（1）由bn=

2n

an

，bn+1=

2n+1

an+1

，得到an=

2n

bn

，an+1=

2n+1

bn+1

，b1=

2

a1

=1．

代入an+1=

2n+1an

(n+ 1 2 )an+2n

，化为bn+1-bn=n+

1

2

．

∴b_n=（b_n-b_n-1）+（b_n-1-b_n-2）+…+（b₂-b₁）+b₁

=（n-1）+

1

2

+（n-2）+

1

2

+…+1+

1

2

+1

=

n(n-1)

2

+

n-1

2

+1

=

n2+1

2

．

（2）由（1）可得an=

2n

bn

=

2n+1

n2+1

，

∴cn=

2n+1

n2+1

×(n2+1)-1=2ⁿ⁺¹-1．

∴dn=

2n

cncn+1

=

2n

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

=

1

2

(

1

2n+1-1

-

1

2n+2-1

)，

∴S_n=

1

2

[(

1

22-1

-

1

23-1

)+(

1

23-1

-

1

24-1

)+…+(

1

2n+1-1

-

1

2n+2-1

)]

=

1

2

(

1

3

-

1

2n+2-1

)

=

1

6

-

1

2n+3-2

．