Question

对任意整数n>2 ,求证：[n(n+1)/4n-2] =[n+1/4]

Answer 1

令A=n(n+1)/(4n-2),B=(n+1)/4,则
A=n(n+1)/(4n-2)=(2n+3)/8+3/(16n-8)=n/4+3/8+3/(16n-8)
B=(n+1)/4=n/4+1/4
令P=3/(16n-8),则当n≥3时,P≤3/(16×3-8)=3/40=0.075
n被4除的余数无非为0、1、2、3共四种,分别讨论：
当n=4k时(k≥1)
[A]=[(4k)/4+3/8+P]=[K+3/8+P]=K+[3/8+P],由于0＜3/8+P≤0.375+0.075＜1,故[A]=k
[b]=[(4k)/4+1/4]=[k+1/4]=k
故[A]=[B]
当n=4k+1时(k≥1)
[A]=[(4k+1)/4+3/8+P]=[K+5/8+P]=K+[5/8+P],由于0＜5/8+P≤0.625+0.075＜1,故[A]=k
[b]=[(4k+1)/4+1/4]=[k+1/2]=k
故[A]=[B]
当n=4k+2时(k≥1)
[A]=[(4k+2)/4+3/8+P]=[K+7/8+P]=K+[7/8+P],由于0＜7/8+P≤0.875+0.075＜1,故[A]=k
[b]=[(4k+2)/4+1/4]=[k+3/4]=k
故[A]=[B]
当n=4k+3时(k≥0)
[A]=[(4k+3)/4+3/8+P]=[K+1+1/8+P]=K+1+[1/8+P],由于0＜1/8+P≤0.175+0.075＜1,故[A]=k+1
[b]=[(4k+3)/4+1/4]=[k+1]=k+1
故[A]=[B]
总之各种情况下都有[A]=[B].