摘要：3． 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)若列数{bn}满足b1=1,bn+1=bn+.求证:bn ·bn+2＜b2n+1. 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. 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

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