n=1,代入验证,省略假设n=k成立,k>=11^3+2^3+3^3+...+k^3=k^2(k+1)^2/4则n=k+11^3+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