用数学归纳法证明1^3+2^3+3^3+...+n^3=n^2(n+1)^2 / 4 = (1+2+3+...+n)^2

n=1,代入验证,省略
假设n=k成立,k>=1
1^3+2^3+3^3+...+k^3=k^2(k+1)^2/4
则n=k+1
1^3+2^3+3^3+...+k^3+(k+1)^3
=k^2(k+1)^2/4+(k+1)^3
=(k+1)^2*[k^2+4(k+1)]/4
=(k+1)^2*(k+2)^2/4
=(k+1)^2*[(k+1)+1]^2/4
综上
1^3+2^3+3^3+...+n^3=n^2(n+1)^2/4