Let Sn be the sum of the first n terms of the sequence an, and the point P (an, Sn) (n ∈ n +, n ≥ 1) be on the straight line y = 2x-2. (I) find the general term formula of the sequence an; (II) mark BN = 2 (1 − 1An), the sum of the first n terms of the sequence BN is TN, and find the minimum value of n such that TN > 2011; (III) let the positive sequence CN satisfy log2an + 1 = (CN) n + 1, and find the maximum term in the sequence CN

Let Sn be the sum of the first n terms of the sequence an, and the point P (an, Sn) (n ∈ n +, n ≥ 1) be on the straight line y = 2x-2. (I) find the general term formula of the sequence an; (II) mark BN = 2 (1 − 1An), the sum of the first n terms of the sequence BN is TN, and find the minimum value of n such that TN > 2011; (III) let the positive sequence CN satisfy log2an + 1 = (CN) n + 1, and find the maximum term in the sequence CN

(1) According to the meaning of the title, Sn = 2an-2, then when n > 1, sn-1 = 2an-1-2  n ≥ 2, sn-sn-1 = 2an-2an-1, that is, when an = 2an-1, (2 points) and N = 1, A1 = 2  sequence {an} is an equal ratio sequence with A1 = 2 as the first term and 2 as the common ratio, an = 2n. (4 points) (2) according to the meaning of the title, BN = 2 − (12) n − 1, ｜ TN = 2n 2 + 2 · (12) n from TN > 2011, N + (12) n > 20132 (6 points) n ≤ 1006, N + (12) n < 2013 2, when n ≥ 1007, N + (12) n > 20132, so the minimum value of n is 1007. (9 points) (3) it is known that (CN) n + 1 = n + 1, that is, lncn (n + 1) = ln (n + 1) ‖ lncn = ln (n + 1) n + 1, (11 points) Let f (x) = lnxx, X ∈ [3, + ∞), then f '(x) = 1 − lnxx2. When x ≥ 3, LNX > 1, that is, f ^ (x) ＜ 0 ‖ when x ∈ [3, + ∞), f (x) is a decreasing function, and {CN} is a decreasing sequence (12 points) ∵ CN > 0, ∵ C1 = 2, C2 = 33, C3 = 44, ∵ C1 < C2 > C3 ∵ C2 is the largest term in the sequence CN. (14 points)