问题 解答题
已知数列{an} (n∈N*)满足:a1=1,an+1-sin2θ•an=cos2θ•cos2nθ,其中θ∈(0,
π
2
)

(1)当θ=
π
4
时,求{an}的通项公式;
(2)在(1)的条件下,若数列{bn}中,bn=sin
πan
2
+cos
πan-1
4
(n∈N*,n≥2)
,且b1=1.求证:对于∀n∈N*,1≤bn
2
恒成立;
(3)对于θ∈(0,
π
2
)
,设{an}的前n项和为Sn,试比较Sn+2与
4
sin2
的大小.
答案

(1)当θ=

π
4
时,sin2θ=
1
2
,cos2θ=0
,∴an+1-
1
2
an=0
,即
an+1
an
=
1
2
.…2

故数列{an}是首项为a1=1,公比为

1
2
的等比数列.

故数列{an}的通项公式为 an=

1
2n-1
.…4分

(2)证明:由(1)得,an=

1
2n-1

∴当n∈N*,n≥2时,有bn=sin

πan
2
+cos
πan-1
4
=sin(
π
2
1
2n-1
)+cos(
π
4
1
2n-2
)=sin
π
2n
+cos
π
2n
=
2
sin(
π
2n
+
π
4
).…6

b1=1也满足上式,故当n∈N*时,bn=

2
sin(
π
2n
+
π
4
).

∵n∈N*

0<

π
2n
π
2
π
4
π
2n
+
π
4
4

1≤

2
sin(
π
2n
+
π
4
)≤
2
,即1≤bn
2
. …8

(3)解法一:由an+1-sin2θ•an=cos2θ•cos2nθ得:an+1-sin2θ•an=(cos2θ-sin2θ)•cos2nθ

an+1-cos2n+2θ=(an-cos2nθ)sin2θ,即

an+1-cos2n+2θ
an-cos2nθ
=sin2θ,

{an-cos2nθ}是首项为a1-cos2θ=1-cos2θ=sin2θ,公比为sin2θ的等比数列,

an-cos2nθ=sin2nθ, ∴ an=cos2nθ+sin2nθ.…9

∴Sn=a1+a2+…+an=(cos2θ+cos4θ+…+cos2nθ)+(sin2θ+sin4θ+…+sin2nθ)

=

cos4θ+sin4θ-cos2n+4θ-sin2n+4θ
sin2θcos2θ
.…11分

因此,Sn+2-

4
sin2
=
cos4θ+sin4θ-cos2n+4θ-sin2n+4θ
sin2θcos2θ
+2-
4
sin2

=

cos4θ+sin4θ-cos2n+4θ-sin2n+4θ+2sin2θcos2θ-1
sin2θcos2θ

=

(cos2θ+sin2θ)2-(cos2n+4θ+sin2n+4θ)-1
sin2θcos2θ

=-

(cosn+2θ)2+(sinn+2θ)2
sin2θcos2θ
<0,

∴Sn+2<

4
sin2
.…(14分)

解法二:同解法一得 an=cos2nθ+sin2nθ.…9

θ∈(0,

π
2
),0<cos2nθ<1,0<sin2nθ<1;…11分

∴Sn=a1+a2+…+an=(cos2θ+cos4θ+…+cos2nθ)+(sin2θ+sin4θ+…+sin2nθ)=

cos2θ(1-cos2nθ)
1-cos2θ
+
sin2θ(1-sin2nθ)
1-sin2θ
cos2θ
1-cos2θ
+
sin2θ
1-sin2θ
=
cos4θ+sin4θ
sin2θcos2θ
=
(cos2θ+sin2θ)2-2sin2θcos2θ
sin2θcos2θ
=
1
sin2θcos2θ
-2=
4
sin2
-2

∴Sn+2<

4
sin2
.…(14分)

(其他解法酌情给分)

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