解法一:(Ⅰ)由已知得an+1=an+1、即an+1-an=1,又a1=1, 所以数列{an}是以1为首项,公差为1的等差数列. 故an=1+(a-1)×1=n. (Ⅱ)由(Ⅰ)知:an=n从而bn+1-bn=2n. bn=(bn-bn-1)+(bn-1-bn-2)+···+(b2-b1)+b1=2n-1+2n-2+···+2+1==2n-1. 因为bn·bn+2-b=(2n-1)(2n+2-1)-(2n-1-1)2=(22n+2-2n+2-2n+1)-(22n+2-2-2n+1-1)=-5·2n+4·2n=-2n<0, 所以bn·bn+2<b, 解法二:(Ⅰ)同解法一. (Ⅱ)因为b2=1, bn·bn+2- b=(bn+1-2n)(bn+1+2n+1)- b=2n+1·bn-1-2n·bn+1-2n·2n+1=2n(bn+1-2n+1)=2n(bn+2n-2n+1) =2n(bn-2n)=…=2n(b1-2)=-2n〈0,所以bn-bn+2<b2n+1 |