It is known that the sum of the first n terms of the sequence {an} is Sn, satisfying an + Sn = 2n. (I) prove that the sequence {An-2} is an equal ratio sequence, and find out an; (II) let BN = (2-N) (An-2), find out the maximum term of {BN}

It is known that the sum of the first n terms of the sequence {an} is Sn, satisfying an + Sn = 2n. (I) prove that the sequence {An-2} is an equal ratio sequence, and find out an; (II) let BN = (2-N) (An-2), find out the maximum term of {BN}

A1 = 1 from a1 + S1 = 2A1 = 2, A1 = 1 from a1 + S1 = 2A1 = 2, A1 = 1 from a1 + S1 = 2A1 = 2; from an + Sn = 2n to an + 1 + Sn + 1 = 2 (n + 1) from the two-way subtraction to get 2An + 1-an = 2, that is, 2An + 1-4 = An-2, that is, an + 1-2 = 12 (An-2) is the first item of A1-2 = -2-1, which is the same number sequence of 12, which is the same ratio of 12. So An-2 = - (12) n - (12) n-1, so an = 2 - (12) n = 2 - (12) - 2 - (12) n-2 - (12) n-2 - (12) n-2 - (12) n-2 - (12) n-2 - (12) n-12) n-2 - (12) n-12) n-2 - (12) n-12) n-12) n-12) n-n-1 − BN = n − 12n − n − 22 N − 1 = n − 1 − 2n + 42N = 3 − n2n ≥ 0 can get n ≤ 3, and BN + 1-bn < 0 can get n > 3, so B1 < B2 < B3 = B4 > B5 > The maximum term of BN is B3 = B4 = 14