Question

已知数列{b_n}中，b1=

11

7

，b_n+1b_n=b_n+2．数列{a_n}满足：an=

1

bn-2

(n∈N*)
（Ⅰ）求证：a_n+1+2a_n+1=0；
（Ⅱ） 求数列{a_n}的通项公式；
（Ⅲ） 求证：（-1）b₁+（-1）²b₂+…+（-1）ⁿb_n＜1（n∈N^*）

Answer 1

证明：（Ⅰ）an+1=

1

bn+1-2

=

1

bn+2 bn -2

=

bn

2-bn

= -1+

2

2-bn

=-2an-1，移向整理得a_n+1+2a_n+1=0
（Ⅱ）∵a_n+1=-2a_n-1∴an+1+

1

3

=-2  (an+

1

3

)
又 a1+

1

3

=-2 ≠0∴{an+

1

3

}为等比数列
∴an+

1

3

=(-2)n∴an=(-2)n-

1

3

证明：（Ⅲ）bn=

1

an

+2=

1

(-2)n- 1 3

+2∴(-1)nbn=2•(-1)n+

1

2n- 1 3 •(-1)n

①当n为奇数时（-1）ⁿb_n+（-1）ⁿ⁺¹b_n+1=

1

2n+ 1 3

+

1

2n+1- 1 3

=

2n+2n+1

(2n+ 1 3 )(2n+1- 1 3 )

＜

2n+2n+1

2n•2n+1

=

1

2n

+

1

2n+1

（-1）b₁+（-1）²b₂+…+（-1）ⁿb_n＜

1

2

+

1

22

+…+

1

2n-2

+

1

2n-1

-2+

1

2n+ 1 3

＜

1 2

1- 1 2

-2+

1

2n+ 1 3

=

1

2n+ 1 3

-1＜1
 ②当n为偶数时，（-1）b₁+（-1）²b₂+…+（-1）ⁿb_n＜

1

2

+

1

22

+…+

1

2n-1

+

1

2n

＜

1 2

1- 1 2

=1
综上所述，（-1）b₁+（-1）²b₂+…+（-1）ⁿb_n＜1