题目

已知{an}是正数组成的数列,a1=1,且点()(nN*)在函数y=x2+1的图象上. (Ⅰ)求数列{an}的通项公式; (Ⅱ)若数列{bn}满足b1=1,bn+1=bn+,求证:bn・bn+2<b2n+1. 答案: 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+(n-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b1=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
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