(2) Prove that: ① When n= 1, A 1 = 2 = 1× 22 can be obtained, and the proposition holds; ? ... (2 points)
② If n=k, the proposition holds, that is, AK = k (k+ 1) 2, …( 1).
Then when n=k+ 1, the sum of inductive hypothesis and (ak? ak？ 1)2=ak？ 1+ak，get【AK+ 1？ k(k+ 1)2]2 = k(k+ 1)2+an+ 1。
Namely (ak+ 1)2? (k2+k+ 1)ak+ 1+[k(k？ 1)2]? [(k+ 1)(k+2)2]=0
The solution is AK+1= (k+1) (k+2) 2 (AK+1= k (k? 1) 2 < AK doesn't matter.
That is, when n=k+ 1, the proposition holds ... (4 points)
In summary, for all n∈N*, an = n (n+ 1) 2...( 1).
(3) solution: bn =1an+1an+2+1an+3 …+1a2n = 2 (n+1) (n+2 22n+1= 2nn2+3n+1= 2 (2n+1n)+3 ... (2 points)
Because the function f (x) = 2x+ 1x monotonically increases in the interval [1, +∞], when n= 1, the maximum bn is 13, that is, BN ≤ 13...(2 points
From the meaning of the question, there is 13 < log8t, so t > 2.
Therefore, t∈(2, +∞). ... (2 points)