已知数列an满足a1=14,an=an-1(-1)nan-1-2(n≥2,n∈N)(1)求数列an的通项公式an;(2)设bn=1a2n,求数列bn的前n项和Sn;(3)设cn=ansin(2n-1)

题目简介

已知数列an满足a1=14,an=an-1(-1)nan-1-2(n≥2,n∈N)(1)求数列an的通项公式an;(2)设bn=1a2n,求数列bn的前n项和Sn;(3)设cn=ansin(2n-1)

题目详情

已知数列an满足a1=
1
4
an=
an-1
(-1)nan-1-2
(n≥2,n∈N)

(1)求数列an的通项公式an
(2)设bn=
1
a2n
,求数列bn的前n项和Sn
(3)设cn=ansin
(2n-1)π
2
,数列cn的前n项和为Tn.求证:对任意的n∈N*Tn
4
7
题型:解答题难度:中档来源:深圳二模

答案

(1)∵class="stub"1
an
=(-1)n-class="stub"2
an-1
,∴class="stub"1
an
+(-1)n=(-2)[class="stub"1
an-1
+(-1)n-1]

又∵class="stub"1
a1
+(-1)=3
,所以数列{class="stub"1
an
+(-1)n}
(n∈N*)是以3为首项,-2为公比的等比数列,
an=
(-1)n-1
2n-1+1

(2)bn=(3×2n-1+1)2
=9•4n-1+6•2n-1+1,
Sn=9•
1•(1-4n)
1-4
+6•
1•(1-2n)
1-2
+n

=3•4n+6•2n+n-9.
(3)证明:由(1)知an=
(-1)n-1
3•2n-1+1
,sin
(2n-1)
2
=(-1)n-1
,∴cn=class="stub"1
3•2n-1+1
,当n≥3时,则Tn=class="stub"1
3+1
+class="stub"1
3•2+1
+class="stub"1
3•22+1
++class="stub"1
3•2n-1+1
<class="stub"1
4
+class="stub"1
7
+class="stub"1
3•22
+class="stub"1
3•23
++class="stub"1
3•2n-1

=class="stub"11
28
+
class="stub"1
12
[1-(class="stub"1
2
)
n-2
]
1-class="stub"1
2
=class="stub"11
28
+class="stub"1
6
[1-(class="stub"1
2
)
n-2
]<class="stub"11
28
+class="stub"1
6
=class="stub"47
84
<class="stub"48
84
=class="stub"4
7

又∵T1<T2<T3,
∴对任意的n∈N*,Tn<class="stub"4
7
.(12分)

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