Question

求下列数列的前n项和S_n：an=

1

n + n+1

．

Answer 1

分析：由题设知Sn=

1

2 + 1

+

1

3 + 2

++

1

n+1 + n

=(

2

-1)+(

3

-

2

)++(

n+1

-

n

)=

n+1

-1．

解答：解：∵an=

1

n + n+1

=

n+1 - n

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

=

n+1

-

n

∴Sn=

1

2 + 1

+

1

3 + 2

++

1

n+1 + n

=(

2

-1)+(

3

-

2

)++(

n+1

-

n

)
=

n+1

-1．

点评：本题考查数列的求和，解题时要注意裂项求和法的合理运用．