已知等比数列{an}满足an+1+an=4×3n﹣1(n∈N*).
(I)求数列{an}的通项公式;
(Ⅱ)若bn=log3an,求Tn=b1b2﹣b2b3+b3b4﹣b4b5+…+b2n﹣1b2n﹣b2nb2n+1.
解:(I)设等比数列{an}的公比为q,由an+1+an=4×3n﹣1(n∈N*).
可得a1q