数列求和(1).1^4+2^4+3^4+.+n^4 (2).1^5+2^5+3^5+.+n^5

1个回答

  • 以下证明利用到:1^2+2^2+3^2+……+n^2=n(n+1)(2n+1)/6和1^3+2^3+3^3+……+n^3=[n(n+1)/2]^2;

    证明:(1)2^5=(1+1)^5=1^5+5×1^4+10×1^3+10×1^2+5×1^1+1

    3^5=(2+1)^5=2^5+5×2^4+10×2^3+10×2^2+5×2^1+1

    ……

    (n+1)^5=n^5+5×n^4+10×n^3+10×n^2+5×n^1+1

    上式相加,相同项消去

    (n+1)^5=1^5+5×(1^4+2^4+……+n^4)+10×(1^3+2^3+……+n^3)+10×(1^2+2^2+……n^2)+5×(1+2+……+n)+(1+1+……+1)

    5×(1^4+2^4+……+n^4)=(n+1)^5-10×[n(n+1)/2]^2-10×n(n+1)(2n+1)/6-5×n(n+1)/2-n-1

    化简得1^4+2^4+……+n^4=n(n+1)(2n+1)(3n^2+3n-1)/30

    (2)方法和上面差不多,把左边的5次方改为6次方即可,在此不再赘述.

    结果为(n+1)^2 * n^2 * (2n^2+2n-1) / 12