求证:1⼀(n+1)*(1+1⼀3+1⼀5+...+1⼀2n-1)>1⼀n*(1⼀2+1⼀4+1⼀6+...+1⼀2n)(n>=2且为正数)

2024-12-20 18:16:51
推荐回答(2个)
回答1:

数学归纳法:
当n=2时,有1/3*4/3=4/9>1/2*3/4=3/8
假设当n=K时成立;则有1/(k+1)(1+1/3+1/5+...+1/2k-1)>1/k*(1/2+1/4+...+1/2k)
则1/(k+2)*(1+1/3+...1/2k+1)
>(k+1)/(k+2)*1/k(1/2+...+1/2k)+1/(k+2)*1/(2k+1)
>1/(k+2)[(k+1)/k(1/2+...+1/2k)+1/(2k+1)]
显然(k+1)/k>1,1/(2k+1)>1/(2k+2)
则上式>1/(k+2)(1/2+...1/(2k+1)+1/(2k+2))
即证当n=k+1时也成立。
故对于所有的n>=2时结论成立!

回答2:

1楼错解