题目内容

已知各项均不为零的数列{an}的前n项和为Sna1=1且Sn=
1
2
anan+1(n∈N*)

(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
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=
1
a2n-1
-
1
a2n
=
1
a2n-1a2n
>0
Tn=
1
a1
-
1
a2
+
1
a3
-
1
a4
+…+
1
a2n-1
-
1
a2n
=b1+b2+…+bnb1=
1
2
.由(2n-1)•2n=(2n)2-2n(2n)2-n-
3
4
=(2n-
3
2
) (2n+
1
2
)
,知bn
1
(2n-
3
2
) (2n+
1
2
)
=
1
4n-3
-
1
4n+1
,由此能够证明对任意n∈N*
1
2
1
a1
-
1
a2
+
1
a3
-
1
a4
+
1
a5
-
1
a6
+…+
1
a2n-1
-
1
a2n
2
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=
1
a2n-1
-
1
a2n
=
1
a2n-1a2n
>0

Tn=
1
a1
-
1
a2
+
1
a3
-
1
a4
+…+
1
a2n-1
-
1
a2n

=b1+b2+…+bnb1=
1
2

∵(2n-1)•2n=(2n)2-2n(2n)2-n-
3
4
=(2n-
3
2
) (2n+
1
2
)

bn
1
(2n-
3
2
) (2n+
1
2
)
=
1
4n-3
-
1
4n+1

T1=
1
2
2
2
T2=
7
12
2
2
T3=
37
60
2
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
=0.7<
2
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
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.
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