已知Tn=1/4-1/(n+1)(n∈正整数)Cn=1/(n+3)(n∈正整数)若对于一切n∈正整数,不等式4mTn>(n+2)Cn恒成立,求实数m的取值范围
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![已知Tn=1/4-1/(n+1)(n∈正整数)Cn=1/(n+3)(n∈正整数)若对于一切n∈正整数,不等式4mTn>(n+2)Cn恒成立,求实数m的取值范围](/uploads/image/z/13344723-27-3.jpg?t=%E5%B7%B2%E7%9F%A5Tn%3D1%2F4-1%2F%28n%2B1%29%28n%E2%88%88%E6%AD%A3%E6%95%B4%E6%95%B0%EF%BC%89Cn%3D1%2F%28n%2B3%29%28n%E2%88%88%E6%AD%A3%E6%95%B4%E6%95%B0%EF%BC%89%E8%8B%A5%E5%AF%B9%E4%BA%8E%E4%B8%80%E5%88%87n%E2%88%88%E6%AD%A3%E6%95%B4%E6%95%B0%2C%E4%B8%8D%E7%AD%89%E5%BC%8F4mTn%EF%BC%9E%EF%BC%88n%2B2%EF%BC%89Cn%E6%81%92%E6%88%90%E7%AB%8B%2C%E6%B1%82%E5%AE%9E%E6%95%B0m%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4)
已知Tn=1/4-1/(n+1)(n∈正整数)Cn=1/(n+3)(n∈正整数)若对于一切n∈正整数,不等式4mTn>(n+2)Cn恒成立,求实数m的取值范围
已知Tn=1/4-1/(n+1)(n∈正整数)Cn=1/(n+3)(n∈正整数)若对于一切n∈正整数,不等式4mTn>(n+2)Cn恒成立,求实数m的取值范围
已知Tn=1/4-1/(n+1)(n∈正整数)Cn=1/(n+3)(n∈正整数)若对于一切n∈正整数,不等式4mTn>(n+2)Cn恒成立,求实数m的取值范围
因为4mTn=m*(n-1)/(n+1),由4mTn>(n+2)Cn可得m>(n+2)(n+1)/(n+3)(n-3),我们只需得到(n+2)(n+1)/(n+3)(n-3)的单调性即可,将n+1代入上式得(n+2)(n+3)/(n+4)(n-2),比较(n+2)(n+1)/(n+3)(n-3)与(n+2)(n+3)/(n+4)(n-2)大小即可,又因为两式都有(n+2)可约去,在(n+1)/(n+3)(n-3)与(n+3)/(n+4)(n-2)中,若n>3,对角线相乘得等价于比较n^3+3n^2-9n-27与n^3+3n^2-6n-8的大小,约去n^3+3n^2即比较-9n-27与-6n-8的大小,显然-9n-27(n+2)(n+3)/(n+4)(n-2),即(n+2)(n+1)/(n+3)(n-3)递减,所以只需令n=4时m>=(4+2)(4+1)/(4+3)=30/7,当n
4mTn>(n+2)Cn
代入得m>1+2/(n^2+3n)
因为n>0所以n^2+3n单增
m>最大值
把n=1代入所以m>3/2
崩溃了,联考考到这题我也不会- -|||