Question

已知数列{a_n}：an=

8

(n+1)(n+3)

，求

∞

n=1

(n+1)(an-an+1)的值．

Answer 1

分析：先找通项的特征，进行裂项，之后进行分组求和即可求解

解答：解：∵(n+1)(an-an+1)=8(n+1)[

1

(n+1)(n+3)

-

1

(n+2)(n+4)

]
=

8

n+3

-

8(n+1)

(n+2)(n+4)

=

8(n+4)

(n+3)(n+4)

-

8(n+1)

(n+2)(n+4)

=

8(2n+5)

(n+2)(n+3)(n+4)

=

8[(n+2)+(n+3)]

(n+2)(n+3)(n+4)

=8•[

1

(n+2)(n+4)

+

1

(n+3)(n+4)

]
=4•(

1

n+2

-

1

n+4

)+8(

1

n+3

-

1

n+4

)
∴

∞

n=1

(n+1)(an-an+1)=4

∞

n=1

(

1

n+2

-

1

n+4

)+8

∞

n=1

(

1

n+3

-

1

n+4

)
=4•(

1

3

+

1

4

)+8•

1

4

=

13

3

点评：本题主要考查了数列求和的应用，解题的关键是寻求数列的通项的规律，数列裂项求和是该题的重点