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)

(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
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