Question

正项数列{an}满足，a1＝1，a2n+1＝a2n+2an+1，则1a1a2+1a2a3+…1anan+1=（　　）A．2?2nB．1?1nC．1?1n+1

正项数列{an}满足，a1＝1，a2n+1＝a2n+2an+1，则1a1a2+1a2a3+…1anan+1=（　　）A．2?2nB．1?1nC．1?1n+1D．2?2n+1

Answer 1

∵an+12＝an2+2an+1=(an+1)2且a_n＞0
∴a_n+1=a_n+1
∵a₁=1
∴数列{a_n}是以1为首项，以1为公差的等差数列
∴a_n=1+n-1=n
∴

1

an+1an

=

1

n(n+1)

=

1

n

?

1

n+1

∴

1

a1a2

+

1

a2a3

+…+

1

anan+1

=

1

1×2

+

1

2×3

+…+

1

n(n+1)

=1-

1

2

+

1

2

?

1

3

+…+

1

n

?

1

n+1

=1-

1

n+1

故选C