张牙舞爪的蚂蚁
幼苗
共回答了18个问题采纳率:88.9% 举报
(1)证明:由已知a n+1 =a n 2 +4a n +2,
∴a n+1 +2=(a n +2) 2
∵a 1 =1⇒a n +2>1,两边取对数,得lg(a n+1 +2)=2lg(a n +2)
∴{lg(a n +2)}是等比数列,公比为2,首项为lg(a 1 +2)=lg3
(2)由(1)得 lg( a n +2)= 2 n-1 lg3=lg 3 2 n-1 ,
∴ a n = 3 2 n-1 -2 ,
∵lgT n =lg[(a 1 +2)(a 2 +2)(a n +2)]=lg(a 1 +2)+lg(a 2 +2)+…+lg(a n +2)=
( 2 n -1)lg3
2-1 =lg 3 2 n -1
∴ T n = 3 2 n -1
(3)
∵ b n =
1
2 (
1
a n +1 +
1
a n +3 )=
1
2 (
1
3 2 n-1 -1 +
1
3 2 n-1 +1 )=
3 2 n-1
3 2 n -1 =
1
3 2 n-1 -1 -
1
3 2 n -1
=
1
a n +1 -
1
a n+1 +1 =
1
a 1 +1 -
1
a n+1 +1 =
1
2 -
1
3 2 n -1
显然b n >0,
∴ S n ≥ S 1 =
3
8 ,
又 S n =
1
2 -
1
3 2 n -1 <
1
2 ,
∴
3
8 ≤ S n <
1
2 .
1年前
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