1^3+2^3+3^3+........n^3=?

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

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

右边的两项各都容易裂为连续两项差(称为差分)
(n-1)n(n+1)=(n-1)n(n+1)(n+2)/4-(n-2)(n-1)n(n+1)/4
n=n(n+1)/2-(n-1)n/2

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

3n*(1+n)/2