(1)∵a n+S n=n,∴n=1时, a 1 =
1
2
n=2时,a 2+S 2=2,∴ a 2 =
3
4
n=3时,a 3+S 3=3,∴ a 3 =
7
8
n=4时,a 4+S 4=4,∴ a 4 =
15
16 ;…(2分)
(2)猜想: a n =1-
1
2 n ,下面用数学归纳法证明:…(3分)
①当n=1时, a 1 =1-
1
2 1 =
1
2 ,猜想成立;
②假设当n=k时猜想成立,即 a k =1-
1
2 k ,
则当n=k+1时, a k+1 = S k+1 - S k =(k+1)- a k+1 -k+ a k =- a k+1 +1+1-
1
2 k ,
即 2 a k+1 =2-
1
2 k ,∴ a k+1 =1-
1
2 k+1 ,即当n=k+1时猜想也成立,
∴由①②知:n∈N *时 a n =1-
1
2 n 都成立.…(8分)
(3)∵b n+1=a n+1-a n,∴ b n = a n - a n-1 =
1
2 n (n≥2),
∵ b 1 = a 1 =
1
2 ,∴ b n =
1
2 n (n∈N *).…(10分)