Question

數(shù)列

1

1+2

+

1

1+2+3

+

1

1+2+…+(n+1)

…前n項(xiàng)和（　�。�

• A、

  n

  n+1

• B、

  n

  n+2

• C、

  2

  n(n+1)

• D、

  4

  n(n+1)

Answer 1

分析：將通項(xiàng)

1

1+2+…+(n+1)

化簡(jiǎn)為

2

(n+1)(n+2)

再裂項(xiàng)成2(

1

n+1

-

1

n+2

)，問(wèn)題即解．

解答：解：a_n=

1

1+2+…+(n+1)

=

2

(n+1)(n+2)

=2(

1

n+1

-

1

n+2

)．
Sn=2[(

1

2

-

1

3

)+(

1

3

-

1

4

)+…+(

1

n+1

-

1

n+2

)]=2(

1

2

-

1

n+2

)=

n

n+2

 
故選B．

點(diǎn)評(píng)：本題考查等差數(shù)列求和，裂項(xiàng)法數(shù)列求和．本題將

1

1+2+…+(n+1)

化為2(

1

n+1

-

1

n+2

)是關(guān)鍵．