Question

请教·一个数学问题？

Answer 1

证明：n，n³-1为整数，所以只需证明3/2×n²+1/2×n是整数

当n为偶数时，设n=2k(k为整数)，则3/2×n²+1/2×n=3/2×4k²+1/2×2k=6k²+k，显然为整数

当n为奇数时，设n=2k+1,则3/2×n²+1/2×n=3/2×(2k+1)²+1/2×(2k+1)=6k²+6k+3/2+k+1/2=6k²+6k+2

显然也为整数。

所以n³+3/2×n²+1/2×n-1对任何整数n都为整数

当n=3k时，n³+3/2×n²+1/2×n-1=27k³+3/2×9k²+1/2×3k-3+2，而27k³+3/2×9k²+1/2×3k-3能被3整除，所以此时n³+3/2×n²+1/2×n-1用3除时余2。

同理，当n=3k+1,n=3k+2时都可以得出n³+3/2×n²+1/2×n-1用3除时余2

故对一切整数n,n³+3/2×n²+1/2×n-1用3除时余2

综上，n³+3/2×n²+1/2×n-1对任何整数n都为整数，且用3除时余2

Answer 2

因为n³+3/2×n²+1/2×n-1=n³+n²+1/2×n²+1/2n=n³+1/2(3n²+n-2)=n³+1/2(n+1)(3n-2)

又因为当n为奇数时n+1能被2整除

当n为偶数时3n-2能被2整除

当n=0时n³+3/2×n²+1/2×n-1=-1

所以n为任何整数n³+3/2×n²+1/2×n-1都为整数

暂时不知道（用3除时余2）怎么解不好意思