设等比数列{a_n}的公比为q，∵a_n＞0，∴q＞0．经验证q=1不成立．

由a2=

1

4

，

S4

S2

=

5

4

，可得

a1q= 1 4 a1(q4-1) q-1 a1(q2-1) q-1 = 5 4

，及q＞0，解得

a1= 1 2 q= 1 2

．

∴an=a1qn-1=(

1

2

)n．

∴

1

a1

-

1

a2

+

1

a3

-

1

a4

+…+(-1)n+1

1

an

=2-2²+2³+…+（-1）ⁿ⁺¹•2ⁿ

=

2[1-(-2)n]

1-(-2)

=

2

3

[1-(-2)n]．

故选D．