已知数列{a n }的各项均为正数,其前n项和为S n ,且满足2S n =a n 2 +a n (n∈N * ).(Ⅰ)求a 1 ,

2025-06-20 09:36:00
推荐回答(1个)
回答1:

(Ⅰ)a 1 =1,a 2 =2,a 3 =3.(3分)
(Ⅱ)2S n =a n 2 +a n ,①2S n-1 =a n-1 2 +a n-1 ,(n≥2)②(5分)
①-②即得(a n -a n-1 -1)(a n +a n-1 )=0,(6分)
因为a n +a n-1 ≠0,所以a n -a n-1 =1,所以a n =n(n∈N * )(8分)
(Ⅲ)(Ⅲ)∵ b n =n(
1
2
) n

T n =
1
2
+2× (
1
2
)
2
+…+n× (
1
2
)
n

1
2
T n = (
1
2
)
2
+2× (
1
2
)
3
+…+n× (
1
2
)
n+1

两式相减得,
1
2
T n  =
1
2
+ (
1
2
)
2
 +…+ (
1
2
)
n
-n× (
1
2
)
n+1

=1-
2+n
2 n+1

所以 T n =2-
2+n
2 n
.(13分)