求证:1/(n+1)*(1+1/3+1/5+...+1/2n-1)>1/n*(1/2+1/4+1/6+...+1/2n)(n>=2且为正数)
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求证:1/(n+1)*(1+1/3+1/5+...+1/2n-1)>1/n*(1/2+1/4+1/6+...+1/2n)(n>=2且为正数)
求证:1/(n+1)*(1+1/3+1/5+...+1/2n-1)>1/n*(1/2+1/4+1/6+...+1/2n)(n>=2且为正数)
求证:1/(n+1)*(1+1/3+1/5+...+1/2n-1)>1/n*(1/2+1/4+1/6+...+1/2n)(n>=2且为正数)
二楼确实不对.此题可以运用数学归纳,也可以采用逆推法.要证1/(n+1)*(1+1/3+1/5+...+1/2n-1)>1/n*(1/2+1/4+1/6+...+1/2n),等价于n*(1+1/3+1/5+...+1/2n-1)>(n+1)(1/2+1/4+1/6+...+1/2n),左右各加上n*(1/2+1/4+```````+1/2n),那么左边就是n(1+1/2+1/3+1/4+``````+1/2n),右边是(2n+1)(1/2+1/4+1/6+...+1/2n)=(n+1/2)(1+1/2+1/3+````+1/n),即等价于证明n*(1+1/2+1/3+1/4+``````+1/2n)>(2n+1)(1/2+1/4+1/6+...+1/2n)=(n+1/2)(1+1/2+1/3+````+1/n),消去相同的项,等价于证明[n/(n+1)+n/(n+2)+````+n/(2n)]>1/2*(1+1/2+1/3+````+1/n),只须证明每一项都是前者大于后者,即n/(n+k)>1/2k,这个可以通过真分数不等式,或者直接通分得证.故本题得证
1楼错解
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+.....
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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!
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