Question

已知數(shù)列{a_n}的通項(xiàng)公式an=

1+2+…+n

n

，bn=

1

anan+1

，則數(shù)列{b_n}的前n項(xiàng)和為

2n

n+2

．

Answer 1

分析：利用等差數(shù)列的前n項(xiàng)和公式、“裂項(xiàng)求和”即可得出．

解答：解：∵an=

1+2+…+n

n

=

n(n+1)

2n

=

n+1

2

，∴bn=

1

anan+1

=

4

(n+1)(n+2)

=4(

1

n+1

-

1

n+2

)，
∴數(shù)列{b_n}的前n項(xiàng)和=4[(

1

2

-

1

3

)+(

1

3

-

1

4

)+…+(

1

n+1

-

1

n+2

)]
=4(

1

2

-

1

n+2

)=

2n

n+2

．
故答案為

2n

n+2

．

點(diǎn)評(píng)：熟練掌握等差數(shù)列的前n項(xiàng)和公式、“裂項(xiàng)求和”是解題的關(guān)鍵．