Question

設a_n是(3-

x

)n（n=2，3，4，5，…）展開式中x一次項系數，則

lim

n→∞

(

32

a2

+

33

a3

+

34

a4

+…+

3n

an

)=______．

Answer 1

展開式的通項為 Tr+1=(-1)r3n-r

C

rn

x

r

2

令

r

2

=1得r=2
∴a_n=3^n-2C_n²．

3n

an

=

3n

C 2n •3n-2

=9×

2

n(n-1)

=

18

n(n-1)

=18×(

1

n-1

-

1

n

)，
∴

lim

n→∞

(

32

a2

+

33

a3

+

34

a4

+…+

3n

an

)
=

lim

n→∞

{18×[(1-

1

2

) +(

1

2

-

1

3

)+(

1

3

-

1

4

)+…+(

1

n-1

-

1

n

)]}
=

lim

n→∞

[18×(1-

1

n

)]
=18．
故答案為：18．