1>n=1时,左边=3,右边=3,成立.
2>n=k成立时,即k+(k+1)+.+2k=3k(k+1)/2,
则当n=k+1时,
(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
故当n=k+1时同样成立
得证.
1>n=1时,左边=3,右边=3,成立.
2>n=k成立时,即k+(k+1)+.+2k=3k(k+1)/2,
则当n=k+1时,
(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
故当n=k+1时同样成立
得证.