Sandy5200
幼苗
共回答了12个问题采纳率:100% 举报
(Ⅰ)解法一:由na n+1 =2S n ①
得当n≥2时,(n-1)a n =2S n-1 ②,
由①-②可得,na n+1 -(n-1)a n =2(S n -S n-1 )=2a n ,
所以na n+1 =(n+1)a n ,
即当n≥2时,
a n+1
a n =
n+1
n ,
所以
a 3
a 2 =
3
2 ,
a 4
a 3 =
4
3 ,
a 5
a 4 =
5
4 ,…,
a n
a n-1 =
n
n-1 ,
将上面各式两边分别相乘得,
a n
a 2 =
n
2 ,
即 a n =
n
2 • a 2 (n≥3),
又a 2 =2S 1 =2a 1 =2,所以a n =n(n≥3),
此结果也满足a 1 ,a 2 ,
故a n =n对任意n∈N + 都成立.…(7分)
解法二:由na n+1 =2S n 及a n+1 =S n+1 -S n ,
得nS n+1 =(n+2)S n ,
即
S n+1
S n =
n+2
n ,
∴当n≥2时, S n = S 1 •
S 2
S 1 •
S 3
S 2 •…•
S n
S n-1 =1×
3
1 ×
4
2 ×
5
3 ×…×
n+1
n-1 =
n(n+1)
2 (此式也适合S 1 ),
∴对任意正整数n均有 S n =
n(n+1)
2 ,
∴当n≥2时,a n =S n -S n-1 =n(此式也适合a 1 ),
故a n =n.…(7分)
(Ⅱ)依题意可得 b n =
4 a n+1
a n 2 a n+2 2 =
4n+4
n 2 (n+2) 2 =
1
n 2 -
1
(n+2) 2
T n =
1
1 2 -
1
3 2 +
1
2 2 -
1
4 2 +
1
3 2 -
1
5 2 +…+
1
n 2 -
1
(n+2) 2
=1+
1
4 -
1
(n+1) 2 -
1
(n+2) 2
=
5
4 -
(n+1) 2 + (n+2) 2
(n+1) 2 (n+2) 2 <
5
4
∴ T n <
5
4 .…(13分)
1年前
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