(1)(an+1-2n+1)-(an-2n)=an+1-an-2n=1 故数列{an-2n}为等差数列,且公差d=1. an-2n=(a1-2)+(n-1)d=n-1,an=2n+n-1; (2)由(1)可知an=2n+n-1,∴bn=log2(an+1-n)=n 设f(n)=(1+)(1+)(1+)…(1+)×,(n≥2) 则f(n+1)=(1+)(1+)(1+)…(1+)×(1+)×, 两式相除可得=(1+)×=>1, 则有f(n)>f(n-1)>f(n-2)>…>f(2)=, 要使(1+)(1+)(1+)…(1+)>k对一切n∈N*且n≥2恒成立, 必有k<; 故k的取值范围是k<. |