问题
解答题
已知数列{an}满足a1=a2=2,a3=3,an+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=
=
+1a 23 a2
=59+1 2
a5=
=
-1a 24 a3
=825-1 3
(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.