Question

數列{a_n}中，a₁=1，a_n，a_n+1是方程x2-(2n+1)x+

1

bn

=0的兩個根，則數列{b_n}的前n項和S_n等于（　　）

A． n 2n+1

B． n n+1

C． 1 2n+1

D． 1 n+1

Answer 1

由題意可得a_n+a_n+1=2n+1
∴a_n=n
∵anan+1=

1

bn

∴bn=

1

n(n+1)

=

1

n

-

1

n+1

S_n=b₁+b₂+…+b_n
=(1-

1

2

+

1

2

-

1

3

+…

1

n

-

1

n+1

)
=1-

1

n+1

=

n

n+1

故選B．