设n为自然数,求证1/(n+1)+1/(n+2)+1/(n+3)+...+1/(3n)>4n/(4n+1)
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/05/18 01:45:20
设n为自然数,求证1/(n+1)+1/(n+2)+1/(n+3)+...+1/(3n)>4n/(4n+1)
用柯西不等式证明
用柯西不等式证明
证明:
由柯西不等式:
[(n+1)+(n+2)+...+(3n)][1/(n+1)+1/(n+2)+...+1/(3n)]>(1+1+...+1)^2=(2n)^2{注,一共有2n个1,而且等号显然不成立}
而由等差数列求和公式有:(n+1)+(n+2)+...+(3n)=n(4n+1)
于是1/(n+1)+1/(n+2)+...+1/(3n)>(4n^2)/[n(4n+1)]=4n/(4n+1)
证毕.
由柯西不等式:
[(n+1)+(n+2)+...+(3n)][1/(n+1)+1/(n+2)+...+1/(3n)]>(1+1+...+1)^2=(2n)^2{注,一共有2n个1,而且等号显然不成立}
而由等差数列求和公式有:(n+1)+(n+2)+...+(3n)=n(4n+1)
于是1/(n+1)+1/(n+2)+...+1/(3n)>(4n^2)/[n(4n+1)]=4n/(4n+1)
证毕.
设n为自然数,求证1/(n+1)+1/(n+2)+1/(n+3)+...+1/(3n)>4n/(4n+1)
设n为自然数,求证n+1分之1+n+2分之1+n+3分之1+...+3n分之1大于4n+1分之4n
当n为正偶数,求证n/(n-1)+n(n-2)/(n-1)(n-3)+...+n(n-2).2/(n-1)(n-3)..
证明不等式:(1/n)^n+(2/n)^n+(3/n)^n+.+(n/n)^n
设n ,n +1,n +2 ,n +3为四个连续的自然数
设n∈N,n>1.求证:logn (n+1)>log(n+1) (n+2)
化简:1/(n+1)(n+2)+1/(n+2)(n+3)+1/(n+3)(n+4)
(n+1)(n+2)/1 +(n+2)(n+3)/1 +(n+3)(n+4)/1
若n为正整数,求1/n(n+1)+1/(n+1)(n+2)+1/(n+2)(n+3)+1/(n+3)(n+4)+.+1/
证明当自然数n>=4时,n^3>3n^2+3n+1
2^n/n*(n+1)
[3n(n+1)+n(n+1)(2n+1)]/6+n(n+2)化简