bluepromiser2010-10-06TA获得超过2万个赞关注an=n²-kna(n+1)=(n+1)²-k(n+1)=n²+(2-k)n+1-k由an<a(n+1)得:n²-kn<n²+(2-k)n+1-k2n+1>k对任意n∈N*成立∴k<2·1+1=3