n=1,n(n+1)(n+2)=1*2*3=6,顯然成立
假設(shè)n=k時(shí),k(k+1)(k+2)能被3整除
當(dāng)n=k+1時(shí),
n(n+1)(n+2)
=（k+1)(k+2)(k+3)
=k(k+1)(k+2)+3(k+1)(k+2),
由假設(shè)知：式中第一項(xiàng)k(k+1)(k+2)能被3整除,
第二項(xiàng)3(k+1)(k+2)也能被3整除
所以當(dāng)n=k+1時(shí),
n(n+1)(n+2)
=（k+1)(k+2)(k+3)能被3整除
綜上可知,n(n+1)(n+2)能被3整除