Question

已知數(shù)列{a_n}的通項(xiàng)公式a_n=2n+1，求{

1

anan+1

}前n項(xiàng)的和．

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=2n+1，可得

1

anan+1

=

1

2

(

1

2n+1

-

1

2n+3

)．利用“裂項(xiàng)求和”即可得出．

解答： 解：∵a_n=2n+1，
∴

1

anan+1

=

1

(2n+1)(2n+3)

=

1

2

(

1

2n+1

-

1

2n+3

)．
∴{

1

anan+1

}前n項(xiàng)的和=

1

2

[(

1

3

-

1

5

)+(

1

5

-

1

7

)+…+(

1

2n+1

-

1

2n+3

)]=

1

2

(

1

3

-

1

2n+3

)=

n

6n+9

．

點(diǎn)評(píng)：本題考查了“裂項(xiàng)求和”方法，考查了推理能力與計(jì)算能力，屬于基礎(chǔ)題．