Question

若a_n是（1+x）ⁿ⁺¹（n∈N^*）展開(kāi)式中含x²項(xiàng)的系數(shù)，則

lim

n→∞

（

1

a1

+

1

a2

+…+

1

an

）=（　�。�

A．2

B．1

C． 1 2

D．0

Answer 1

∵a₁=C₂²=1，a2=

C

23

=

3×2

2×1

=3，a3=

C

24

=

4×3

2×1

=6，…，an=

C

2n+1

=

(n+1)n

2×1

，
∴

lim

n→∞

（

1

a1

+

1

a2

+…+

1

an

）=

lim

n→∞

(

2

2×1

+

2

3×2

+

2

4×3

+…+

2

(n+1)×n

)
=2

lim

n→∞

[(1-

1

2

)+(

1

2

-

1

3

)+(

1

3

-

1

4

)+…+(

1

n

-

1

n+1

)]
=2

lim

n→∞

(1-

1

n+1

)
=2

lim

n→∞

n

n+1

=2．
故選A．