1>n=1時(shí),左邊=3,右邊=3,成立.
2>n=k成立時(shí),即k+(k+1)+.+2k=3k(k+1)/2,
則當(dāng)n=k+1時(shí),
(k+1)+(k+2)+.+2k+（2k+1)+(2k+2)
=3k(k+1)/2 -k+2k+1+2k+2
=3k(k+1)/2+3(k+1)
=(k+1)(3k/2+3)
=3(k+1)(k+2)/2
故當(dāng)n=k+1時(shí)同樣成立
得證.