题目内容

(2012•广东)设数列{an}的前n项和为Sn,满足2Sn=an+1-2n+1+1,n∈N*,且a1,a2+5,a3成等差数列.
(1)求a1的值;
(2)求数列{an}的通项公式;
(3)证明:对一切正整数n,有
1
a1
+
1
a2
+
1
a3
+…+
1
an
3
2
分析:(1)在2Sn=an+1-2n+1+1中,令分别令n=1,2,可求得a2=2a1+3,a3=6a1+13,又a1,a2+5,a3成等差数列,从而可求得a1
(2)由2Sn=an+1-2n+1+1,2Sn+1=an+2-2n+2+1得an+2=3an+1+2n+1①,an+1=3an+2n②,由①②可知{an+2n}为首项是3,3为公比的等比数列,从而可求an
(3)(法一),由an=3n-2n=(3-2)(3n-1+3n-2×2+3n-3×22+…+2n-1)≥3n-1可得
1
an
1
3n-1
,累加后利用等比数列的求和公式可证得结论;
(法二)由an+1=3n+1-2n+1>2×3n-2n+1=2an可得,
1
an+1
1
2
1
an
,于是当n≥2时,
1
a3
1
2
1
a2
1
a4
1
2
1
a3
1
a5
1
2
1
a4
,…,
1
an
1
2
1
an-1
,累乘得:
1
an
(
1
2
)
n-2
1
a2
,从而可证得
1
a1
+
1
a2
+
1
a3
+…+
1
an
3
2
解答:解:(1)在2Sn=an+1-2n+1+1中,
令n=1得:2S1=a2-22+1,
令n=2得:2S2=a3-23+1,
解得:a2=2a1+3,a3=6a1+13
又2(a2+5)=a1+a3
解得a1=1
(2)由2Sn=an+1-2n+1+1,
2Sn+1=an+2-2n+2+1得an+2=3an+1+2n+1
又a1=1,a2=5也满足a2=3a1+21
所以an+1=3an+2n对n∈N*成立
∴an+1+2n+1=3(an+2n),又a1=1,a1+21=3,
∴an+2n=3n
∴an=3n-2n
(3)(法一)
∵an=3n-2n=(3-2)(3n-1+3n-2×2+3n-3×22+…+2n-1)≥3n-1
1
an
1
3n-1

1
a1
+
1
a2
+
1
a3
+…+
1
an
≤1+
1
3
+
1
32
+…+
1
3n-1
=
1×(1-(
1
3
)
n
)
1-
1
3
3
2

(法二)∵an+1=3n+1-2n+1>2×3n-2n+1=2an
1
an+1
1
2
1
an
,,
当n≥2时,
1
a3
1
2
1
a2
1
a4
1
2
1
a3
1
a5
1
2
1
a4

1
an
1
2
1
an-1

累乘得:
1
an
(
1
2
)
n-2
1
a2

1
a1
+
1
a2
+
1
a3
+…+
1
an
≤1+
1
5
+
1
2
×
1
5
+…+(
1
2
)
n-2
×
1
5
7
5
3
2
点评:本题考查数列与不等式的综合,考查数列递推式,着重考查等比数列的求和,着重考查放缩法的应用,综合性强,运算量大,属于难题.
练习册系列答案
相关题目

违法和不良信息举报电话:027-86699610 举报邮箱:58377363@163.com

精英家教网