数学证明下列命题

显示全部楼层 · 2012-7-17 22:26:06

1)相邻两个奇数, 令2n+1, 2n+3平方差为(2n+3)2 - (2n+1)2 = [(2n+3)+(2n+1)][(2n+3)-(2n+1)] = (4n+4)*2=8(n+1) 一定能被8整除2) 令三个连续整数分别为 n, n+1, n+2则平方和为 n2+(n+1)2+(n+2)2= n2+n2+2n+1+n2+4n+4 = 3n2+6n+5 = 3(n2+2n+1)+2被3除余23)令奇数2n+1(2n+1)2 -1 = 4n2+4n = 4n(n+1), n和n+1有一个...

千问 · 2012-7-17 22:26:06

1、(2k+1)^2-(2k-1)^2=8k2、(k-1)^2+k^2+(k+1)^2=3k^2 + 23、(2k+1)^2 -1=4k^2+4k=4k(k+1)，而k和k+1中必然有一个是偶数...

千问 · 2012-7-17 22:26:06

（2n+1)^2-(2n-1)^2=4n^2+4n+1-(4n^2-4n+1)=8n(n-1)^2+n^2+(n+1)^2=n^2-2n+1+n^2+n^2+2n+1=3n^2+2(2n+1)^2-1=4n^2+4n+1-1=4n(n+1)n和n+1总有一个偶数...

数学 证明下列命题

数学证明下列命题