已知:数列{a n }前n项和为S n ,a n +S n =n,数列{b n }中b 1 =a 1 ,b n+1 =a

1个回答

  • (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)