题目内容
【题目】(本小题满分12分)已知数列
的首项![]()
(1)求
的通项公式;
(2)证明:对任意的
.
【答案】解:(1)
∴
∴ ![]()
又 ∵
∴
是以
为首项,
为公比的等比数列·········································4分
∴
∴
································································6分
(2) 由 (1) 知
···························································7分
![]()
![]()
![]()
∴ 原不等式成立··························································12分
【解析】略
练习册系列答案
相关题目