 an 是不同正数的等差数列
故 a4 = a1+3d a2 =a1+d (d>0,a1>0)
(a1+d)^2 = a1*(a1+3d)
a1d = d*d 即 a1 = d
an = nd
bn = 1 / [(2^n)d](这里面是2的N次方然后再乘以d，下面都一样，百度这些符号我编辑不了~~)
bn-1 = 1/2(n-1)d
bn+1 = 1/2(n+1)d
bn*bn = bn-1*bn+1 又 b1=1/(2d)
即 bn 为等比数列
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lga4-lga2=lga2-lga1
a4/a2=a2/a1
a4=a2+2d,a1=a2-d,d为等差
得a2=2d,(d>0),an=nd,a2n=2nd
bn=1/a2n=1/2nd,
bn/b(n-1)=n/n-1,不是等比数列