题目内容
已知各项均不为零的数列{an}的前n项和为Sn,a1=1且Sn=1 |
2 |
(I)求数列{an}的通项公式;
(II)求证:对任意n∈N*,
1 |
2 |
1 |
a1 |
1 |
a2 |
1 |
a3 |
1 |
a4 |
1 |
a5 |
1 |
a6 |
1 |
a2n-1 |
1 |
a2n |
| ||
2 |
分析:(I)由2Sn=anan+1,2Sn+1=an+1an+3,知2an+1=an+1(an+2-an),an+2-an=2,由此能求出an=n(n∈N+).
(II)令bn=
-
=
>0,Tn=
-
+
-
+…+
-
=b1+b2+…+bn≥b1=
.由(2n-1)•2n=(2n)2-2n>(2n)2-n-
=(2n-
) (2n+
),知bn<
=
-
,由此能够证明对任意n∈N*,
≤
-
+
-
+
-
+…+
-
<
.
(II)令bn=
1 |
a2n-1 |
1 |
a2n |
1 |
a2n-1a2n |
1 |
a1 |
1 |
a2 |
1 |
a3 |
1 |
a4 |
1 |
a2n-1 |
1 |
a2n |
1 |
2 |
3 |
4 |
3 |
2 |
1 |
2 |
1 | ||||
(2n-
|
1 |
4n-3 |
1 |
4n+1 |
1 |
2 |
1 |
a1 |
1 |
a2 |
1 |
a3 |
1 |
a4 |
1 |
a5 |
1 |
a6 |
1 |
a2n-1 |
1 |
a2n |
| ||
2 |
解答:解:(I)由题设知2Sn=anan+1,2Sn+1=an+1an+3,
∴2an+1=an+1(an+2-an),
∵an≠0,∴an+2-an=2,
∵a1=1,a2=2,
∴an=n(n∈N+).
(II)令bn=
-
=
>0,
Tn=
-
+
-
+…+
-
=b1+b2+…+bn≥b1=
.
∵(2n-1)•2n=(2n)2-2n>(2n)2-n-
=(2n-
) (2n+
),
∴bn<
=
-
,
T1=
<
,T2=
<
,T3=
<
,
n≥4时,Tn=T3+b4+b5+…+bn
<
+(
-
) +(
-
) +…+(
-
)
=
-
<
=0.7<
,
∴对任意n∈N*,
≤
-
+
-
+
-
+…+
-
<
.
∴2an+1=an+1(an+2-an),
∵an≠0,∴an+2-an=2,
∵a1=1,a2=2,
∴an=n(n∈N+).
(II)令bn=
1 |
a2n-1 |
1 |
a2n |
1 |
a2n-1a2n |
Tn=
1 |
a1 |
1 |
a2 |
1 |
a3 |
1 |
a4 |
1 |
a2n-1 |
1 |
a2n |
=b1+b2+…+bn≥b1=
1 |
2 |
∵(2n-1)•2n=(2n)2-2n>(2n)2-n-
3 |
4 |
3 |
2 |
1 |
2 |
∴bn<
1 | ||||
(2n-
|
1 |
4n-3 |
1 |
4n+1 |
T1=
1 |
2 |
| ||
2 |
7 |
12 |
| ||
2 |
37 |
60 |
| ||
2 |
n≥4时,Tn=T3+b4+b5+…+bn
<
37 |
60 |
1 |
13 |
1 |
17 |
1 |
17 |
1 |
21 |
1 |
4n-3 |
1 |
4n+1 |
=
541 |
780 |
1 |
4n+1 |
546 |
780 |
| ||
2 |
∴对任意n∈N*,
1 |
2 |
1 |
a1 |
1 |
a2 |
1 |
a3 |
1 |
a4 |
1 |
a5 |
1 |
a6 |
1 |
a2n-1 |
1 |
a2n |
| ||
2 |
点评:第(I)题考查数列通项公式的求法,解题时要注意迭代法的合理运用;第(II)题考查数列与一不等式的综合运用,解题时要注意裂项求和法和放缩法的合理运用.
练习册系列答案
相关题目
已知各项均不为零的数列{an},定义向量
=(an,an+1),
=(n,n+1),n∈N*.下列命题中真命题是( )
cn |
bn |
A、若?n∈N*总有
| ||||
B、若?n∈N*总有
| ||||
C、若?n∈N*总有
| ||||
D、若?n∈N*总有
|