在数列{an}中,a1=0,且对任意k∈N+,a2k-1,a2k,a2k+1成等差数列,其公差为2k.(Ⅰ)证明a4,s5,a6成等比数列;(Ⅱ)求数列{an}的通项公式.

问题描述:

在数列{an}中,a1=0,且对任意k∈N+,a2k-1,a2k,a2k+1成等差数列,其公差为2k.
(Ⅰ)证明a4,s5,a6成等比数列;
(Ⅱ)求数列{an}的通项公式.

(I)由题设可知,a2=a1+2=2,a3=a2+2=4,a4=a3+4=8,a5=a4+4=12,a6=a5+6=18从而a6a5=a5a4=32,所以a4,s5,a6成等比数列;(II)由题设可得a2k+1-a2k-1=4k,k∈N*,所以a2k+1-a1=(a2k+1-a2k-1)+(a2k-1-a2k-3...
答案解析:(I)由题设可知,对任意k∈N+,a2k-1,a2k,a2k+1成等差数列,分别取前几个值,可得

a6
a5
a5
a4
3
2
;(II)由题设可得:分n为奇数和偶数分别来求,可得答案.
考试点:等比关系的确定;等差数列的性质.
知识点:本题考查等差数列的知识,涉及等比数列的定义和分类讨论的思想,属中档题.