So 2sn-1= 3an-1-1,(n≥2)

Subtract the two expressions to get 2an=3an-3an- 1,

therefore

an=3an- 1，

So the sequence {an} is the common ratio of geometric progression q=3.

When n= 1, 2a 1=3a 1- 1, the solution is a 1= 1.

Then an=3n- 1.

(ⅱ)bn = an+(- 1)nlog 3 an = 3n- 1+(- 1)nlog 33n- 1 = 3n- 1+(- 1)n(n- 1)，

Then the sum of the first 2n terms of the sequence {bn} is t2n = (1+3+32+…+32n-1)+[-0+1-2+3-…+(2n-1)] =

1-32n

1-3

+n=

32n

2

+n-

1

2

.