1^3+2^3+...+n^3=?怎么推导?

1个回答

  • 1^2+2^2+3^2+……+n^2=n(n+1)(2n+1)/6

    利用立方差公式

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

    =n^2+(n-1)^2+n^2-n

    =2*n^2+(n-1)^2-n

    2^3-1^3=2*2^2+1^2-2

    3^3-2^3=2*3^2+2^2-3

    4^3-3^3=2*4^2+3^2-4

    .

    n^3-(n-1)^3=2*n^2+(n-1)^2-n

    各等式全相加

    n^3-1^3=2*(2^2+3^2+...+n^2)+[1^2+2^2+...+(n-1)^2]-(2+3+4+...+n)

    n^3-1=2*(1^2+2^2+3^2+...+n^2)-2+[1^2+2^2+...+(n-1)^2+n^2]-n^2-(2+3+4+...+n)

    n^3-1=3*(1^2+2^2+3^2+...+n^2)-2-n^2-(1+2+3+...+n)+1

    n^3-1=3(1^2+2^2+...+n^2)-1-n^2-n(n+1)/2

    3(1^2+2^2+...+n^2)=n^3+n^2+n(n+1)/2=(n/2)(2n^2+2n+n+1)

    =(n/2)(n+1)(2n+1)

    1^2+2^2+3^2+...+n^2=n(n+1)(2n+1)/6

    1^3+2^3+3^3+……+n^3=[n(n+1)/2]^2

    (n+1)^4-n^4=[(n+1)^2+n^2][(n+1)^2-n^2]

    =(2n^2+2n+1)(2n+1)

    =4n^3+6n^2+4n+1

    2^4-1^4=4*1^3+6*1^2+4*1+1

    3^4-2^4=4*2^3+6*2^2+4*2+1

    4^4-3^4=4*3^3+6*3^2+4*3+1

    .

    (n+1)^4-n^4=4*n^3+6*n^2+4*n+1

    各式相加有

    (n+1)^4-1=4*(1^3+2^3+3^3...+n^3)+6*(1^2+2^2+...+n^2)+4*(1+2+3+...+n)+n

    4*(1^3+2^3+3^3+...+n^3)=(n+1)^4-1+6*[n(n+1)(2n+1)/6]+4*[(1+n)n/2]+n

    =[n(n+1)]^2

    1^3+2^3+...+n^3=[n(n+1)/2]^2