（I）证：由

bn+1

bn

=q，有

an+1an+2

anan+1

=

an+2

an

=q，∴a_n+2=a_nq²（n∈N*）．

（ II）证：∵a_n=q_n-2q²，∴a_2n-1=a_2n-3q²=…=a₁q^2n-2，a_2n=a_2n-2q²=…=a₂q^n-2，

∴c_n=a_2n-1+2a_2n=a₁q^2n-2+2a₂q^2n-2=（a₁+2a₂）q^2n-2=5q^2n-2．

∴{c_n}是首项为5，以q²为公比的等比数列．

（ III）由（ II）得

1

a2n-1

=

1

a1

q2-2n，

1

a2n

=

1

a2

q2-2n，于是

1

a1

+

1

a2

+…+

1

a2n

=(

1

a1

+

1

a3

+…+

1

a2n-1

)+(

1

a2

+

1

a4

+…+

1

a2n

)=

1

a1

(1+

1

q2

+

1

q4

+…+

1

q2n-2

)+

1

a2

(1+

1

q2

+

1

q4

+…+

1

q2n-2

)=

3

2

(1+

1

q2

+

1

q4

+…+

1

q2n-2

)．

当q=1时，

1

a1

+

1

a2

+…+

1

a2n

=

3

2

(1+

1

q2

+

1

q4

+…+

1

q2n-2

)=

3

2

n．

当q≠1时，

1

a1

+

1

a2

+…+

1

a2n

=

3

2

(1+

1

q2

+

1

q4

+…+

1

q2n-2

)=

3

2

(

1-q-2n

1-q-2

)=

3

2

[

q2n-1

q2n-2(q2-1)

]．

故

1

a1

+

1

a2

+…+

1

a2n

=

3 2 n，q=1 3 2 [ q2n-1 q2n-2(q2-1) ]，q≠1.