1、數(shù)學(xué)歸納法：
1）當n=1時,顯然成立
2）設(shè)n=k時成立,則1^3+2^3+.+k^3=[k(k+1)/2]^2
則,當n=k+1時,
1^3+2^3+.+k^3+(k+1)^3
=[k(k+1)/2]^2+(k+1)^3
=(k+1)^2[(k/2)^2+k+1]
=(k+1)^2[(k^2+4k+4)/4]
=(k+1)^2[(k+2)/2]^2
=(k+1)^2{[(k+1)+1]/2}^2
即n=k+1時也成立
綜上,1^3+2^3+.+n^3 = [n(n+1)/2]^2
2、導(dǎo)數(shù)和微積分：
令1^3+2^3+...n^3=S(n),兩邊取導(dǎo)數(shù),
3(1^2+2^2+...+n^2)=S '(n)
已知1^2+2^2+.+n^2=n(n+1)(2n+1)/6
則有S '(n)= n(n+1)(2n+1)/2=(2n^3+3n^2+n)/2
對(2n^3+3n^2+n)/2 求積分,得到(n^4+2n^3+n^2)/4 + C = [n(n+1)/2]^2 + C(C為常數(shù))
將n=1代入,可得得C=0
即S(n)= [n(n+1)/2]^2