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 ∵bn•bn+2-bn+12=(2n-1)(2n+2-1)-(2n+1-1)2 =(22n+2-2n-2n+2+1)-(22n+2-2•2n+1+1) =-2n<0 ∴bn•bn+2<bn+12
解法二: (Ⅰ)同解法一. (Ⅱ)∵b2=1 bn•bn+2-bn+12=(bn+1-2n)(bn+1+2n+1)-bn+12 =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 ∴bn•bn+2<bn+12 |