天堂手指
春芽
共回答了8个问题采纳率:100% 举报
(1)证明:当n=1时,a 1 =S 1 =(m+1)-ma 1 ,解得a 1 =1.
当n≥2时,a n =S n -S n-1 =ma n-1 -ma n .
即(1+m)a n =ma n-1 .
∵m为常数,且m>0,∴
a n
a n-1 =
m
1+m (n≥2)
∴数列{a n }是首项为1,公比为
m
1+m 的等比数列.
(2)由(1)得,q=f(m)=
m
1+m ,b 1 =2a 1 =2.
∵ b n =f( b n-1 )=
b n-1
1+ b n-1 ,
∴
1
b n =
1
b n-1 +1 ,即
1
b n -
1
b n-1 =1 (n≥2).
∴ {
1
b n } 是首项为
1
2 ,公差为1的等差数列.
∴
1
b n =
1
2 +(n-1)•1=
2n-1
2 ,即 b n =
2
2n-1 (n∈N * ).
(3)证明:由(2)知 b n =
2
2n-1 ,则 b n 2 =
4
(2n-1) 2 .
所以T n =b 1 2 +b 2 2 +b 3 2 ++b n 2 = 4+
4
9 +
4
25 ++
4
(2n-1) 2 ,
当n≥2时,
4
(2n-1) 2 <
4
2n(2n-2) =
1
n-1 -
1
n ,
所以 T n =4+
4
9 +
4
25 ++
4
(2n-1) 2 <4+
4
9 +(
1
2 -
1
3 )+(
1
3 -
1
4 )++(
1
n-1 -
1
n ) =
40
9 +
1
2 -
1
n <
89
18 .
1年前
4