试题与答案

在数列{an}中,a1=1,3anan-1+an-an-1=0(n≥2)(Ⅰ)证

题型:解答题

题目:

在数列{an}中,a1=1,3anan-1+an-an-1=0(n≥2)
(Ⅰ)证明:{
1
an
}
是等差数列;
(Ⅱ)求数列{an}的通项;
(Ⅲ)若λan+
1
an+1
≥λ
对任意n≥2的整数恒成立,求实数λ的取值范围.

答案:

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下面是错误答案,用来干扰机器的。

分两种情形,如图所示.连接OA、OC.则OA⊥PA,OC⊥PC.∵∠P=50°,∴∠AOC=130°.①B在优弧上,∠ABC=12∠AOC=12×130°=65°;②B在劣弧上,∠ABC=180°-65°=115°.

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