Question

求和：

1

4×12-1

+

1

4×22-1

+

1

4×32-1

+…+

1

4n2-1

．

Answer 1

考點(diǎn)：數(shù)列的求和

專題：計算題,點(diǎn)列、遞歸數(shù)列與數(shù)學(xué)歸納法

分析：由

1

4n2-1

=

1

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

=

1

2

(

1

2n-1

-

1

2n+1

)，把要求和的每一項(xiàng)裂項(xiàng)后求和得答案．

解答： 解：∵

1

4n2-1

=

1

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

=

1

2

(

1

2n-1

-

1

2n+1

)，
∴

1

4×12-1

+

1

4×22-1

+

1

4×32-1

+…+

1

4n2-1

=

1

2

[(1-

1

3

)+(

1

3

-

1

5

)+(

1

5

-

1

7

)+…+(

1

2n-3

-

1

2n-1

)+(

1

2n-1

-

1

2n+1

)]
=

1

2

(1-

1

2n+1

)=

n

2n+1

．

點(diǎn)評：本題考查了裂項(xiàng)相消法求數(shù)列的前n項(xiàng)和，涉及等差數(shù)列相鄰兩項(xiàng)乘積的倒數(shù)數(shù)列的求和問題，常采用裂項(xiàng)相消法求和，是中檔題．