题目内容
(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
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