(1)根据题意,n+2个数构成递增的等比数列,
设为b 1,b 2,b 3,…,b n+2,其中b 1=1,b n+2=2,
可得A n=b 1•b 2•…•b n+1•b n+2,…①;A n=b n+2•b n+1•…•b 2•b 1,…②
由等比数列的性质,得b 1•b n+2=b 2•b n+1=b 3•b n=…=b n+2•b 1=2,
∴①×②,得
A 2n =( b 1 b n+2 )•( b 2 b n+1 )•…•( b n+1 b 2 )•( b n+2 b 1 ) =2 n+2.
∵A n>0,∴ A n = 2
n+2
2 .
因此,可得
A n+1
A n =
2
n+3
2
2
n+2
2 =
2 (常数),
∴数列{A n}是首项为 A 1 =2
2 ,公比为
2 的等比数列.
∴数列{A n}的前n项和S n=
2
2 [1- (
2 ) n ]
1-
2 = (4+2
2 )[ (
2 ) n -1] .
(2)由(1)得 a n =lo g 2 A n =lo g 2 2
n+2
2 =
n+2
2 ,
∵ tan1=tan[(n+1)-1]=
tan(n+1)-tann
1+tan(n+1)tann ,
∴ tan(n+1)tann=
tan(n+1)-tann
tan1 -1,n∈ N * .
从而tana 2n•tana 2n+2=tan(n+1)tan(n+2)=
tan(n+2)-tan(n+1)
tan1 -1,n∈ N *
∴
T n =tan a 2 •tan a 4 +tan a 4 •tan a 6 +…+tan a 2n •tan a 2n+2
=tan2•tan3+tan3•tan4+…+tan(n+1)tan(n+2)
=(
tan3-tan2
tan1 -1)+(
tan4-tan3
tan1 -1)+…+(
tan(n+2)-tan(n+1)
tan1 -1)
=
tan(n+2)-tan2
tan1 -n.
即T n=tana 2•tana 4+tana 4•tana 6+…+tana 2n•tana 2n+2
=
tan(n+2)-tan2
tan1 -n .