Question

求和：S_n=

1

1×3

+

1

3×5

+

1

5×7

+…+

1

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

結(jié)果為（　�。�

• A．

  n

  2n+1

• B．

  2n

  2n+1

• C．

  2n-3

  4n-2

• D．

  2n-3

  2n-1

Answer 1

分析：可得

1

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

=

1

2n-1

-

1

2n+1

，裂項(xiàng)相消可得．

解答：解：由題意可得S_n=

1

1×3

+

1

3×5

+

1

5×7

+…+

1

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

=

1

2

[（1-

1

3

）+（

1

3

-

1

5

）+（

1

5

-

1

7

）+…+（

1

2n-1

-

1

2n+1

）]
=

1

2

（1-

1

2n+1

）=

n

2n+1

故選A

點(diǎn)評(píng)：本題考查數(shù)列的求和，涉及裂項(xiàng)相消法求和的應(yīng)用，屬中檔題．