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)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
≤
,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,
<
•
,于是当n≥2时,
<
•
,
<
•
,
<
•
,…,
<
•
,累乘得:
<(
)n-2•
,从而可证得
+
+
+…+
<
.
(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 |
(法二)由an+1=3n+1-2n+1>2×3n-2n+1=2an可得,
1 |
an+1 |
1 |
2 |
1 |
an |
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 |
1 |
an |
1 |
2 |
1 |
a2 |
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+
+
+…+
=
<
;
(法二)∵an+1=3n+1-2n+1>2×3n-2n+1=2an,
∴
<
•
,,
当n≥2时,
<
•
,
<
•
,
<
•
,
…
<
•
,
累乘得:
<(
)n-2•
,
∴
+
+
+…+
≤1+
+
×
+…+(
)n-2×
<
<
.
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 |
3 |
1 |
32 |
1 |
3n-1 |
1×(1-(
| ||
1-
|
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 |
1 |
a2 |
∴
1 |
a1 |
1 |
a2 |
1 |
a3 |
1 |
an |
1 |
5 |
1 |
2 |
1 |
5 |
1 |
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.
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