Question

若（1+2^x）⁷展开式的第三项为168，则

lim

n→∞

(

1

x

+

1

x2

+…+

1

xn

)= ．

Answer 1

由题意，C₇²2^2x=168，解得x=

3

2

∴

lim

n→∞

(

1

x

+

1

x2

+…+

1

xn

)=

lim

n→∞

[

2

3

+(

2

3

)2+…+(

2

3

)n]=

lim

n→∞

2 3 ×(1-( 2 3 )n)

1- 2 3

=

lim

n→∞

2×(1-(

2

3

)n)=2
故答案为2