问题 解答题
已知数列{an}满足a1=a2=2,a3=3,an+2=
a2n+1
(-1)n
an
(n≥2)
(Ⅰ)求a4,a5
(Ⅱ)是否存在实数λ,使得数列{an+1-λan}(n∈N*)是等差数列?若存在,求出所有满足条件的λ的值;若不存在,说明理由;
(Ⅲ)写出数列{an}中与987相邻的后一项(不需要过程)
答案

(I)a4=

a23
+1
a2
=
9+1
2
=5

a5=

a24
-1
a3
=
25-1
3
=8

(II)假设存在实数λ,使得数列{an+1-λan}(n∈N*)是等差数列,则

2(a3-λa2)=(a2-λa1)+(a4-λa3),解得λ=1

由a3=3,a4=5,a5=8,a6=13得2(a5-a4)≠(a4-a3)-(a3-a2)与数列{an+1-an}(n∈N*)是等差数列矛盾

故不存在实数λ,使数列{an+1-λan}(n∈N*)是等差数列

(III)a2=2,a3=3,a4=5,a5=8,a6=13,猜想an+2=an+1+an(n≥2)

∴数列{an}中与987相邻的后一项为1597.

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