发布人:繁体字网(www.fantiz5.com) 发布时间:2016-03-06 07:30:00
试题原文 |
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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 =
∵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 |
经过对同学们试题原文答题和答案批改分析后,可以看出该题目“已知{an}是正数组成的数列,a1=1,且点(an,an+1)(n∈N*)在函数y=..”的主要目的是检查您对于考点“高中等差数列的通项公式”相关知识的理解。有关该知识点的概要说明可查看:“高中等差数列的通项公式”。