(1) when q=2 and n=3, m = {0, 1}, a = {x | x = x 1+x2? 2+x3？ 2 2，xi∈M，I = 1，2，3}。 Set a can be obtained.

(2) Because ai, bi∈M, i= 1, 2, …, N. An < BN, an-BN ≤- 1. S-T =(a 1-b 1)+ can be obtained from the meaning of the question. 1? bn？ 1)q^(n？ 2)+ (Ann? bn)q^(n？ 1) ≤-[1+q+…+q (n-2)+q (n-1)], and then we can get it by using the first n terms and formulas of geometric series.

Answer:

(1) solution: when q=2 and n=3, m = {0, 1}, a = {x | x = x 1+x2? 2+x3？ 2 2，xi∈M，i= 1，2，3}。 We can get A={0, 1, 2, 3, 4, 5, 6, 7}.

(2) proof: let s, t∈A, s = a1+a2q+…+Anq (n-1), t = b1+b2q+…+bnq (n-/kloc-0 bn？ 1)q^(n-2)+(an？ bn)q^(n- 1)≤-[ 1+q+…+q^(n-2)+q^(n- 1)]=？ [q^(n]？ 1/q？ 1]