Question

數(shù)列{

2

4n2-1

}的前n項(xiàng)之和為

 

．

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：由

2

4n2-1

=

2

(2n+1)(2n-1)

=

1

2n-1

-

1

2n+1

，利用裂項(xiàng)求和法能求出數(shù)列{

2

4n2-1

}的前n項(xiàng)之和．

解答： 解：∵

2

4n2-1

=

2

(2n+1)(2n-1)

=

1

2n-1

-

1

2n+1

，
∴數(shù)列{

2

4n2-1

}的前n項(xiàng)之和為：
S_n=1-

1

3

+

1

3

-

1

5

+…+

1

2n-1

-

1

2n+1

=1-

1

2n-1

=

2n-2

2n-1

．
故答案為：

2n-2

2n-1

．

點(diǎn)評：本題考查數(shù)列的前n項(xiàng)之和的求法，是中檔題，解題時(shí)要認(rèn)真審題，注意裂項(xiàng)求和法的合理運(yùn)用．