已知数列{an}满足a1=1,an+1=(1+cos2nπ2)an+sin2nπ2,n∈N*.(1)求a2,a3,a4,并求出数列{an}的通项公式;(2)设bn=a2na2n-1,Sn=b1+b2+

题目简介

已知数列{an}满足a1=1,an+1=(1+cos2nπ2)an+sin2nπ2,n∈N*.(1)求a2,a3,a4,并求出数列{an}的通项公式;(2)设bn=a2na2n-1,Sn=b1+b2+

题目详情

已知数列{an}满足a1=1,an+1=(1+cos2
2
)an+sin2
2
,n∈N*
(1)求a2,a3,a4,并求出数列{an}的通项公式;
(2)设bn=
a2n
a2n-1
,Sn=b1+b2+…+bn,求证:Sn≤n+
5
3
题型:解答题难度:中档来源:不详

答案

(1)a2=(1+0)a1+1=2,a3=(1+1)a2+0=4,a4=(1+0)a3+1=5,
∵an+1=
an+1 (n=2m-1,m∈N+
2an(n=2m,m∈N+)
,∴
a2m+1=2a2m
a2m=a2m-1+1

∴a2m+1=2a2m-1+2,∴a2m+1+2=2(a2m-1+2),∴
a2m+1+2
a2m-1+2
=2
∴数列{a2m-1+2}是公比为2的等比数列,∴a2m-1+2=(a1+2)2m-1,
∴a2m-1=-2+3•2m-1(m∈N+),a2m=class="stub"1
2
a2m+1=-1+3•2m-1(m∈N+),
∴an=
-2+3•2class="stub"n+1
2
-1
-1+3•2class="stub"n
2
-1
=
-2+3•2class="stub"n+1
2
-1
(n为奇数)
-1+3•2class="stub"n
2
-1
  (n为偶数)
=
-2+3•2class="stub"n-1
2
(n为奇数)
-1+3•2class="stub"n-2
2
(n为偶数)

(2)bn=
-1+3•2n-1
-2+3•2n-1
=1+class="stub"1
-2+3•2n-1
=1+class="stub"1
2(-1+3•2n-2)

①当n=1时,S1=b1=2≤1+class="stub"5
3
,不等式成立;
②当n≥2时,-1+3•2n-2≥2,∴0<class="stub"1
-1+3•2n-2
<1,
∵0<class="stub"1
-1+3•2n-2
class="stub"1+1
(-1+3•2n-2)+1
=class="stub"2
3•2n-2

class="stub"1
2(-1+3•2n-2)
class="stub"1
3•2n-2

∴bn<1+class="stub"1
3•2n-2
=1+class="stub"4
3•2n

∴Sn<2+(1+class="stub"4
3•22
)+(1+class="stub"4
3•23
)+…+(1+class="stub"4
3•2n

=n+1+class="stub"4
3
×
class="stub"1
4
1-class="stub"1
2
(1-class="stub"1
2n-1
)=n+1+class="stub"2
3
(1-class="stub"1
2n-1

=n+class="stub"5
3
-class="stub"4
3•2n
<n+class="stub"5
3

由①②知:Sn≤n+class="stub"5
3

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