求证,对任意正整数n,N=1/5n^5+1/3n^3+7/15n的值恒为整数

2个回答

  • N=1/5*n^5+1/3*n^3+7/15*n

    =1/15*(3n^5+5n^3+7n)

    =1/15*n(3n^4+5n^2+7)

    =1/15*n(3n^4-10n^2+7+15n^2)

    =n/15*[(3n^2-7)(n^2-1)+15n^2]

    =(n-1)n(n+1)(3n^2-7)/15+n^3

    因为(n-1),n,(n+1)是3个连续的自然数,一定有个是3的倍数.

    如果(n-1),n,(n+1)里有5的因子,则(n-1)n(n+1)是15的倍数,得证.

    如果(n-1),n,(n+1)里没有5的因子,

    则只能是n=5k+2,n=5k+3.(k是自然数)

    n=5k+2时,

    3n^2-7=3(5k+2)^2-7=75k^2+60k+5,是5的倍数.

    n=5k+3时,

    3n^2-7=3(5k+3)^2-7=75k^2+90k+20,是5的倍数.

    所以(n-1)n(n+1)(3n^2-7)是15的倍数.

    得证.