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问题 解答题
已知数列{an}满足2n-1a1+2n-2a2+2n-3a3+…+an=n•2n，记所有可能的乘积aiaj（1≤i≤j≤n）的和为Tn．
（1）求{an}的通项公式；
（2）求Tn的表达式；
（3）求证：
n-1
4
＜
T1
T2
+
T2
T3
+
T3
T4
+…+
Tn
Tn+1
＜
n
4
．
答案

（1）∵数列{an}满足2n-1a1+2n-2a2+2n-3a3+…+an=n•2n，

∴

a1
2
+
a2
22
+
a3
23
+…+
an
2n
=n，

当n=1时，a1=2．

当n≥2时，

a1
2
+
a2
22
+
a3
23
+…+
an
2n
=n，

a1
2
+
a2
2 2
+
a3
23
+…+
an-1
2n-1
=n-1，

两式相减，得

an
2n
=1，

∴an=2n．

（2）∵aiaj（1≤i≤j≤n）的和为Tn，

∴Tn=a1a1+（a1a2+a2a2）+（a1a3+a2a3+a3a3）+…+（a1an+a2an+a3an+…+anan）

=（23-22）+（25-23）+（27-24）+…+（22n+1-2n+1）

=

23-22n+3
1-4
-（2n+2-4）

=

4
3
(2n+1-1)(2n-1)．

（3）∵

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

=

2n-1
2n+2-1

=

2n-1
4(2n-1)+3
＜
1
4
，

∴

T1
T2
+
T2
T3
+
T3
T4
+…+
Tn
Tn+1
＜
1
4
+
1
4
+
1
4
+…+
1
4
=
n
4
，

∵

Tn
Tn+1
=
2n-1
2n+2-1
=
1
4
-
3
4
•
1
2n+2-1

=

1
4
-
3
4
•
1
3•2n+2n-1

＞

1
4
-
3
4
•
1
3•2n

=

1
4
-
1
2 n+2
，

∴

T1
T2
+
T2
T3
+
T3
T4
+…+
Tn
Tn+1
＞（
1
4
-
1
2 3
）+（
1
4
-
1
2 4
）+…+（
1
4
-
1
2 n+2
）

=

n
4
-(
1
2 3
+
1
2 4
+… +
1
2 n+2
)

=

n
4
-(
1
4
-
1
2 n+2
)＞
n-1
4
，

∴

n-1
4
＜
T1
T2
+
T2
T3
+
T3
T4
+…+
Tn
Tn+1
＜
n
4
．

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