ajingaialing
幼苗
共回答了22个问题采纳率:90.9% 举报
(Ⅰ)∵sin(2α+β)=3sinβ,
∴sin2αcosβ+cos2αsinβ=3sinβsin2αcosβ=sinβ(3-cos2α)
tanβ=
sin2α
3-cos2α =
2sinαcosα
3-2 cos 2 α+1 =
2sinαcosα
4 sin 2 α+2 cos 2 α =
tanα
2 tan 2 α+1
∴ f(x)=
x
2 x 2 +1
(Ⅱ)∵
a 2n+1 =2 a n f(n)=2 a n •
a n
2
a 2n +1 =
2
a 2n
2
a 2n +1
∴
1
a 2n+1 =1+
1
2
a 2n
∴
1
a 2n+1 -2=
1
2 (
1
a 2n -2)
∴数列 {
1
a 2n -2} 是以2为首项,
1
2 为公比的等比数列.
(Ⅲ)∵ b n =
1
a 2n - 2 n a 1 =
1
2
∴ S n =
2[1- (
1
2 ) 2 ]
1-
1
2 =4[1- (
1
2 ) 2 ]
又 S n >
31
8 即4[1- (
1
2 ) n ]>
31
8
∴ (
1
2 ) n <
1
32 ∴n>5
∴满足 S n >
31
8 的最小n为6 .
1年前
8