Question

已知數(shù)列{a_n}滿足a1=2，an+1=

2n+1an

(n+ 1 2 )an+2n

，n∈N*
（1）設bn=

2n

an

，求數(shù)列bn的通項公式．
（2）設cn=an•(n2+1)-1，dn=

2n

cn•cn+1

，求數(shù)列{d_n}的前n項和S_n．

Answer 1

（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

，化為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

．