Question

已知數(shù)列{a_n} 是通項(xiàng)a_n和公差都不為零的等差數(shù)列，設(shè)S_n=

1

a1a2

+

1

a2a3

+…+

1

anan+1

，則S_n=（　�。�

• A．

  n

  a1an+1

• B．

  n

  a1an

• C．

  n-1

  a1an

• D．

  n-1

  a1an+1

Answer 1

分析：由等差數(shù)列的通項(xiàng)公式可得a_n+1-a_n=d，可得

1

anan+1

=

1

d

(

1

an

-

1

an+1

)，從而可得S_n=

1

a1a2

+

1

a2a3

+…+

1

anan+1

，=

1

d

(

1

a1

-

1

a2

+

1

a2

-

1

a3

+…+

1

an

-

1

an+1

)

解答：解：∵

1

anan+1

=

1

d

(

1

an

-

1

an+1

)
∴S_n=

1

a1a2

+

1

a2a3

+…+

1

anan+1

，
=

1

d

(

1

a1

-

1

a2

+

1

a2

-

1

a3

+…+

1

an

-

1

an+1

)
=

1

d

(

1

a1

-

1

an+1

)
=

1

d

×

an+1-a1

a1an+1

=

nd

da1an+1

=

n

a1an+1

故選：A

點(diǎn)評(píng)：本題主要考查了等差數(shù)列的通項(xiàng)公式的應(yīng)用，數(shù)列求和的裂項(xiàng)求和方法的應(yīng)用，要注意本題裂項(xiàng)時(shí)要乘以

1

d

是解題中容易漏掉的，這也是本題的解題關(guān)鍵與易錯(cuò)點(diǎn)．