考點(diǎn):數(shù)列的求和
專(zhuān)題:等差數(shù)列與等比數(shù)列
分析:(1)由題意可得2a
n=
+s
n,令n=1可求a
1,n≥2時(shí),s
n=2a
n-,s
n-1=2a
n-1-,兩式相減可得遞推式,由遞推式可判斷該數(shù)列為等比數(shù)列,從而可得a
n;
(2)表示出b
n,進(jìn)而可得
,并拆項(xiàng),利用裂項(xiàng)相消法可求和,由和可得結(jié)論.
解答:
解:(1)∵
,a
n,S
n成等差數(shù)列,
∴2a
n=
+s
n,
當(dāng)n=1時(shí),2a
1=
+a
1,解得a
1=
;
當(dāng)n≥2時(shí),s
n=2a
n-,s
n-1=2a
n-1-,兩式相減得:a
n=S
n-S
n-1=2a
n-2a
n-1,
∴
=2,
所以數(shù)列{a
n}是首項(xiàng)為
,公比為2的等比數(shù)列,
∴a
n=
•2
n-1=2
n-2.
(2)b
n=(log
2a
2n+1)×(log
2a
2n+3)
=log222n+1-2×log222n+3-2
=(2n-1)(2n+1),
∴
=
=
(
-),
∴
+
+
+…+
=
(1
-+-+…+
-)=
(1-
)=
-
<
.
點(diǎn)評(píng):本題考查數(shù)列與不等式的綜合,考查裂項(xiàng)相消法對(duì)數(shù)列求和,考查等比數(shù)列的通項(xiàng)公式,屬中檔題.