在数列{an}中,a1=1,an+1=(1-1/n+1)an.若对一切n>1的自然数,不等式an+1+an+2+...+a2n>1/12loga (a-1)+2/3恒成立,试求a的取值范围
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![在数列{an}中,a1=1,an+1=(1-1/n+1)an.若对一切n>1的自然数,不等式an+1+an+2+...+a2n>1/12loga (a-1)+2/3恒成立,试求a的取值范围](/uploads/image/z/1740740-68-0.jpg?t=%E5%9C%A8%E6%95%B0%E5%88%97%7Ban%7D%E4%B8%AD%2Ca1%3D1%2Can%2B1%3D%281-1%2Fn%2B1%29an.%E8%8B%A5%E5%AF%B9%E4%B8%80%E5%88%87n%3E1%E7%9A%84%E8%87%AA%E7%84%B6%E6%95%B0%2C%E4%B8%8D%E7%AD%89%E5%BC%8Fan%2B1%2Ban%2B2%2B...%2Ba2n%3E1%2F12loga+%28a-1%29%2B2%2F3%E6%81%92%E6%88%90%E7%AB%8B%2C%E8%AF%95%E6%B1%82a%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4)
在数列{an}中,a1=1,an+1=(1-1/n+1)an.若对一切n>1的自然数,不等式an+1+an+2+...+a2n>1/12loga (a-1)+2/3恒成立,试求a的取值范围
在数列{an}中,a1=1,an+1=(1-1/n+1)an.若对一切n>1的自然数,不等式an+1+an+2+...+a2n>1/12loga (a-1)+2/3恒成立,试求a的取值范围
在数列{an}中,a1=1,an+1=(1-1/n+1)an.若对一切n>1的自然数,不等式an+1+an+2+...+a2n>1/12loga (a-1)+2/3恒成立,试求a的取值范围
an+1=(1-1/n+1)an
则an+1=(n/n+1)an
则an+1=(n/n+1)an
=(n/n+1)*(n-1/n)an-1
=...
=n/n+1*(n-1/n)*..1/2*a1
=1/n+1
所以an+1=1/n+1
则an+1+an+2+...+a2n=1/n+1+1/n+2+..+1/n+n
对于1/n+1+1/n+2+..+1/n+n,
令Sn=1/n+1+1/n+2+..+1/n+n.则
Sn+1=1/n+2+1/n+3+..+1/n+n+1/n+n+1+1/n+n+2
Sn+1-Sn=1/n+n+1+1/n+n+2-1/n+1
=1/2n+1-1/2n+2
=1/(2n+1)*(2n+2)>0
可知随着n的增加值是增加的.
n>1,所以n=2是取得最小值
所以要an+1+an+2+...+a2n>1/12loga (a-1)+2/3恒成立
只要最小值大于1/12loga (a-1)+2/3,那么所有的都满足了
则n=2时,1/3+1+1/4=7/12
所以1/12loga (a-1)+2/3