证明:
1当n=1时
1^3=[1(1+1)/2]^2=1
原式成立
2假设当n=k,k属于N+时成立
即1^3+2^3+.+k^3=[k(k+1)/2]^2
则当n=k+1时
原式左边=1^3+2^3+.+k^3+(k+1)^3
=[k(k+1)/2]^2+(k+1)^3
=k^2(k+1)^2/4+(k+1)^3
=(k+1)^2/4*[k^2+4(k+1)]
=(k+1)^2/4*[k^2+4k+4]
=(k+1)^2(k+2)^2/4
=[(k+1)(k+1+1)/2]^2
=右边
所以原式成立
看不懂请HI我