Question

數(shù)列{

1

4n2-1

}（n∈N^*）的前n項(xiàng)的和S_n=

 

．

Answer 1

考點(diǎn)：數(shù)列的求和

專題：等差數(shù)列與等比數(shù)列

分析：a_n=

1

4n2-1

=

1

2

(

1

2n-1

-

1

2n+1

)，利用“裂項(xiàng)求和”即可得出．

解答： 解：a_n=

1

4n2-1

=

1

2

(

1

2n-1

-

1

2n+1

)，
∴S_n=

1

2

[(1-

1

3

)+(

1

3

-

1

5

)+…+(

1

2n-1

-

1

2n+1

)]
=

1

2

(1-

1

2n+1

)
=

n

2n+1

．
故答案為：

n

2n+1

．

點(diǎn)評(píng)：本題考查了“裂項(xiàng)求和”求數(shù)列的前n項(xiàng)和，屬于基礎(chǔ)題．